The Definite Integral


 Egbert Hardy
 1 years ago
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1 CHAPTER 3 The Definite Integrl Key Words nd Concepts: Definite Integrl Questions to Consider: How do we use slicing to turn problem sttement into definite integrl? How re definite nd indefinite integrls relted? We now enter the lst phse of our course, in which we consider the definite integrl. Note tht we hve lredy seen the indefinite integrl s nother nme for ntiderivtive ( function whose derivtive equls some given function). When we introduce the definite integrl, it will look nothing like the indefinite integrl: it will be number (not function), nd it will rise from thought process we will cll slicing. But fer not: before long, the two types of integrls will be connected by the Fundmentl Theorem of Clculus. 3.. Definition of b f(x) dx Are under curve We will introduce the definite integrl by seeking the nswer to: Question: Wht is the re under curve in the plne? For exmple, let the curve be the grph of the function f(x) = + 4x x 2 from x = to x = 3. The re we seek is below this curve nd bove the xxis, s shded in the imge below: 95
2 3.. DEFINITION OF b f(x) dx AREA UNDER A CURVE 96 This question cn not be nswered using methods seen prior to clculus. If you think bck to re formuls you hve lerned prior to clculus, they were exclusively for regions whose boundries were stright lines (tringles, rectngles, trpezoids, etc.) or rcs of circles. As the prbolic curve in our region is neither stright nor n rc of circle, none of those re formuls will nswer our question Approximting n re by slicing. Fced with dunting problem like this, we revert to the ide of pproximtion, s we did with Euler s Method to pproximte solution to DE. Reclling our pproximtion philosophy, we im first to devise resonble method of pproximting the re, nd then second to devise wy to improve the pproximtion to mke it s ccurte s we my desire. Our pproximtion ide will be slicing : to divide the problem into smller pieces. In this cse, we will divide the region into smller regions by slicing perpendiculr to the xxis, e.g., one region with x <.5, nother with.5 x <, nother with x.5, etc. through 2.5 x 3:
3 3.. DEFINITION OF b f(x) dx AREA UNDER A CURVE 97 We still hve more work to do, s we still cnnot compute the res of our six slices, since their upper boundries re curved. However, now tht we hve mde the regions skinny, it is resonble to pproximte the slices by regions whose res we do know. For exmple, we could pproximte ech region by rectngle : Our pproximtion to the originl re is simply the re of the six rectngles. To compute tht pproximtion, we need to know the rectngle heights (ech is clerly.5 units wide). The wy the figure ws drwn, ech rectngle height is the height of the curve y = f(x) t the left endpoint of the rectngle. So, the heights (nd res) of the rectngles re: Rectngle Height Are x.5 f() = (.5) () =.5.5 x f(.5) = 2.75 (.5) (2.75) =.375 x.5 f() = 4 (.5) (4) = 2.5 x 2 f(.5) = 4.75 (.5) (4.75) = x 2.5 f(2) = 5 (.5) (5) = x 3 f(2.5) = 4.75 (.5) (4.75) = which mkes our pproximtion of the originl region: Are under curve = Improving the pproximtion. Recll our pproximtion strtegy: we wnt not only n ide for how to pproximte something, but lso scheme for improving our pproximtion. How could we improve the sixrectngle pproximtion we just computed? Use more rectngles! 2 Here s wht it would look like to use twelve rectngles insted of six: If you sid to yourself Rectngles? Surely it is smrter to use trpezoids, well done! Lter we will use trpezoids s smrter pproximtions thn rectngles. But for now we opt for rectngles s simpler to work with. 2 If you sid Use trpezoids, well done! Tht s nother ide, nd we ll come bck to it. If you switched from rectngles to trpezoids, you would indeed likely improve the pproximtion. But then if you wnted to improve it even further, you d need new ide, like using more trpezoids.
4 3.. DEFINITION OF b f(x) dx AREA UNDER A CURVE 98 In the interest of voiding (your) boredom, we will only show some of the twelve rectngleres involved in our improved pproximtion: Rectngle Height Are x.25 f() = (.25) () = x.5 f(.25) =.9375 (.25) (.9375) = x.75 f(.5) = 2.75 (.25) (2.75) = x 2.5 f(2.25) = (.25) (4.9375) = x 2.75 f(2.5) = 4.75 (.25) (4.75) = x 3 f(2.75) = (.25) (4.4375) =.9375 Summing over the res of ll twelve rectngles, we would find: Are under curve = Riemnn sum nottion. This pproximtion ide we hve just seen is sometimes clled Riemnn Sum (nmed fter the mthemticin Bernhrd Riemnn). If we wnt to write out the pproximtion s formul, we will need some nottion for the sum tht we tke s the lst step. By trdition, mthemticins often use Greek letter Σ (cpitl sigm) to indicte sum. For exmple, for our first pproximtion, we found six rectngle res: A =.5, A 2 =.375, A 3 = 2, A 4 = 2.375, A 5 = 2.5 nd A 6 = The sigmnottion for the finl summtion would be: 6 Are under curve (our first pprox) The symbols below nd bove the Σ indicte tht there is counting vrible (or index ) i tht runs from to 6, nd for ech i, we hve rectngle re A i, nd the Σ sys to dd up those six A i. So the ide of tking more slices lets you repetedly improve the pproximtion, nd tht s wht we re fter here. i= A i
5 3.. DEFINITION OF b f(x) dx AREA UNDER A CURVE 99 Then in our improved pproximtion, we hd twelve rectngle res: A =.25, A 2 = , through A 2 =.9375, nd 2 Are under curve (our second pprox) In generl, of course, we need not hve used six, or twelve, rectngles, but could choose to use n rectngles for ny n, in which cse: n Are under curve This formul does not spell out how to compute the res A i. Let s do so. Suppose tht our curve is the grph of y = f(x) from x = to x = b, s in our exmple. (We re lso supposing tht our curve is bove the xxis, so it is sensible to tlk bout re under the curve. We will return lter in the course to how to hndle curves below the xxis.) Ech rectngle re is height times width. The formul for width is pretty strightforwrd. We hd the segment from x = to x = b on the xxis, nd we sliced it into n equl segments. Since the overll width ws b before slicing, the width of ech slice is b n. As for the height, we see from our exmples tht ech rectngle height ws the vlue of f t some point. Note, however, tht the point we plug into f chnges from rectngle to rectngle. To indicte this dependence, we will write x i for the xvlue tht gets plugged into f for the ith rectngle. In tht cse, we would write the height s f(x i ). Now tht we hve formuls for the width nd height of ech rectngle, we cn use them to compute A i, nd hence the pproximtion for the re under the curve: n ( Are under curve f(x i ) b ). n This expression is thus the generl formul for Riemnn sum: i= i= A i i= A i Definition 3. (Riemnn Sum). Given function f(x), nd given endpoints x = nd x = b to be divided into n slices, the sum n f(x i ) b n i= is clled Riemnn sum, if x i is point in the ith segment (when we subdivide the segment [, b] into n equl pieces). Notice tht this definition of Riemnn sum is bit more flexible thn the specific computtions we did, in tht it llows wide vriety of choices of x i. In our exmple, we decided to mke x i the left endpoint of the ith segment, but this definition lso llows us to mke other choices, like the right endpoint, or even the midpoint. Ech
6 3.. DEFINITION OF b f(x) dx AREA UNDER A CURVE 2 set of choices of x i (for i = through i = n) yields different pproximtion to the re under the curve. To distinguish mong this different choices, people sometimes use terminology like left endpoint Riemnn sum, right endpoint Riemnn sum, or midpoint Riemnn sum. If you wnt to mke this formul even more specific, you cn work out expressions for x i for ech choice (left endpoint, right endpoint, midpoint). Let s derive n expression for the left endpoint of the ith segment. First we note tht the i = left endpoint is x =. Next, we note tht the i = 2 left endpoint is b n to the right of x, which mens x 2 = + b b n. Similrly, the i = 3 left endpoint is n to the right of x 2, which mens x 3 = x 2 + b n = + 2 b n. From here we cn see the generl pttern x i = + (i ) b (if choosing left endpoints) n By similr logic, x i = + (i) b (if choosing right endpoints), n nd ( x i = + i ) b (if choosing midpoints). 2 n The Definite Integrl tking the limit n. So fr we hve n pproximtion ide tht involves choice of number n of slices (or rectngles), nd we hve the ide tht choosing lrger vlue of n should mke the pproximtion more ccurte. Given tht sitution, it is t lest plusible tht if we continue the process of incresing n indefinitely, the results will converge to limit, nd tht limit will be the exct vlue of the re under the curve. Here is nother instnce of the key clculus ide of lim n, whose precise definition we sw in the DDS phse of the course. Now we re not tking the limit of terms in DDS, but rther the limit of sequence of pproximtions, but the notion of limit is the sme s before: if the exct re is A, nd we re given some smll tolernce ɛ, we would wnt to be ble to choose n lrge enough so tht our pproximtion for tht n would be within ɛ of A. It is outside the scope of our course to prove under wht circumstnces the limit of our pproximtion exists 3, so we will just ssume tht it does, in which cse the exct vlue of the re under the curve is lim n of our Riemnn sum formul. This limit is exctly wht we cll the definite integrl: 3 For ny function f(x) tht is continuous on [, b], the limit exists. It lso exists for some functions with discontinuities.
7 3.2. ANOTHER USE OF b f(x) dx DISTANCE TRAVELED GIVEN VELOCITY 2 Definition 3.2 (Definite Integrl). Given function f(x), nd given endpoints x = nd x = b, the definite integrl b f(x) dx is defined by: [ b n ( f(x) dx = lim f(x i ) b ) ], n n where x i is point in the ith segment (when we subdivide the segment [, b] into n equl pieces). To mke this limit more concrete, consider tht in our exmple bove, we hd Riemnn sums equl to.25 for n = 6 nd.5938 for n = 2. If we continued to increse n, we would compute Riemnn sums equl to.85 for n = 24 nd.955 for n = (detils not shown). At this point, we might guess tht the limit is 2, so tht 3 ( + 4x x2 )dx = 2, since the pproximtions seem to be pproching tht vlue. 4 Tken s whole, the definition bove is lot to bsorb: it involves function f, four vribles (, b, n, nd x i ), summtion symbol, nd limit! But, in fct, the formul itself is often less importnt thn the process tht led to the formul, i.e., the exmple tht we worked through t the strt of this section. Sid nother wy, if you red this formul from the inside out, it relly just recretes the exmple tht we did. At the very inside, we hve the product f(x i ) b n, which is rectngle re: you re given f, nd b, then you pick n (the number of rectngles), nd tht defines bunch of x i (once you decide if you re using left endpoints or something else), nd then you compute rectngle res. Continuing outwrd, the next item in the definition is the Σ, which sys to dd up the rectngle res (to get the pproximtion to the re under the curve). Finlly, ll the wy on the outside is the limit, which sys tht you tke more nd more rectngles to improve the pproximtion, nd, in fct, tke the limit s n to get the exct re under the curve Another use of b f(x) dx Distnce trveled given velocity Next we ll see nother use of the definite integrl, both to expnd our set of pplictions, but lso (snekily) to pve the wy for the Fundmentl Theorem. Our new gol is to nswer: Question: How cn we compute the position of n object if we know its velocity s function of time? Agin it will be convenient to consider specific exmple. Suppose cr is trveling forwrd, nd slowly brking, on stright rod with velocity (speed, in ft/sec) given by the function v(t) = 6 4e t/8 where t is the time in seconds. Our gol is to determine how fr the cr trvels between the times t = nd t =. 4 Of course we hve not proven tht the limit is 2 (nd it won t lwys be true tht definite integrl equls some tidy guessble number like 2). i=
8 3.2. ANOTHER USE OF b f(x) dx DISTANCE TRAVELED GIVEN VELOCITY 22 Just s with the reundercurve problem where we hd no bsic geometric fct to solve the problem t hnd, here there is no bsic physics principle tht gives us the distnce trveled. We do hve one bsic physics principle tht involves the quntities present in our problem Distnce = (Velocity) (Time) but tht principle is only vlid if the velocity is constnt. To overcome this obstcle, we use the sme strtegy s in the reunderthecurve problem: we slice time into smll timeintervls, so tht the velocity is t lest pproximtely constnt on ech intervl, in which cse we cn pply the bsic physics principle. For exmple, suppose we slice t into five intervls, nd plug the resulting times into the velocity function v(t): t (sec) velocity (ft/sec) In the first 2 seconds, we could estimte the cr s velocity s being constnt t 2 ft/sec, in which cse it would trvel (2 sec)(2 ft/sec) = 4 ft. Then in the next 2 seconds, we could estimte the cr s velocity s being constnt t ft/sec, in which cse it would trvel (2 sec)(28.8 ft/sec) = 57.6 ft. Continuing in this mtter, we would estimte: Distnce Trveled (2)(2) + (2)(28.8) + (2)(35.7) + + (2)(45.2) = 34.8 ft. Hopefully tht process felt fmilir, since it followed exctly the pttern we sw with reundercurve: we took the totl time intervl t nd sliced it into five pieces, then for ech piece we multiplied the width of the time subintervl by vlue of the velocity function in tht subintervl, nd dded up the results. This mtches exctly wht we clled Riemnn sum; insted of n intervl on the xxis we now hve time intervl, nd insted of the height function f(x) we now hve the velocity function v(t), but the process is precisely the sme. In our computtion bove, we used the left endpoint Riemnn sum, but of course we could just s plusibly used the right endpoint or midpoint sum. Wht if we wnt more ccurte pproximtion? As with reundercurve, nturl wy to improve the ccurcy is to hve more slices. For exmple, we could consider slices, yielding these velocity vlues: t velocity Using left endpoint Riemnn sum gin, we would sy: Distnce Trveled = 357. ft. Note tht ech timeslice is now second in durtion, s compred to 2 seconds in the first computtion, so we multiply ech velocity vlue by to crete the sum. If we continue this process, repetedly incresing the number n of timeslices so tht n, then our pproximtions will converge to the exct distnce trveled
9 3.3. WHAT IF THE INTEGRAND IS NEGATIVE? 23 between t = nd t =. 5 So fr we hve pproximtions 34.8 for n = 5 nd 357. for n =. If we continue to lrger n, we would find pproximtions for n = 2 nd 37.3 for n =, so we might guess tht the exct distnce trveled is round 372, which ppers to be the limit of these pproximtions. Note tht in this cse, the pproximtions do not pper to be converging to round number, nd re not converging so quickly s to give us much confidence in our guessed limit of 372. In the ner future, we will lern pproximtion methods tht converge more quickly thn this leftendpoint Riemnn sum. Since this slicingthenlimit process mtches our reundercurve exmple precisely (only with vrible nmes chnged), we know tht the finl n limit will be definite integrl, s defined in the previous section. Thus, Theorem 3.3. If v(t) is function tht gives the velocity of n object moving on line, with v(t) for ll times between t = nd t = b, then Distnce Trveled from (t = ) to (t = b) = b v(t) dt Wht if the integrnd is negtive? Thus fr we restricted ourselves to the sitution tht our integrnd (the function f(x) in b f(x) dx or v(t) in b v(t) dt) ws positive. This llowed us to void the complictions of negtive re or n object chnging direction. Now it s time to ddress those complictions. Our definition of definite integrl works perfectly well if the function f(x) (or v(t)) is negtive: the sum n i= f(x i) b n would just involve some (or ll) negtive terms. How would llowing the integrnd to be negtive impct our two pplictions (re under curve, distnce trvelled given velocity)? Exmple 3.4 (Areundercurve with sometimesnegtive integrnd). For reundercurve, the definite integrl simply counts ny portion of the re tht lies below the xxis s negtive re (since the terms in the Riemnn sum will be negtive if f(x i ) < ). For exmple, if we consider the grph of f(x) = 2 x 2 : 5 As with the reundercurve exmple, we will not in this course see the theorem tht gurntees tht the Riemnn sums converge s n. Be ssured tht for most physiclly relistic v(t) (including continuous v(t)), the Riemnn sums do converge to limit s n.
10 3.3. WHAT IF THE INTEGRAND IS NEGATIVE? 24 (the blck digonl line is the grph of f), then the definite integrl 6 f(x) dx will equl 3, becuse the tringle tht lies bove the xxis (from x = to x = 4) hs re 2 bh = 2 (4)(2) = 4 nd will be counted positively, wheres the tringle tht lies below the xxis (from x = 4 to x = 6) hs re 2 bh = 2 (2)() = nd will be counted negtively, so overll 6 f(x) dx = (Positive Are) (Negtive Are) = 4 = 3. Exmple 3.5 (Distncegivenvelocity with sometimesnegtive velocity). For distncetrveledgivenvelocity, the definite integrl similrly counts distnce s negtive if the velocity is negtive, so tht overll the definite integrl will compute forwrd distnce minus bckwrd distnce. In physics, this difference is often clled the totl displcement. To clrify this ide, consider the velocity function below: This velocity function describes n object tht moves forwrd on line for 3 seconds (from t = to t = 3) with velocity v(t) = 2 ft/sec, then instntly switches (t t = 3)
11 3.3. WHAT IF THE INTEGRAND IS NEGATIVE? 25 to moving bckwrds for 2 seconds (from t = 3 to t = 5) with velocity v(t) = ft/sec. Thus, in the first 3 seconds, the object trvels distnce of (2 ft/sec) (3 sec) = 6 ft, in the forwrd direction, nd in the next 2 seconds, the object trvels distnce of ( ft/sec) (2 sec) = 2 ft, in the bckwrd direction. Since the definite integrl counts bckwrd distnce s negtive, we will hve: 5 v(t) dt = (Positive Distnce) (Negtive Distnce) = 6 2 = 4 ft. So wht does this 4 ft represent in terms of the problem? It is simply the distnce between the object s finl position nd its initil position. If we choose to lbel its initil position s s =, then t time t = 3 sec, the object will be t s = 6 ft, given its forwrd distnce of 6 ft in those 3 seconds, nd then t the finl time t = 5 sec, the object will be t s = 4 ft, given its bckwrd distnce of 2 ft in the finl 2 seconds. Thus, indeed, over the whole 5 seconds, the object is 4 ft forwrd reltive to where it strted. This is exctly wht physicists cll displcement Wht if you wnt to count ll re or ll distnce s positive? Summrizing, the definite integrl will count re below the xis s negtive, nd similrly will count distnce trveled bckwrds s negtive. Wht if tht is not wht you wnt to compute? For exmple, in our first exmple in this section, wht if you wnted to count both tringle res s positive, so tht you would compute the totl re s you literlly see it in the figure: 4 + = 5. Similrly, in the second exmple, wht if you wnt to compute the true distnce trveled of the object: 8 ft, to reflect the 6 ft of forwrd progress followed by the 2 ft of bckwrd progress. One wy to compute these perfectly interesting quntities is to mnully count the negtive contributions s positive. To do this, work out where the integrnd is positive nd where it is negtive, then seprtely compute the definite integrl for the positive prts nd for the negtive prts, nd then simply erse the negtive sign in the second definite integrl. Thus, in the first exmple, we would look t the grph of f, see tht it is positive for < x < 4 nd negtive for 4 < x < 6, so consider 4 f(x) dx nd 6 4 f(x) dx seprtely. For the first, we would hve 4 f(x) dx = 4 (the re), nd for the second, we would hve 6 4 f(x) dx = (the re, but counted negtively), so then our true totl re (ll counted positively) would be 4 + (+) = 5. A mthemticl wy to express this process is to tke the bsolute vlue of the integrnd before you integrte. The bsolute vlue will tke ll plces where the integrnd is negtive nd flip them to positive, nd then the definite integrl will only compute positive contributions. Thus, b f(x) dx computes the totl re (where res under the xxis re counted positively) nd b v(t) dt computes the distnce trveled (where forwrd distnce nd bckwrd distnce re both counted positively). 6 6 These formuls in which you integrte the bsolute vlue re tidy to write down, but not relly timesver in prctice. If you integrte f(x), your first step will typiclly be to replce
12 3.4. PROPERTIES OF DEFINITE INTEGRALS Properties of Definite Integrls Using our ide of definiteintegrlsre, we cn mke plusible some fundmentl properties of the definite integrl (whose proofs re beyond the scope of this course). Theorem 3.6 (Sum nd Difference Rules for Definite Integrls). b b (f(x) + g(x)) dx = (f(x) g(x)) dx = b b f(x) dx + f(x) dx b b g(x) dx, g(x) dx. For the first eqution, the lefthnd side sys dd the heights f(x) nd g(x), then compute the re under the combinedheight curve, wheres the righthnd side sys compute ech re first, then dd the res. The rule sys we get the sme nswer either wy. If you think bout the Riemnn sums tht underlie the definite integrl, the lefthnd side would involve rectngles with heights f(x i )+g(x i ), ech of which could be viewed s rectngle with height g(x i ) stcked on top of rectngle with height f(x i ), so tht the rectngle sum for f + g would involve the rectngle sum for f plus the rectngle sum for g. (The justifiction for the second eqution is similr, with now heights being subtrcted rther thn dded). Theorem 3.7 (ConstntMultiple Rule for Definite Integrls). If k is constnt, then b k f(x)dx = k b f(x) dx. The lefthnd side sys to scle the heights f(x) by the constnt k nd then compute the re under this new curve, wheres the righthnd side sys compute the re, nd then scle by k. Since k is constnt, the rule sys we cn do the scling either before or fter the re nd we get the sme nswer, e.g., doubling every height will double the overll re. These lst two theorems mirror properties we sw for indefinite integrls. In ddition, now tht our integrl hve integrtion limits nd b, there re some new properties involving them. f(x) by f(x) in the regions where f(x) > nd replce it by f(x) in the regions where f(x) <. If you think bout it, this is exctly the sme process we went through in the workedout exmples bove.
13 3.5. THE FUNDAMENTAL THEOREM OF CALCULUS 27 Theorem 3.8 (Integrting from point to itself). f(x) dx = If b =, then our region of integrtion hs zero width, so it stnds to reson tht the re is zero, no mtter wht f(x) is. Theorem 3.9 (Splitting up the integrtion region). If < c < b, then b f(x) dx = c f(x) dx + b c f(x) dx. This rule sys tht if we wnt to compute the re of region betwen x = nd x = b, we cn choose ny c betwen nd b, nd split up the re into two prts, one to the left of c nd one to the right. Then we cn seprtely compute the res of the two prts nd dd them to get the totl re Integrting Bckwrds. It my seem to be weird thing to wnt to do, but every once in while, it is hndy to be ble to integrte from lrger x vlue to smller one (integrting righttoleft insted of lefttoright). Our notion of Riemnn sum does not relly hndle this odd sitution, so we will think of this s new definition: Definition 3. (Integrting righttoleft). If < b, then we define b f(x) dx = b f(x)dx. Thus, our new definition sys tht if you integrte from righttoleft, the nswer picks up negtive sign. Thus, in this cse, res bove the xxis get counted s negtive, nd those below the xxis get counted s positive, the opposite of our usul interprettion. All the other properties in this section hold even if the integrls involve righttoleft integrtion The Fundmentl Theorem of Clculus We will now meet one of the most remrkble theorems in clculus. It s often clled the Fundmentl Theorem of clculus, becuse it connects two mjor topics in clculus derivtives nd definite integrls tht seem t first to hve no connection. Specificlly, the Fundmentl Theorem gives n eqution involving definite integrl (which we derived by long process involving pproximting n re by rectngles nd then tking limit) nd ntiderivtives (the reverse of differentition, which reltes to rtes of chnge or slopes, neither of which seems to relte to reunderthecurve):
14 3.5. THE FUNDAMENTAL THEOREM OF CALCULUS 28 Theorem 3. (Fundmentl Theorem of Clculus). If f(x) is continuous function on [, b], nd F (x) is n ntiderivtive of f(x), then b f(x) dx = F (b) F (). Another perspective on the power of this theorem is tht it provides prcticl wy to compute some definite integrls. Without this theorem, we do hve definition of definite integrls, nd we do hve wy to pproximte definite integrl by picking Riemnn sum with some vlue of n, but the lst step of tking n to get the exct vlue of definite integrl is hrd to work through in prctice. Even if you hve some fcility with computing lim n, you would first need tidy symbolic formul for n i= f(x i) b n nd tht is hrd to get, since there re few symbolic formuls for sums n i=. For few simple cses (like f(x) = x), you cn work through ll the lgebr nd get the nswer, but for more complicted situtions, this pproch is not prcticl. But the Fundmentl Theorem of Clculus provides powerful lterntive, by connecting the definite integrl to the indefinite integrls we hve seen erlier in the course. Since we re ble to compute some indefinite integrls (vi the combintion of the tble of bsic ntiderivtives nd methods like substitution nd integrtion by prts), the Fundmentl Theorem gives us wy to compute b f(x) dx if we cn find n ntiderivtive of f. A creful proof of this key theorem is beyond the scope of our course, but we cn mke it t lest plusible if we reflect bck on our distncefromvelocity exmple. Recll tht if v(t) is velocity function (with t equl to time), then the definite integrl b v(t)dt equls the displcement. But notice tht displcement the difference between initil position nd finl position, so it equls s(b) s(), where s is the position function (it sys where the object is t ech time). Thus, for the velocity exmple, b v(t) dt = s(b) s(), which mtches the Fundmentl Theorem exctly, since the position function is n ntiderivtive of the velocity function (becuse the derivtive of position is velocity). 7 Shorthnd Nottion: Becuse the expression F (b) F () rises so frequently, it is sometimes bbrevited s F (x) x=b x=, or simply F (x) b. Exmple 3.2 (Are under curve, revisited). Let s now compute exctly the re under the grph of f(x) = + 4x x 2 between x = nd x = 3. This ws our first reundercurve exmple, nd we used Riemnn sums to pproximte the 7 Thus, the Fundmentl Theorem rises nturlly in the velocity exmple, nd becuse we connected the velocity exmple to reundercurve, it lso pplies there, nd hence to ll definite integrls, since reundercurve ws used to define the definite integrl.
15 3.5. THE FUNDAMENTAL THEOREM OF CALCULUS 29 re, finding.25 s our first pproximtion, nd then.5938 s our (supposedly) improved pproximtion. We know tht the re equls 3 (+4x x2 ) dx, nd the Fundmentl Theorem sys tht we cn compute this definite integrl exctly if we cn find n ntiderivtive F (x) of +4x x 2. But from our previous experience with ntiderivtives, we know tht F (x) = x + 4 x2, from the tble of bsic ntiderivtives. Thus, by the Fundmentl Theorem 3 2 x3 3 ( + 4x x 2 ) dx = F (3) F () = ) (3 + 2(3) 2 33 ( + ) = 2. 3 (Notice tht the pproximtions we computed erlier.25,.594,.85,.955 for n = 6, 2, 24, pproch this exct vlue of 2.) Exmple 3.3 (Distnce trveled given velocity, revisited). Let s now compute the distnce trveled between t = nd t = for cr whose velocity is v(t) = 6 4e t/8. This ws our first velocity exmple, nd we used Riemnn sums to pproximte the nswer s 34.8 ft, nd then 357. ft (supposedly closer to correct). We cn get the exct distnce trveled by computing ( 6 4e t/8 ) dt. We will compute this definite integrl using the Fundmentl Theorem, but the ntiderivtive of 6 4e t/8 does not come directly from the tble of bsic ntiderivtives. Insted, we compute the ntiderivtive by substitution u = t/8, suggested by seeing the function t/8 inside the e to the function. For tht substitution, we hve du dt = 8 so tht ( 6 4e t/8) dt = = du = 8 dt, (6 4e u ) ( 8du) (solving the du eqution for dt nd substituting tht in for dt). Thus, ( 6 4e t/8) dt = ( e u ) du = 48u + 32e u + C = 48( t/8) + 32e t/8 + C = 6t + 32e t/8 + C. Thus, now tht we hve found the ntiderivtive F (t) = 6t + 32e t/8 for the integrnd 6 4e t/8, we know by the Fundmentl Theorem of Clculus, ( (Distnce Trveled) = 6 4e t/8) dt = F () F () ( ) = 6() + 32e /8 + C ( + 32e + C ) = e 5/4 32 = ft
16 3.5. THE FUNDAMENTAL THEOREM OF CALCULUS 2 (Notice tht the pproximtions we computed erlier 34.8, 357., 364.5, 37.3 for n = 5,, 2, pproch this exct vlue of , though in this cse we could hrdly hve guessed this exct vlue from the pproximtions.) Some possible efficiencies when using the Fundmentl Theorem. This lst exmple illustrtes some wys we could mke computtions like this bit more efficient. Skip the +C In the previous exmple, when we computed the ntiderivtive, we hd (of course) +C fctor t the end, but then when we plugged the ntiderivtive into F (b) F (), those +C terms cncelled. This will lwys hppen. So, when you re computing n ntiderivtive for the purpose of plugging it into the Fundmentl Theorem, it is sfe to just drop the +C. (Sid nother wy, the Fundmentl Theorem sks for n ntiderivtive, not ll ntiderivtives.) Perform substitution on the definite integrl Another slight inefficiency is tht we hd both definite integrls nd indefinite integrls in our writeup. This mde perfect sense, becuse the definite integrl ws the number tht nswered our question (the distnce trveled), wheres the indefinite integrl ws the ntiderivtive tht we needed to use the Fundmentl Theorem. However, people relized they could shorten the computtion by performing the substitution on the definite integrl, rther thn on the indefinite integrl s we did bove. This would look like this: We need to compute ( 6 4e t/8 ) dt, which we will do vi the substitution u = t/8. For this choice, we hve du dt = = du = 8 8 dt. In ddition this preliminry computtion, we will lso convert the integrtion limits t =, to corresponding limits for u using the rule u = t/8, yielding u =, /8. Then, the definite integrl version of the substitution computtion would look like: ( 6 4e t/8) u= /8 dt = (6 4e u ) ( 8du) = u= u= 5/4 u= ( e u ) du = ( 48u + 32e u ) u= 5/4 u= = ( 48( 5/4) + 32e 5/4) ( + 32e ) = e 5/4 32 = ft Notice tht this version of the computtion, in which we performed the substitution on the definite integrl, follows mny of the sme steps s the previous computtion, but is bit shorter, prtly becuse it voids the need to convert the vrible u bck to t. However, in order for this method to work, it is crucil tht when you convert the integrnd nd dt to u nd du, you lso convert the t limits to their corresponding
17 3.5. THE FUNDAMENTAL THEOREM OF CALCULUS 2 u vlues notice this hppened fter the first equlity. If you fil to do this, you will plug in t vlues for u nd get the wrong nswer. Perform integrtion by prts on the definite integrl Here is n exmple tht shows how: Exmple 3.4. Let s find the re under the curve y = x sin x from x = to x = π, by computing π x sin x dx. Becuse we re integrting product, nd one of the terms (the x) becomes nicer when we differentite it, this is good cndidte for u in n integrtion by prts. Tht would mke v = sin x, resonble choice, since its ntiderivtive is not too bd. Following the usul first step of implementing integrtion by prts, we compute: u = x v = sin x u = v = cos x. Now we pply the integrtion by prts formul directly to the definite integrl: π π x sin x dx = x( cos x) x=π x= ()( cos x)dx = ( x cos x) x=π x= + π So, the re under the curve is π. cos x dx = ( x cos x) x=π x= + (sin x) x=π x= = (( π cos π) ) + (sin π sin ) = (π ) ( ) = π Notice tht in the first equlity, when the integrtion by prts formul outputs prt of the nswer (x( cos x) in this exmple), tht term must get evluted t b, nd t, nd then subtrcted. Unlike with substitution, there is no need to convert the limits, since with integrtion by prts, there is no chnge of vrible (in our exmple, the vrible ws x throughout).
18 CHAPTER 4 Numericlly Approximting Definite Integrl Key Words nd Concepts: Left Endpoint sum Right Endpoint sum Midpoint sum Trpezoid sum Simpson s Rule Questions to Consider: How do different pproximtions of b f(x) dx relte to ech other? How do they relte to the exct nswer? How do we estimte the vlue of definite integrl to some desired level of ccurcy? We begn our unit on definite integrls by pproximting quntity of interest, specificlly pproximting n reundercurve by rectngle sums. In this chpter, we return to this theme of pproximtion, first improving our understnding of the methods we hve lredy seen, nd then introducing more ccurte pproximtions. 4.. More on Left nd Right Endpoint Sums Recll from the beginning of this unit the ide of pproximting n re under the curve by rectngle sum: n ( Are under grph of f(x) from x = to x = b f(x i ) b ). n In this pproximtion, we must choose vlue of n (the number of intervls tht the segment [, b] on the xxis is split into) nd then, for ech segment i, we must choose vlue x i in tht segment. Two reltively strightforwrd choices re to pick x i to be the left endpoint of segment i, or pick it to the right endpoint. So tht we cn explore these pproximtions bit more, let s give them nmes: i= 22
19 4.. MORE ON LEFT AND RIGHT ENDPOINT SUMS 23 Definition 4. (Left nd Right Endpoint sums). Given function f(x), rnge x = to x = b, nd positive integer n, let n ( L n = f(x i ) b ) for x i the left endpoint of segment i, n nd let R n = i= n i= ( f(x i ) b ) n for x i the right endpoint of segment i. We hve sid tht incresing n probbly mkes these pproximtions closer to the exct vlue of b f(x) dx, but tht obviously is pretty imprecise sttement. Cn we do better? Here re severl wys to think bout it Brcketing the exct solution. In one specil cse, we cn know for sure tht the exct vlue of b f(x) dx is between L n nd R n : Theorem 4.2. If f(x) is n incresing function of x for x b, then (for ny choice of n), L n b f(x) dx R n. If f(x) is decresing function of x for x b, then (for ny choice of n), R n b f(x) dx L n. A picture mkes the reson behind this theorem pretty cler. Consider the cse of n incresing function (the logic for decresing function is very similr):
20 4.. MORE ON LEFT AND RIGHT ENDPOINT SUMS 24 The grph of f is shown in red, nd the left imge shows the rectngles corresponding to L 8, while the right imge shows the rectngles corresponding to R 8. The re L n is smller thn the reunderthecurve, becuse the height of ech rectngle f(x i ) is the lowest possible vlue of f in the intervl (since it is tken t the left endpoint nd the grph of f is rising). In contrst, the re R n is lrger thn the reunderthecurve, becuse the height of ech rectngle f(x i ) is the highest possible vlue of f in the intervl (since it is tken t the right endpoint nd the grph of f is rising). The bility to brcket the exct solution between two numericllycomputed pproximtions is quite useful. It s certinly more precise informtion thn just being ble to sy you hve computed n pproximtion. Even better, the brcket llows you to sy something specific bout the error in ech of your pproximtions: Theorem 4.3. If f(x) is n incresing function of x for x b, then (for ny choice of n), Error in the pproximtion L n L n R n nd Error in the pproximtion R n L n R n We get the sme two error bounds if f(x) is decresing function of x for x b. You might quite resonbly be sking So wht do I do if f is not n incresing function nd not decresing function, but rther hs some segments where it is incresing nd some where it is decresing? In tht cse, by the property of definite integrls tht we clled Splitting up the integrtion region, you cn split up b into subintegrls so tht on ech piece, f is either lwys incresing or lwys decresing, nd then you cn pply this theorem to ech piece.
21 4.. MORE ON LEFT AND RIGHT ENDPOINT SUMS 25 The logic of this theorem is pretty strightforwrd: if you know the exct solution is between L n nd R n, then the exct solution is definitely closer to L n nd to R n thn the two pproximtions re to ech other (nd L n R n is how fr the two pproximtions re from ech other). Exmple 4.4 (Approximting the re inside qurtercircle). We know tht the eqution of the unit circle is x 2 + y 2 =. If we solve for y, we find y 2 = x 2, so tht y = ± x 2. If we choose the + sign, we will get semicircle (the top hlf of the full circle). If we only consider x, we will get qurtercircle (the righthlf of our semicircle). Thus, the definite integrl x 2 dx computes the re inside the qurtercircle. Let s pproximte tht re using our brcketing theorems. Note tht the theorems do pply becuse the curve y = x 2 is decresing with x s x. We will choose n = 8 (firly rbitrrily, to mke the computtions of L n nd R n not too burdensome). For L n, the left endpoints re x =, x 2 = 8, x 3 = 2 8,, x 8 = 7 8. Plugging these into f(x) = x 2, we find f(x ) =., f(x 2 ) =.9922, f(x 3 ) =.9682,, f(x 8 ) =.484. (At this point we hve no knowledge of the ccurcy of L 8. I recorded 4 deciml point in ech nswer, in cse the ccurcy is tht high, though it will not turn out to be nerly tht high.) Since b n = 8 =.25, the pproximtion is L 8 = (.)(.25)+(.9922)(.25)+(.9682)(.25)+ +(.484)(.25) = For R n, the right endpoints re x = 8, x 2 = 2 8, x 3 = 3 8,, x 8 =. Plugging these into f(x) = x 2, we find so tht f(x ) =.9922, f(x 2 ) =.9682, f(x 3 ) =.927, f(x 8 ) =, R 8 = (.9922)(.25) + (.9682)(.25) + (.927)(.25) + ()(.25) =.799. Thus, our conclusion (by the first theorem) is:.799 Are.8349 If you prefer to express the result s n pproximtion with n error bound, since L n R n = =.25, we cn use our second theorem to sy: or, if we prefer: Are.8349, with Error.25, Are.799, with Error.25.
22 4.2. MORE ACCURATE METHODS FOR APPROXIMATING A DEFINITE INTEGRAL How do the ccurcy of L n nd R n improve with n? OK, so now we hve wy of bounding the error in the pproximtions L n nd R n (by using the brcketing theorems) nd we hve the vgue sense tht if we increse n, the pproximtions will improve. It would sure be nice to hve more specific sense of how much improvement we will get when we increse n. Here you go: Rule of Thumb: Once you hve chosen n lrge enough tht L n or R n is resonbly close to the exct solution, multiplying n by will divide the error by pproximtely, thereby giving you one dditionl digit of ccurcy in your nswer. This rule of thumb is reminiscent of the previous rule of thumb we sw erlier in the course, when we were discussing Euler s Method. Let s see it in ction: Exmple 4.5 (Are inside qurtercircle, continued). In our work on this exmple bove, we lerned tht the re ws pproximtely L 8 =.8349, with error.25. Let s sy tht we would like to pproximte the re to three deciml plces, which we could define s hving the error be less thn.. Thus, we wnt to divide our current error bound by 25. The rule of thumb sys tht in order to chieve tht gol, we must multiply n by 25, so tht we use n = 8 25 =. Whew! Tht does not sound fun to do by hnd, but with computer, it s no big problem. For L n, the left endpoints re x =, x 2 =, x 3 = 2,, x = 999. Plugging these into f(x) = x 2, we find f(x ) =, f(x 2 ) = , f(x 3 ) = ,, f(x ) = Since b n = =., the pproximtion is L = ()(.)+( )(.)+( )(.)+ +(.4478)(.) = Under the ssumption tht the Rule of Thumb holds, this pproximtion should be correct to within.. If we wnt to be sure, we cn compute R nd pply the brcketing theorem. The computtion (detils not shown) yields R =.7849, nd then the brcketing theorem tells us tht.7849 Are.7859, confirming tht our new pproximtion is ccurte to within More ccurte methods for pproximting definite integrl At this point, you my hve sense of despir tht it tkes so much effort (so mny rectngles!) to pproximte b f(x) dx modertely well ( few deciml plces). Cheer up: we will now lern some pproximtions tht re not much hrder to implement thn L n nd R n but re much more ccurte. 2 Becuse we know from geometry tht the re of qurtercircle is π/4, we cn use the results of our exmple to pproximte π. If we multiply L nd R by 4, we know tht these will brcket the number π, which yields the inequlities π
23 4.2. MORE ACCURATE METHODS FOR APPROXIMATING A DEFINITE INTEGRAL Midpoint nd Trpezoid pproximtions. The first two improved pproximtions re firly intuitive extensions of the rectnglesum ide. The first is, in fct, lso rectngle sum, but it uses midpoints of ech intervl to determine the rectngle height, rther thn the left endpoint or right endpoint. The formul for this method is very much like our formuls for L n nd R n : Definition 4.6 (Midpoint sum). Given function f(x), rnge x = to x = b, nd positive integer n, let n ( M n = f(x i ) b ) for x i the midpoint of segment i. n i= Here is visuliztion of the rectngles being used: The second pproximtion involves shifting from using rectngles to using the trpezoids you get if you replce the curve by sequence of digonl lines. Here s the picture first:
24 4.2. MORE ACCURATE METHODS FOR APPROXIMATING A DEFINITE INTEGRAL 28 We hve included zoomin (on the right) of the curviest portion of the grph, since otherwise it is hrd to distinguish the trpezoids from the re under the curve. The computtion of this sum involves the formul for the re of ech trpezoid 3 : Definition 4.7 (Trpezoid sum). Given function f(x), rnge x = to x = b, nd positive integer n, let n ( ) f(ci ) + f(d i ) b T n = 2 n i= for c i the left endpoint of segment i nd d i the right endpoint of segment i. From the pictures, it seems likely tht M n nd T n re better pproximtions thn L n nd R n, since the midpointrectngles seem to hve roughly s much overhng s undershoot reltive to the curve, nd the trpezoids re brely distinguishble from the reunderthe curve. But, let s im for more thn just hope nd plusibility. We will see tht there is wy to use M n nd T n to brcket the exct vlue of b f(x) dx, s well s Rule of Thumb bout how much we need to increse n to gin dditionl digits of ccurcy when using M n or T n A brcket involving M n nd T n. Recll tht we sw tht the rectngle sums L n nd R n were gurnteed to brcket the exct solution b f(x) dx if 3 As reminder, trpezoid is qudrilterl such tht one pir of opposite sides is prllel. The prllel sides re often clled bses nd the seprtion between them is clled the height, in which cse the re of the trpezoid is b +b 2 h. In our exmple, for ech gry region in the 2 figure, the two verticl lines re prllel, nd their lengths re f(c i) nd f(d i), while the seprtion between them is b. n
25 4.2. MORE ACCURATE METHODS FOR APPROXIMATING A DEFINITE INTEGRAL 29 f ws either lwys incresing for x b, or lwys decresing for x b. In contrst, our new pproximtions M n nd T n re gurnteed to brcket the exct solution if f hs consistent concvity throughout x b: Theorem 4.8. If f(x) is concve down for x b, then (for ny choice of n), T n b f(x) dx M n. If f(x) is concve up for x b, then (for ny choice of n), M n b f(x) dx T n. To see why this theorem is true, let s consider the concvedown cse (the concveup cse follows by similr logic). We cn lredy see from the zoom in our lst figure tht for f concve down, the trpezoids re bit below the curve, so tht T n b f(x) dx. To justify the other hlf of the inequlity, let s zoom in on portion of our midpoint figure: If you look crefully, you cn see tht the downwrd concvity of the curve mens tht for ech midpointrectngle, the overhng (the prt of the rectngle bove the red curve) is bit lrger thn its undershoot (the gp between the rectngle nd the red curve), mking b f(x) dx M n. You my be more convinced by version of the figure tht dds in one of the intervls segment AE tht is tngent to the curve t the midpoint (point C):
26 4.2. MORE ACCURATE METHODS FOR APPROXIMATING A DEFINITE INTEGRAL 22 The downwrd concvity of the grph of f (the red curve) mkes it lie below the segment AE, so tht, for this intervl Notice lso tht Are under curve Are under AE. Are under AE = Are of rectngle Are of ABC + Are of CDE. nd since ABC is congruent to CDE, we hve Are under AE = Are of rectngle. Putting these fcts together, we hve Are under curve Are of rectngle, nd summing over the intervls, we find b f(x)dx M n How do the ccurcy of M n nd T n improve with n? The sttement below (though not precise theorem) gives substnce to our clim tht M n nd T n re better pproximtions thn L n nd R n. Rule of Thumb: Once you hve chosen n lrge enough tht M n or T n is resonbly close to the exct solution, multiplying n by will divide the error by pproximtely, thereby giving you two dditionl digits of ccurcy in your nswer. Exmple 4.9 (Are inside qurtercircle, continued some more). Let s see how to use M n nd T n to pproximte x 2 dx, the re inside qurtercircle. We ll choose n = 8, to mtch our first choice of n when we used L n nd R n. For M n, the midpoints re x = 6, x 2 = 3 6, x 3 = 5 6,, x 8 = 5 6.
27 4.2. MORE ACCURATE METHODS FOR APPROXIMATING A DEFINITE INTEGRAL 22 Plugging these into f(x) = x 2, we find f(x ) =.998, f(x 2 ) =.9823, f(x 3 ) =.9499,, f(x 8 ) =.348. (At this point we hve no knowledge of the ccurcy of M 8. I recorded 4 deciml point in ech nswer, in cse the ccurcy is tht high.) Since b n = 8 =.25, the pproximtion is M 8 = (.998)(.25)+(.9823)(.25)+(.9499)(.25)+ +(.348)(.25) = For T n, we use the left nd right endpoints, which re,.25,.25,.375,,.875,. Plugging these into f(x) = x 2, we find f() =, f(.25) =.9922, f(.25) =.9682, The pproximtion is T 8 = (.25) + 2 f(.375) =.927,, f(.875) =.484, f() = (.25) (.25) 2 (.25) = Bsed on the brcket theorem for our concvedown function, we know tht.7724 Are.7892 If you prefer to express the result s n pproximtion with n error bound, since M n T n = =.68, we cn sy: or, if we prefer: Are.7724, with Error.68, Are.7892, with Error.68. Notice tht this error bound is ten times smller thn the ccurcy we got when we used L n or R n with the sme n = 8. If we now sy tht we strive for threedecimlplce ccurcy (error less thn.), our Rule of Thumb sys tht if we multiply n by, we should divide the error by pproximtely, which will be more thn enough to chieve our gol. Thus, either T 8 or M 8 should give us threedecimlplce ccurcy (in fct, pretty close to four). We ll compute them both, so tht we hve brcket nd get gurnteed error bound. For M n, the midpoints re x = 6, x 2 = 3 6, x 3 = 5 6,, x 8 = Plugging these into f(x) = x 2, we find f(x ) =.99998, f(x 2 ) = , f(x 3 ) =.99952,, f(x 8 ) =.629.
28 4.2. MORE ACCURATE METHODS FOR APPROXIMATING A DEFINITE INTEGRAL 222 (We recorded 6 deciml points in ech nswer, since we now hve reson to believe we hve firly high ccurcy.) Since b n = 8 =.25, the pproximtion is M 8 = (.99998)(.25) + ( )(.25) + (.99952)(.25) + + (.629)(.25) = For T n, we use the left nd right endpoints, which re,.25,.25,.375,,.9875,. Plugging these into f(x) = x 2, we find f() =, f(.25) = , f(.25) = , f(.375) = , f(.9875) =.5769, f() =. The pproximtion is T 8 = (.25) + (.25) (.25) (.25) = Bsed on the brcket theorem for our concvedown function, we know tht Are so we hve indeed pproximted the re to within three deciml plces (in fct, our error bound is =.6). 4 We needed mny fewer intervls to chieve threedecimlplce ccurcy thn we needed with L n or R n (n = 8 s compred to n = ) An even more ccurte method: Simpson s Rule. Once you lern tht M n nd T n brcket the exct solution, you might nturlly think tht n even better pproximtion ide is to verge them. Tht would be pretty smrt. It turns out it is even smrter to use prticulr weighted verge: Definition 4. (Simpson s Rule). Given function f(x), rnge x = to x = b, nd positive integer n, let S n = 2M n + T n. 3 This Simpson s Rule pproximtion is thus betwen M n nd T n, but becuse of the coefficients, it is 2/3 of the wy from T n to M n. Why this prticulr weighting? It turns out to be the re you get when you replce the grph of f(x) by set of prbols tht pss through the grph of f(x) t the intervl endpoints nd midpoints. As we sw with T n, the line segments tht pss through the grph t 4 Notice tht the Rule of Thumb is not precise; it suggested the error bound would be divided by, but in this cse, it ws divided by bout 3. The Rule of Thumb turns out to be bit off in this cse due to the verticl tngent t x =. If we use M 8 nd T 8 to estimte the integrl just from x = to x = /2, thereby voiding the verticl tngent, the two pproximtions re ccurte to six deciml plces. So, there re some subtleties to these pproximtions tht re beyond the scope of our course but could be found in more dvnced course in numericl nlysis.
29 4.2. MORE ACCURATE METHODS FOR APPROXIMATING A DEFINITE INTEGRAL 223 the intervl endpoints re often indistinguishble (to the eye) from the grph itself. When we replce those line segments by prbolic segments, the fit becomes even better. Becuse S n is just single pproximtion, it cnnot possibly brcket the exct solution, but it does hve n impressive sttement of its expected ccurcy: Rule of Thumb: Once you hve chosen n lrge enough tht S n is resonbly close to the exct solution, multiplying n by will divide the error by pproximtely, thereby giving you four dditionl digits of ccurcy in your nswer. Exmple 4. (Are inside qurter circle, continued one lst time). We cn use our computed vlues of M 8 nd T 8 to compute the Simpson s Rule pproximtion S 8 for x 2 dx. However, becuse Simpson s Rule is expected to be very ccurte, we should mke sure we hve very ccurte versions of T 8 nd M 8, wheres previously we computed them to only 4 deciml plces. If we go bck nd recompute them, we find T 8 = nd M 8 = Thus, S 8 = 2M 8 + T 8 3 = 2( ) = Given tht we know tht the exct solution is between nd.78558, we cn see tht S 8 is within.2 of the exct solution, lmost three digits of ccurcy with only n = 8. Exmple 4.2 (Are under n rch of the sin curve). To see the full power of Simpson s Rule, we will shift to n exmple tht does not hve verticl tngent (which will interfere with the Rule of Thumb for S n s it did with M n nd T n ). Our new exmple will be to compute the re under y = sin x from x = to x = π.
30 4.2. MORE ACCURATE METHODS FOR APPROXIMATING A DEFINITE INTEGRAL 224 With the benefit of the Fundmentl Theorem, we cn compute the exct solution π Approximtions with n = 4 sin x dx = cos x x=π x= = cos(π) ( cos()) = + = 2. For M n, the midpoints re Since b n = π 4, the pproximtion is ( π ) ( π 3π M 4 = sin sin 8 x = π 8, x 2 = 3π 8, x 3 = 5π 8, x 4 = 7π 8. ) π 4 + sin ( ) 5π π sin ( ) 7π π 8 4 = For T n, we use the left nd right endpoints, which re, π 4, π 2, 3π 4, π. Plugging these into f(x) = sin x, we find f() =, f(π/4) = 22, f(π/2) =, f(3π/4) = 2/2, f(π) =. The pproximtion is T 4 = + 2/2 (π/4)+ 2 2/2 + 2 (π/4)+ + 2/2 2/2 + (π/4)+ (π/4) = For only hving used n = 4 intervls, these re decent pproximtions to the exct solution of 2. Next we compute S 4 = 2M 4 + T 4 = 2.269, 3 giving quite impressive error of.269. Approximtions with n = 4 Now we ll check our Rules of Thumb, by multiplying n by. Since we hve shown detils of the methods in previous exmples, we will only report the results: M 4 = , T 4 = , S 4 = 2.26 Sure enough, The error in M 4 (.5) is bout times smller thn the error in M 4 (.5). The error in T 4 (.) is bout times smller thn the error in T 4 (.). The error in S 4 (2 8 ) is bout times smller thn the error in S 4 (.2). At this point, the populrity of Simpson s Rule is probbly quite sensible: s this exmple shows, with only few hundred computtions (bout hundred for T 4 nd nother hundred for M 4, nd then simple weighted verge of the two), we got eight digits of ccurcy in the definite integrl.
31 4.2. MORE ACCURATE METHODS FOR APPROXIMATING A DEFINITE INTEGRAL Automted lgorithms. The exmples in this section often involve step requiring humn thought/intervention, e.g., nlyzing the integrnd f(x) to determine whether the pir L n, R n, or the pir T n, M n formed brcket for the exct vlue of b f(x)dx. Over the yers, reserchers hve worked hrd to develop lgorithms tht re utomted, in the sense tht you provide the integrnd f(x) nd trget ccurcy nd the lgorithm determines the vlue of n nd n pproximtion (mybe M n, T n, S n, or something even fncier) tht chieves the desired ccurcy. Understnding how these utomted lgorithms work is beyond the scope of this courses (consult n introductory text in numericl nlysis if you re interested in lerning bout them), but it is certinly worth knowing of their existence, nd indeed we will rely on them in some future exmples s we shift to studying pplictions of definite integrls. Technology Tool 4.3. It is quite possible tht your clcultor includes n utomted numericlpproximtion of b f(x)dx tht llows you to input the integrnd f(x) nd the limits nd b nd outputs numericl pproximtion to b f(x)dx tht chieves some desired ccurcy; for exmple, on TI clcultors, you cn use the commnd fnint to get ccurcy of t lest 5 (or you cn set tolernce to get different level of ccurcy). In Mthemtic (or Wolfrm Alph), if you input n integrl, it will first ttempt to compute it symboliclly (s we hve seen), but if it is unble to do so, it will echo bck to you the definite integrl unchnged, with link you cn click to get numericl pproximtion: Once you click the link, Mthemtic uses n utomted lgorithm to produce numericl pproximtion: Typiclly, this numericl pproximtion will be ccurte to t lest 6, or you cn set n AccurcyGol to get different level of ccurcy.
32 CHAPTER 5 Applictions of the Definite Integrl: Geometry Key Words nd Concepts: Arclength of curve in the plne Slicing vrible Solid of revolution Questions to Consider: How cn we use slicing to rrive t definite integrl tht computes some geometric quntity? We hve lredy seen tht certin res cn be computed using definite integrls; indeed, it ws the ppliction tht led to the definition of b f(x) dx. In this chpter, we will see tht b f(x) dx cn be used to compute not just res, but lso the length of curve, nd even (in some cses) the volume of 3D solid. By the end of the chpter, we will hve seen severl exmples of such definite integrls, ech with different integrnd f(x) depending on wht geometric quntity is being computed. Our focus will be on the process of deriving those formuls by slicing rgument, since mstering tht process will llow you to solve much wider rry of problems thn just the prticulr ones we show here. 5.. Arclength Let us consider once gin curve y = f(x) in the plne between x = nd x = b, but rther thn find the re under tht curve, let us seek the length of the curve itself (often clled the rclength to indicte tht the object is curved nd not stright). Like the re, this is problem tht typiclly will not be solvble by bsic geometry, becuse we only know formule for lengths of line segment or circulr rcs. The clue to solving this problem is in tht lst sentence: if we chop the curve up into n pieces, then ech piece will be very close to stright line, nd then we will be ble to find the length of ech piece, nd dd up those lengths to get n pproximtion to the length of the curve. Furthermore, it is t lest plusible tht s we increse n, this pproximtion gets better nd better, nd in the limit n, the pproximtion converges to the exct length of the curve. This process mirrors precisely the slicing ide tht led to our first two pplictions of the definite integrl (reunderthecurve, nd distncefromvelocity). 226
33 5.. ARCLENGTH 227 But this does not men tht the rclength is b f(x) dx: to see which integrl we get, we must spell out the slicing in more detil Derivtion of the Arclength Integrl. Strting with the curve y = f(x) from x = to x = b, let us slice it in exctly the sme wy we did for the re ppliction: tke the segment on the xxis from x = to x = b nd brek it into n equlwidth intervls. For ech of these intervls on the xxis, we will consider the piece of the curve y = f(x) tht lies bove the intervl. For exmple, in the imge below (for which f(x) = x + x 2, =, nd b = 2): we chose n = (note the dots on the xxis) nd highlight the 2nd slice of the curve in red nd the 8th slice in blue. The next step in the slicing process is to solve (pproximtely) the problem on ech slice. In this cse, the problem is to find the length of the curve, so our gol would be to find the lengths of ech slice of the curve, like the red piece nd the blue piece in the bove figure. Given tht imge, it seems resonble to pproximte ech piece by stright line; indeed, to the eye the blue piece looks stright nd the red piece only slightly curved. Let s zoom in on the red piece: We hve dded to the figure (s green segment) n pproximtion of the red curve by stright line segment, nmely one tngent to the curve t the left endpoint. Our gol is now to write down n expression for the length of the green segment, nd to tht end, we hve lbeled the leftendpoint of the intervl s x i nd the width of the intervl s x (which will just equl (b )/n s we hve seen in erlier exmples). In order to compute the length of the green segment, we hve drwn
34 5.. ARCLENGTH 228 right tringle: its legs re dshed lines, one of which hs length x, nd the other of which hs length tht we hve clled y (since it is length in the ydirection). But note tht we cn relte y to x if we recll tht the slope of the curve t x = x i is f (x i ), nd since we re pproximting the red curve by the green segment, we would sy y x = slope of green segment = f (x i ). Now the Pythgoren Theorem tells us tht: Length of green segment = ( x) 2 + ( y) 2, nd if we solve the slope eqution for y, this becomes Length of green segment = ( x) 2 + (f (x i ) x) 2 = ( x) 2 ( + [f (x i )] 2 ) = x + [f (x i )] 2. Now tht we hve solved the problem on the slice, most of our work is over. The next step is to dd up our solutions on the slices: n Arclength of curve = Sum of lengths of slices + [f (x i )] 2 b n, (where we hve filled in the formul for x). The finl step is to tke n, t which point the pproximtion should become n exct equlity: Arclength of curve = lim n [ n i= i= ] + [f (x i )] 2 b n But now we cn see tht we hve n expression tht mtches exctly the definition of the definite integrl: lim n outside, inside which is n i=, nd then inside tht is some function of x i times (b )/n. The only difference is tht when we were computing re, the some function ws f(x i ), wheres now the some function is + [f (x i )] 2. Thus, we hve the definite integrl b + [f (x)] 2 dx, rther thn b f(x) dx we got when computing re. This derivtion hs sketched the resons behind the following theorem: Theorem 5.. If f(x) is differentible function from x = to x = b, nd its derivtive function f (x) is continuous from x = to x = b, then b Arclength of the curve y = f(x) from x = to x = b is + [f (x)] 2 dx Shorthnd Derivtion of the Integrl. The derivtion we just completed fills in ll the steps to connect given quntity (the rclength of the curve) to the definition of the Riemnn integrl, thus showing us wht integrl needs to be computed. Once you hve some experience with this process, you cn shorten
35 5.. ARCLENGTH 229 the writeup of the derivtion bit, minly by dropping the lst few steps which involve writing out the sum, writing down lim n, nd compring the result to the Riemnn integrl definition. Insted, we focus on the first hlf of the definition, nd stop t the point where we solve the problem on the slice. One slight difference is tht we write the solution to the problem on the slice in terms of x rther thn writing out x = (b )/n s we did bove to connect to the definition of definite integrl. Our derivtion would look just like the first hlf of the derivtion bove: we would sketch the sme two figures, nd work through the sme geometric rgument to get to this result: Arclength of slice + [f (x i )] 2 x At this point, we merely condense the second hlf of the rgument: becuse we re slicing, we know we will be dding up the slices, nd tking n, nd tht will crete definite integrl. The expression multiplying x will become the integrnd, nd we will surround it by b nd dx to get the finl nswer: b Arclength of entire curve = + [f (x)] 2 dx. If you compre the lst two equtions, you will see tht there is simple procedure to go from solve the problem on the slice to solve the originl problem : You put b out front. The x i becomes x. The x becomes dx. For wide rnge of problems, if you figure out wy to slice so tht you cn solve the problem on the slice nd get the nswer expressed in terms of x i nd x, the bove three steps will generte the definite integrl tht solves the problem Arclength Exmples. Exmple 5.2 (Find the rclength of the curve y = x 3/2 from x = to x =.). Ech of these opertions mkes t lest intuitive sense: we know we will hve sum tht becomes n integrl when we tke the limit; the expression x i for point in the ith intervl becomes the expression x for point in [, b]. And you my well hve seen the trnsition from x ( finite width) to dx (n infinitesiml width) when you sw the connection between n verge rte of chnge y/ x nd n instntneous rte of chnge dy/dx.
36 5.. ARCLENGTH 23 The problem fits the form of the Theorem exctly, with f(x) = x 3/2, =, nd b =, so we know tht Arclength = + [f (x)] 2 dx. We hve f (x) = 3 2 x/2, which mens tht [f (x)] 2 = 9 4x, so Arclength = x dx. To compute this integrl, we note the function + 9 4x inside the squreroot function suggests substitution u = + 9 4x would be helpful. For this choice, we hve: so tht du dx = 9 4 = du = 9 4 dx = dx = 4 9 du, Arclength = u=3/4 u= u ( 4 9 du ). (Note we re opting to use the method in which we pply the substitution to the definite integrl, so we re obliged to convert the xlimits to ulimits, s ws done in this expression.) At this point, we cn complete the computtion using the tble of bsic ntiderivtives: u=3/4 ( ) 4 Arclength = u= 9 u/2 du = 4 u 3/2 u=3/4 9 3/2 = u= [ (3/4) 3/2 3/2] = 8 27 [ ] (3/4) 3/2 =.4397 Truth be told, this lst exmple ws bit typicl, in tht the curve y = f(x) ws crefully chosen to llow the definite integrl b + [f (x)] 2 dx to be computed symboliclly without too much difficulty. In relity, for most functions f(x), symbolic solution of the rclength integrl will be impossible, or t lest very hrd, lrgely becuse of the squre root tht ppers in the formul. But do not despir! At this point, we hve (from the lst chpter) robust method to compute ny definite integrl to ny level of ccurcy tht we my wnt. When computing rclength integrls (nd, rgubly, most definite integrls tht pper in rel pplictions), you should expect to use those numericl pproximtions once you hve set up the integrl. Let s try this by returning to the exmple from the strt of the chpter. Exmple 5.3. Find the rclength of the prbolic curve y = x + x 2 from x = to x = 2. We hve f(x) = x + x 2, so tht f (x) = + 2x. The theorem in
37 the lst section tells us tht Arclength = 5.. ARCLENGTH ( + 2x) 2 dx You my be tempted to compute the integrl symboliclly, but in fct, tht would require methods beyond wht we hve covered. Insted, let s use M n nd T n to compute this rclength to three deciml plces. As strt, we choose n = 8. For M 8, the midpoints re x = 8, x 2 = 3 8, x 3 = 5 8,, x 8 = 5 8. Plugging these into f(x) = + ( + 2x) 2, we find (recording 4 deciml plces): Since b n f(x ) =.25, f(x 2 ) =.38, f(x 3 ) =.38,, f(x 8 ) = = 8 =.25, the pproximtion is M 8 = (.25)(.25) + (.38)(.25) + (.38)(.25) + + (2.962)(.25) = For T 8, we use the left nd right endpoints, which re,.25,.5,.75,,.75, 2. Plugging these into f(x) = + ( + 2x) 2, we find f() =.442, f(.25) =.8, f(.5) =, The pproximtion is T 8 = f(.75) =.8,, f(.75) = , f(2) = (.25) (.25) (.25) (.25) = Bsed on the brcket theorem for our concveup function, we know tht Arclength Thus, by choosing n = 8, we obtined pproximtions M 8 nd T 8 tht re within.26 of the exct nswer. Bsed on the Rule of Thumb for M n nd T n, we expect tht if we multiply n by, we will decrese the error by fctor of, which should bring it below.. Thus, we compute M 8 nd T 8 (detils not shown): M 8 = 3.434, T 8 = Since these two pproximtions re within.3 of ech other, nd they brcket the exct solution, we know tht they re ech within.3 of the exct solution. So, Arclength 3.4 (with error less thn.)
38 5.. ARCLENGTH Swpping the slicing vrible. Thus fr in our pplictions, we hve consistently sliced the xxis into segments, but there isn t ny reson we couldn t contemplte insted slicing the yxis. For exmple, if we re computing n rclength of curve, we could perfectly well express the curve in the form x = g(y) nd then repet ll the steps of our slicing derivtion with x nd y trding roles. This would led to: Theorem 5.4. If g(y) is differentible function from y = c to y = d, nd its derivtive function g (y) is continuous from y = c to y = d, then d Arclength of the curve x = g(y) from y = c to y = d is + [g (y)] 2 dy. If you think bout it, this isn t relly new theorem; it s exctly the sme s the old theorem with some vrible nmes chnged. Is there ny reson to ever wnt to do this? How bout two? First, if you re computing the integrl symboliclly, the yintegrl could be esier to solve thn the xintegrl. Second, if you re numericlly pproximting the integrl, swpping to the yintegrl could llow you to void verticl tngents, which cn cuse poor performnce in our numericl pproximtions (nlogously to wht we hve seen with res). Exmple 5.5 (Find the rclength of the curve y = x from x = to x = 4.). c First let s try the integrl the old wy, using x s the slicing vrible. We hve f(x) = x = x /2, so tht f (x) = 2 x /2, nd thus: 4 ( ) 2 4 Arclength = + 2 x /2 dx = x dx = + 4x dx. If you were to tckle this integrl symboliclly, it would be dunting. The function inside function ppers to be u = + 4x, but when you compute du you will find du = dx, nd tht /x 2 fctor is not present in the integrl to mke the 4x 2 substitution go through esily. If insted you try numericl pproximtion, you will lso fce some chllenges, due to the verticl tngent t x =. Indeed, the integrnd is not even defined t x =, which mens the formuls for L n, T n, nd S n do not mke sense. We could
39 5.2. AREAS 233 try M n, which never plugs x = into the integrnd, so t lest it will generte n nswer. We find tht M 8 = nd M 8 = (computtionl detils not shown), nd if we tke n, we will get convergence to the correct nswer. However, the convergence is quite slow. Typiclly, when we went from n = 8 to n = 8, we would expect M n to gin two digits of ccurcy, but tht did not hppen here. The exct nswer turns out to be , so M 8 hd error of bout.2 nd M 8 hd error of bout.7, not close to following the Rule of Thumb tht sys the error should be divided by bout. Let s see wht hppens if we switch to using y s the slicing vrible. We must reexpress our curve in the form x = g(y), which is pretty strightforwrd: we squre both sides of the eqution y = x to find y 2 = x, so tht x = g(y) = y 2. Then we compute g (y) = 2y nd plug it into the yslicing version of the rclength formul: y=2 Arclength = + (2y) 2 dy. y= (Note tht in ddition to plugging g (y) into the formul, we hd to work out the integrtion limits for y: since x rnged from to 4, nd y = x on the curve, we hve tht y rnges from to 2.) This integrl is bit tidier thn the xslicing integrl for this problem, so switching to yslicing mkes the symbolic pproch more imginble (though it turns out to be beyond the scope of wht we hve lerned in our course). How bout if we try our numericl pproximtions? Now the integrnd + 4y 2 is welldefined everywhere in our rnge y 2, so we could use ny of our numericl pproximtions (s compred to only M n being defined for the xslicing integrl). The roleswpping of x nd y turns the verticl tngent t x = to horizontl one (t y = ), which is not problem. To compre to our results from the xslicing cse, we compute M 8 = nd M 8 = (computtionl detils not shown). These results re much better thn in the xslicing cse: the n = 8 pproximtion is within.5 of the exct solution , nd the n = 8 pproximtion is within.5, dhering to the Rule of Thumb tht multiplying n by will decrese the error by fctor of. Thus, the bility to slice either xis gives us quite useful flexibility, whether our im is to compute solution symboliclly or vi numericl pproximtion Ares We hve, of course, lredy used definite integrls to compute res, specificlly the re between curve y = f(x) nd the xxis from x = to x = b. Now we will extend these ides to compute res of other sorts of regions in the plne Are between two curves. It tkes just minor dpttion of our previous slicing process to compute the re between two curves (s compred to the re between one curve nd the xxis). Exmple 5.6 (Find the re between the curves y = x 2 nd y = x /3 ). To even understnd the question, we need picture:
40 5.2. AREAS 234 Apprently these two curves hve welldefined region between them from x = to x =, nd tht is the region whose re we wnt to compute. Since now we hve region whose top nd bottom re curved, clerly we cnnot use simple geometry formuls to find the re; insted, we slice the region into n rectngles: Here we hve used n = 8 rectngles, with the top nd bottom loctions of ech rectngle coming from the midpoint of the curves. (Choosing the midpoint mkes this nd future pictures look nice, but you could certinly choose insted the leftendpoints or rightendpoints) As before, our min tsk is to solve the problem on the slice, which in this cse mens finding the re of typicl rectngle in this picture. The width of ech rectngle is x, which equls 8 in this exmple. The height of ny of the rectngles would be the height of the top (ornge) curve minus the height of the bottom (blue) curve. Given tht the equtions of these curves re y = x /3 nd y = x 2 respectively, the height of typicl rectngle is (x i ) /3 (x i ) 2. Thus, Are of slice [ (x i ) /3 (x i ) 2] x,
41 which mens tht the totl re is Are = 5.2. AREAS 235 x= x= [ x /3 x 2] dx. (Recll our shorthnd pproch to constructing definite integrl: if you solve the problem on slice, nd express the nswer in terms of the slice width x nd vlue x i of x in the slice, then to build the definite integrl, you prepend b, chnge ech x i to x, nd chnge x to dx. If you prefer, you could work through the entire process of summing over the contributions of the slices using n i=, then tke lim n, nd then recognize the definition of definite integrl.) To finish the problem, we evlute the integrl: [ ] x= x 4/3 [ Are = 4/3 x3 = 3 4/3 3 x= ] [ ] = = Swpping the slicing vrible. In the rclength ppliction, we found it useful to be ble to use y insted of x s the slicing vrible. The sme is true here. Exmple 5.7 (Find the re between the curves x = y 2 nd y = x 2.). This exmple is wkwrd to set up by slicing the xxis. To see why, we need to drw the region, nd visulize wht the xslice rectngles (for n = 6) would look like: Notice tht the first four rectngles hve their bottom edge on the prbol, wheres the rest hve their bottom edge on the line. Thus, there is no consistent rule for wht the height of the rectngle is, which mens we would not be ble to write single formul for the Are of slice. This doesn t mke it impossible to proceed, but it does tke more work. Becuse x = is the point where our rectngles trnsition from prboltoprbol (for x < ) to linetoprbol (for x > ), we would need to divide the problem into two prts t x =. For < x <, the bottom edge
42 5.2. AREAS 236 of ech rectngle is on the curve y = x (solving the prbol eqution for y nd tking the negtive squre root to get the lower brnch) wheres the top edge is on the curve y = x, so tht Are of slice (for < x < ) [ x i ( x i )] x. For < x < 4, the bottom edge is on the curve y = x 2 wheres the top edge is gin on the curve y = x, so tht Are of slice (for < x < 4) [ x i (x i 2)] x. Thus, dding up the rectngles, we will get the sum of two definite integrls: Are = [ ] 4 [ ] x ( x) dx + x (x + 2) dx nd then we would need to compute ech integrl to compute the problem. Rther thn do this, we will shift to slicing the yxis nd find tht the setup nd computtion re esier. If insted we visulize yslice rectngles (for n = 2): we cn see tht every rectngle follows the sme rule, in tht its left edge is on the prbol nd its right edge is on the line. This mkes the setup of the integrl quite bit esier. The height of ech rectngle is y (which in this cse is 2 ( ) 2 ). The width of rectngle is the x vlue of the right edge minus the x vlue of the left edge. Since the line hs eqution x = y +2 (solving for x, since we need to know wht the x vlue is on the right edge) nd the prbol hs eqution x = y 2, we know tht the width of the rectngle is (y i + 2) (y i ) 2 (where y i is our chosen point the ith intervl of the yxis; in our figure y i is the midpoint of the ith intervl). Thus, Are of slice [ (y i + 2) (y i ) 2] y,
43 5.3. VOLUMES 237 which mens tht the totl re is Are = y=2 y= [ y + 2 y 2 ] dy. We complete the problem by evluting the integrl: [ y 2 Are = 2 ] y=2 [ y y = 3 y= ] [ ] = 3 2. Thus, the process for setting up n re integrl vi slicing the yxis is nlogous to the process when slicing the xxis: we compute the re of the slice in terms of y i ( point chosen in the yintervl) nd y (the width of the yintervl), nd then we prepend the integrl sign (with limits set ccording to the minimum nd mximum vlues of y in the region), nd chnge ll y i to y nd y to dy Volumes In some cses, we cn use the ide of slicing to compute volumes of 3dimensionl solid objects. Anlogously rclength or re, we need to be ble to find wy to slice the 3dimensionl object into thin slbs whose volumes we cn compute (relly, pproximte quite ccurtely) using bsic geometric formuls Solids of revolution. One type of 3dimensionl object for which this technique cn be effective is solid of revolution, in which 2dimensionl region in plne is rotted round some xis in the plne to sweep out 3dimensionl solid. Exmple 5.8. Let R be the region between the curve y = 3x( x) nd the xxis. If we revolve the region R bout the xxis, we get 3dimensionl solid. Wht is the volume of tht solid? Now tht we hve entered the world of 3 dimensions, drwing pictures cn be chllenge. One helpful top is to look for useful 2dimensionl pictures you cn drw in order to help you understnd the 3dimensionl sitution. 2 In this problem, let s drw the 2D region R, nd lso set of rectngles tht pproximte R from slicing the xxis: 2 Mybe you cn drw well in 3D, or mybe not. But since you hve thrived in mth to rech this course, you probbly cn drw pretty good 2D pictures.
44 5.3. VOLUMES 238 Notice tht one of the rectngles is highlighted (in light green), so tht we cn focus on it s typicl rectngle. Now we will rotte this 2D picture s described in the instructions:
45 5.3. VOLUMES 239 Notice tht the rotted curve cretes sort of pointed footbll (in red), nd inside it the rotted rectngle cretes solid disk, circulr in yz nd skinny in x. The sum of the volumes of these disks is n pproximtion to the volume of our 3D solid, nd the pproximtion becomes more nd more ccurte s the number n of disks goes to infinity, so in the end our volume will be definite integrl. To construct tht definite integrl, we solve the problem on slice, which in this cse mens we compute the volume of the typicl disk, iming s lwys to express the nswer in terms of x i ( point in the xintervl of our slice) nd x (the width of our slice). Since the disk hs circulr crosssection in yz, nd width x in the x dimension, its volume is πr 2 x, where r is the rdius of the circle. Notice from our two pictures tht the rdius of the disk (in the 3D picture) is the sme s the height of the curve (in the 2D picture), so tht r = f(x i ) = 3x i ( x i ). Therefore, Volume of slice π [3x i ( x i )] 2 x. As with our previous pplictions, the process of summing over the slices nd tking n converts this to definite integrl: Volume of 3D solid = x= x= π [3x( x)] 2 dx. To complete the problem we compute the integrl, by expnding the squre to crete polynomil in the integrnd: Volume of 3D solid = = = = 9πx 2 ( 2x + x 2 )dx (9πx 2 8πx 3 + 9πx 4 )dx ( ) 9πx 3 x= 8πx4 + 9πx x= ( 3π 9π 2 + 9π ) ( + ) = 3π 5 The technique we used here generlizes to mny solids of revolution: if we tke the 2D region nd divide it into rectngles perpendiculr to the xis of revolution, their rottions will generte slices of the 3D solid whose volume we cn work out using simple geometric formuls. Exmple 5.9 (Volume of cone). Let s derive the formul for the volume of cone with rdius r (of the circulr bse of the cone) nd height h (from the circulr bse to the tip). 3 3 You my hve seen the formul V = 3 πr2 h in geometry clss, but we clim it does not qulify s simple geometric formul. The volume πr 2 h of disk of rdius r nd height h is intuitive since the disk involves the circle of re πr 2 swept unchnged through the third dimension of width h, hence we get the product of those two expressions. With cone, ech crosssection is circle of re πr 2, but tht rdius r is constntly chnging s you sweep in the third dimension, so it is not obvious how to get the finl result. There is wy to derive the formul without clculus
46 5.3. VOLUMES 24 In order to use the method from the previous exmple, we seek 2D region R, nd n xis of revolution, so tht the rottion of R cretes the solid cone. Here is such region: It will be useful to hve n eqution for the digonl line in the figure. We know it psses through the points (, h) nd (r, ), so its slope is rise run = h r = h r. Using the slopeintercept formul, the eqution of the line is then y = h r x + h. To compute the volume, we pproximte the 2D region R by collection of rectngles perpendiculr to the xis of revolution (see the discussion t the end of the lst exmple). In this cse, these rectngles would be horizontl: (the ncient Egyptins knew the formul... ), but it is tricky. formul. We will use slicing to derive the
47 5.3. VOLUMES 24 We hve highlighted typicl rectngle, which we now imgine rotting bout the xis of revolution. It becomes disk tht is circulr in the xz dimensions nd skinny (with height y, the width of our slicing intervls) in the y dimension: Thus, Volume of slice = π ( rdius of disk ) 2 y. Notice tht rdius of disk in this formul is not r, the rdius of the bse of the cone. From the figure, it is cler tht the rdius of typicl disk is quite bit smller thn the rdius of the circle t the bottom of the cone. Once gin, the connection to the 2D picture is the key: the rdius of the disk in the 3D figure is the xvlue on the line in the 2D figure. To find formul for tht xvlue, we merely solve the eqution for tht line for the vrible x: Thus, y = h r x + h = y h = h r x = x = r (y h). h ( Volume of slice π r 2 h (y i h)) y. (In our figures, y i would be the midpoint of the ith intervl in y.) Therefore, y=h ( Volume of cone = π r ) 2 h (y h) dy. y= We complete the problem by computing this integrl: Volume of cone = h π r2 = πr2 h 2 = πr2 h 2 h 2 (y2 2hy + h 2 )dy ( ) y 3 y=h 3 hy2 + h 2 y y= [( ) ] [ ] h 3 3 h3 + h 3 ( + ) = πr2 h 3 h 2 = 3 3 πr2 h.
48 5.3. VOLUMES 242 Exmple 5. (Axis of revolution is not coordinte xis). Let R be the region in the plne bounded by the curve y = x 2, the line x = 2, nd the xxis. If we revolve R bout the line x = 2, wht is the volume of the resulting solid? We strt by sketching the 2D region R, nd rectngles perpendiculr to the xis of revolution: As in the previous exmple, when we rotte the typicl rectngle bout the xis of revolution, we get disk tht is circulr in the xz dimensions nd with height y in the y dimension: Thus, Volume of slice π ( rdius of disk ) 2 y. To relte the rdius of the disk to y i, vlue of y in the ith intervl of the yxis, we refer bck to the 2D picture nd notice tht the rdius of the disk in the 3D picture is the sme s the width of the rectngle in the 2D picture. Tht rectngle extends from the curve y = x 2 (its left edge) to the line x = 2 (its right edge). The width of the rectngle is difference in xvlues, so we solve y = x 2 for x to find x = y.
49 5.3. VOLUMES 243 Thus, the width of the rectngle is x right x left = 2 y, which mens tht Volume of slice π (2 y i ) 2 y. This result for the slice implies tht the totl volume is Volume of solid = y=4 y= π(2 y) 2 dy. In order to compute the integrl, we expnd the squre: Volume of solid = = = π 4 4 ( [( = π = π π(4 4 y + y)dy π(4 4y /2 + y)dy 4y 4y3/2 3/2 + y2 2 6 (4)(8) 3/ ] [ = 8π 3 ) y=4 y= ) ( + ) In summry, this exmple involved very similr thought process s the previous exmple, with the one difference being tht the typicl rectngle did not go from coordinte xis to curve, but rther from one curve (y = x 2 ) to nother curve (x = 2). We expressed ech curve in the form x = some function of y, nd then the width of the rectngle (nd hence the rdius of the disk) ws the difference between these two functions of y. But there is no rel need to think of these two exmples s seprte cses or formuls; if you consistently drw the 2D picture, nd use it to sketch the 3D picture 4, the relevnt formuls should rise logiclly. Exmple 5. (Gp between 2D region nd xis of revolution). Let R be the region in the plne between the curves y = x nd y = x 2 /2. If we revolve R bout the xxis, wht is the volume of the resulting solid? We strt by sketching the 2D region R, nd rectngles perpendiculr to the xis of revolution: ] 4 In our exmples in this section, we re hving the computer drw the 3D solid, but in prctice, to solve the problem ll you need is sketch of typicl 3D slice. The full 3D solid could be complicted nd hrd to visulize, but typicl slice is just disk.
50 5.3. VOLUMES 244 Here we see significnt chnge reltive to the lst three exmples: the rectngles to be rotted re not up ginst the xis of revolution. Sid nother wy, in the first three exmples, the xis of revolution ws one of the boundries of the 2D region R, but in this exmple, it is not. As result, when we rotte rectngle, to get 3D slice of the full 3D solid, tht slice is not pproximtley disk, but rther wsher with hole in the middle: (The 3D solid is the region inside the gry surfce nd outside the pink surfce, vguely like sidewys icecrem cone, though with curved inside surfce.) Thus, in order to pproximte the volume of the slice, we need new bsic geometric fct, nmely the formul for the volume of wsher with outer rdius r out, inner rdius r in, nd height h. Once we relize tht this wsher is relly the sme s the cylinder with rdius r out with cylinder of rdius r in removed from its center, we cn see tht V wsher = π [ (r out ) 2 (r in ) 2] h. In our exmple, the height of the wsher is x, nd we then hve to express r in nd r out in terms of n xvlue x i in the ith intervl. As usul, we refer bck to
51 5.3. VOLUMES 245 the 2D picture nd note tht r in is the sme s the height of the lower curve, so r in = (x i ) 2 /2, nd r out is the sme s the height of the upper curve, so r out = x i. Thus, Volume of slice π [ (x i ) 2 ((x i ) 2 /2) 2] x. Following our usul method to convert the volume of the slice to the full volume, Volume of solid = x=2 x= π [ (x) 2 (x 2 /2) 2] dx. We finish the problem by computing the integrl: 2 Volume of solid = π [x 2 x4 4 [ x 3 = π 3 x5 2 [ 8 = π 3 32 ] 2 ] x=2 x= ] dx = 6π 5. Exmple 5.2 (Volume of doughnut). Consider doughnut: 5 with circulr crosssection of rdius cm (the length of the green tube in the figure), nd whose hole hs rdius 2.5 cm (the length of the blue tube in the figure). We cn use the techniques of this section to compute the volume of the doughnut, becuse we cn express the doughnut s the result of rotting the interior of circle bout n xis: 5 Oddly, mthemticins often insist on clling doughnut torus.
52 5.3. VOLUMES 246 Following the technique of this section, we fill the region with rectngles perpendiculr to the xis of revolution: When we rotte typicl rectngle, we get wsher:
53 5.3. VOLUMES 247 whose volume we know to equl π[(r out ) 2 (r in ) 2 ]h, for r in the inner rdius of the wsher, r out the outer rdius, nd h the height. Referring bck to the 2D picture, we cn see tht h = y, nd r out nd r in re xvlues for points on the circle. We need to express these x vlues in terms of chosen y i on the y xis, which mens we tke the eqution of the circle nd solve it for x: (x 3.5) 2 + y 2 = = (x 3.5) 2 = y 2 = x 3.5 = ± y 2 = x = 3.5 ± y 2. We need r out to be the lrger x vlue on the circle, so r out = 3.5+ (y i ) 2, which mkes r in = 3.5 (y i ) 2, the smller vlue. Thus, [ ( Volume of slice π ) 2 (y i ) 2 (3.5 ) ] 2 (y i ) 2 y, which mens the totl volume is Volume of doughnut = [ ( π ) 2 y 2 (3.5 ) ] 2 y 2 dy. The integrnd looks pretty messy, but if we expnd the squres, some terms cncel: Volume of doughnut = = π [{(3.5) 2 + (2)(3.5) y 2 + ( y 2 ) 2{ { (3.5) 2 2(3.5) y 2 + ( y 2 ) 2}] dy [(2)(3.5) y 2 + 2(3.5) y 2 ] dy = 4 y 2 dy
54 5.3. VOLUMES 248 The symbolic clcultion of this integrl would be beyond the scope of our course, so we will opt to pproximte it numericlly. We choose n = 2 rbitrrily nd compute M 2 nd T 2 (detils not shown) to find M 2 = nd T 2 = Since the integrnd 4 y 2 is concve down for y, these two pproximtions re gurnteed to brcket the exct nswer, so we cn sy Volume of doughnut = with error t most.334 (where.334 is the difference between M 2 nd T 2 ). If we wnted better pproximtion, we could, of course, increse n, though it turns out in this cse tht the improvement in ccurcy would not be s good s the Rule of Thumb for M n nd T n predicts, becuse the integrnd y 2 hs verticl tngents t y = nd y =. Still, eventully we will get quite ccurte solution, e.g., M 2 = nd T 2 = yielding threedigit ccurcy Solids tht re not solids of revolution. We hve lrgely focused on solids of revolution in prt becuse there re mny 3D objects tht hve circulr symmetry nd thus re solids of revolution. However, our slicing method is not limited to solids of revolution; indeed, we cn hndle ny 3D solid s long s we cn express the volume of the slice in terms of w i nd w (where w is whtever vrible we choose for the slicing, i.e., w is either x, y, or z). Exmple 5.3 (A slug, oversimplified). Suppose tht slug hs bse tht is the semicirculr region x 2 + y 2 < nd y >, nd its cross sections perpendiculr to the xxis re semicircles with dimeters long the bse: (In the figure, the bse is colored blue, the top surfce of the slug is colored pink, nd you cn see some of the semicircle cross sections s thin blck lines. With just cursory glnce, it my be tempting to think the pink surfce is hemisphere, but note tht the bse of hemisphere is full circle, not hlfcircle.) We will compute the volume of the slug. If you re gifted t 3D visuliztion, you my hve been ble to conjure in your mind n imge like the one bove of the slug. Fortuntely, tht skill is not 6 If you re clever, you cn ctully compute the integrl 4 y 2 dy without doing ny work t ll! We know tht this integrl is 4 times the re under the curve x = y 2 from y = to y =, nd tht curve is semicircle of rdius. Thus, the integrl 4 y 2 dy equls 4(π/2) = 7π = 2.994, where tht π/2 is the re inside semicircle. Tht s weird (nd cool): we set up definite integrl for volume, nd then once we sw the integrl, we relized we could compute it by insted thinking of it s n re.
55 5.3. VOLUMES 249 necessry to compute the slug s volume; rther, we will rely primrily on 2D imges nd the method of slicing. Indeed, the description of the slug is in terms of crosssections, which connects directly to the volume of typicl slice tht is our centrl computtion. First, s in ll the previous exmples, we drw 2D imge of the bse nd fill it with pproximting rectngles: Ech of these rectngles is the bse of slice of the slug. The shpe of tht slice is governed by the problem s description, which sys tht crosssections re semicircles. This tells us tht the slice corresponding to our highlighted rectngle looks like this: In order to figure out the volume of this typicl slice, we mke our usul connections bck to the 2D picture. First, the thickness of the 3D slice is the width x of the 2D rectngle. Since the slice hs semicirculr shpe, we hve Volume of slice = 2 π ( rdius of semicircle )2 x,
56 5.3. VOLUMES 25 using the bsic geometric formul tht the re of semicircle is 2 πr2. We now need to express the rdius of the semicircle in terms of point x i on the xxis for the ith intervl. Note tht the dimeter of the semicircle in the 3D slice is equl to the height of the rectngle in the 2D imge. Thus, the rdius of the semicircle is equl to hlf of the height of the rectngle in the 2D imge. But the height of the rectngle is the yvlue of the curve y = x 2. Thus, Rdius of semicircle = 2 (xi ) 2, so tht Volume of slice 2 π ( 2 (xi ) 2 ) 2 x = π 8 ( (x ) 2 ) x. When we dd up the contributions of the slices nd tke n, we get Volume of slug = x= x= π 8 ( x2 )dx, which we cn compute symboliclly: Volume of slug = π ) x= (x x3 8 3 x= = π [( ) ( )] = π [ ( )] = π 3 6. Note tht the success of this computtion relied on the fct tht we chose to use rectngles to be perpendiculr to the xxis, bsed on the fct tht the problem describes crosssections perpendiculr to the xxis. This is nlogous to the fct tht for solids of revolution, we lwys chose rectngles in our 2D imge to be perpendiculr to the xis of revolution. Exmple 5.4. The Pyrmid of the Sun in Mexico is pproximtely squre pyrmid, which mens its bse is squre, nd it tpers uniformly s you go up to point. The squre bse is pproximtely 22 meters on ech side, nd the height of the pyrmid is bout 65 meters:
57 5.3. VOLUMES 25 Wht is the volume of this pyrmid? Since we know tht the bse is squre, nd the pyrmid tpers smoothly s we go up, it follows tht if we slice the zxis (the upwrd direction), the slices will hve (pproximtely) squre crosssections: Ech slice is thus pproximtely squre box with height z nd some side length s, so we cn compute its volume using simple geometry: Volume of slice s 2 z. Our next job is to relte the sidelength s to some point z i in the ith intervl. Fced with tht geometry tsk, it would be helpful to hve 2D picture to guide us. The problem tlks bout the squre bse, so tht s 2D picture, but not helpful one, since it does not include the z direction, nor typicl slice. So, we will conjure up our own 2D picture tht does include the zdirection: let s consider wht we would see if we cut the pyrmid in hlf with the xzplne:
58 5.3. VOLUMES 252 (We drew both the pyrmid surfce nd the plne s opque in this cse to better visulize their intersection, so now you cn only see the edges of the slice.) This cut right down the middle of the pyrmid will intersect the pyrmid surfce t n isosceles tringle: Hopefully this picture conjures the notion of similr tringes s geometric technique for relting s to z i. The region bove the slice (in the 2D picture) is n isosceles tringle tht is similr to the overll isosceles tringle. The smller tringle hs bse s nd height 65 z i, wheres the overll tringle hs bse 22 nd height 65, so by similrity: which mens tht Plugging this into our erlier expression: Volume of slice = s = z i 65 s = (65 z i). [ ] (65 z i) z,
59 5.3. VOLUMES 253 which mens tht z=65 [ ] 22 2 [ ] Volume of pyrmid = (65 z) dz = (65 z) 2 dz. z= This integrl cn be computed using the substitution u = 65 z, for which we hve du = = du = dz, dz so tht [ ] 22 2 u= Volume of pyrmid = u 2 ( du) 65 u=65 [ ] 22 2 u= = u 2 du 65 u=65 [ ] 22 2 ( ) u 3 u= = = 65 [ ] 2 ( u=65 ) = (22)2 (65) 3 =.5 6 m 3 (Notice tht in the first equlity we converted the zlimits to corresponding u limits, which creted n oddlooking righttoleft integrl, but we persevered with the computtion nd the negtive nswer coming from tht righttoleft integrl ws cncelled from the minus sign coming from the usubstitution.)
60 CHAPTER 6 Applictions of the Definite Integrl: Density nd Averging Key Words nd Concepts: Density function Weighted Averge Now we shift to new sort of ppliction of definite integrls. First we tckle the ide of density function, which is function tht defines weighting of n object, so tht different regions of the object hve more or less weight. For exmple, if the object is thin metl rod, we could hve mss density tht hs different vlue t ech point on the rod, nd the definite integrl of the mssdensity function gives you the totl mss of the rod. Once we estblish the ide of weighting vi density function, it is nturl to compute verges tht use this weighting, which gin requires certin type of definite integrl. 6.. Density Functions At firly generl level, density is conversion fctor, i.e., quntity whose units re X per Y, typiclly with Y being either length or re or volume, wheres X cn be pretty wide vriety of things, e.g., 7 dollrs per yrd: tht s the cost of duplioni silk 37 people per squre mile: tht s the popultion density of Brooklyn 2 kg/m 3 : tht s the mss density of wter 3 Given constnt density like one of these, it s pretty strightforwrd to use it to compute n mount of X given size Y: you just use the density s multipliction fctor, e.g., 7 dollrs If you buy 5 yrds of silk, it will cost you (5 yrds) = 85 dollrs. If voting district in Brooklyn is.5 squre miles in re, you cn expect it will contin (.5 squre miles) = 85 people. 37 people squre mile yrd If pond contins m 3 of wter, then the the mss of tht wter is ( m 3 ) = 8 kg. kg m 3 Tht s the cost density of silk ccording to howmuchisit.org. For some reson, it s not so common to use the term density when X is cost or price, but it does fit the pttern. 2 In 24, ccording to worldpopultionreview.com 3 At 4 Celsius, the temperture t which wter is its most dense. 254
61 6.. DENSITY FUNCTIONS 255 Of course, those three clcultions require the density to be constnt throughout the object in question. Tht is likely to be true in the first cse, but it is quite possible tht the popultion density of Brooklyn vries with loction, nd tht the mss density of wter increses in the deeper regions of the pond. When density is not constnt, we need definite integrls to do the computtion. To build the pproprite integrl, we use, once gin, the method of slicing. In this setting, we slice the object in such wy tht the density is constnt on the slice (or t lest pproximtely constnt). Then we cn use simple multipliction s bove to do the computtion on the slice, nd the definite integrl will tke cre of summing over the slices nd tking the number of slices to (to mke the pproximte computtion n exct one). Exmple 6.. Suppose tht the weightdensity of thin iron rod extending from x = 2 to x = 2 (in meters) is 5 x2 6 pounds per meter (so tht it is bit hevier in the middle thn t the ends). Wht is the totl weight of the rod? We slice the segment from 2 to 2 into n equlsized pieces of length x, nd let x i be point in the ith segment. Then, if n is lrge, the segments re short, so the density is pproximtely constnt on ech segment. If we tke the ith segment, its length is x nd its density is 5 (x i) 2 6, so Weight of slice (5 (x i) 2 ) x. 6 Thus, Weight of rod = = 2 2 (5 x2 (5x x3 8 6 ) x=2 ) dx x= 2 = 72 9 pounds In this lst exmple, the slices were segments; now let s see n exmple in which the slices hve more interesting geometry. Exmple 6.2. Suppose tht the mss density ρ ( rho ) of wter depends on its depth h below the surfce ccording to: ρ(h) = + h kg/m 3, (so the wter is bit denser the further down you go). If we hve cylindricl tnk of wter with rdius meter nd height 8 meters, wht is the mss of the wter it contins when full? We need to slice the wter in the tnk in such wy tht the density is pproximtely constnt on ech slice. Since the density depends on the depth, we need the slices to be t constnt depth, i.e., circulr slbs:
62 6.. DENSITY FUNCTIONS 256 Note tht ccording to the mening of depth s described in the problem, the vrible h is equl to z in the figure, since s z goes down the depth goes up. Given some typicl slice, its shpe is cylindricl with rdius nd height h, so its volume is π() 2 h. If we let h i be vlue of the depth somewhere in the slice, then we hve Mss of wter in the slice ( + h i ) ( π() 2 h ). Hving expressed the quntity of interest (mss) of the slice in terms of h i nd h, we cn write down the definite integrl tht computes mss on the whole object by the usul process: Mss of wter in the tnk = 8 π( + h)dh = π ( h + 5h 2) h=8 h= = π [(8 + 32) ( + )] = 832π kg Notice from these exmples tht there is no one formul t ply here, since the setup of the integrl involves n interply between wht the density depends on nd the geometry of the object in question. Here is nother vrint: Exmple 6.3. A glss window is rectngle 3 ft wide nd 6 ft high. The glss is /2 inch thick. The weightdensity of the glss is ρ(z) = (5 z) pounds/ft 3 s function of the height z from the bottom of the window (so the glss is slightly hevier t the bottom of the window, due to slight settling of the mteril becuse of grvity). Wht is the totl weight of the window?
63 6.. DENSITY FUNCTIONS 257 We need to slice the window in such wy tht the density is pproximtely constnt on ech slice. Since the density depends on z, we need the slices to be t constnt z: Given some typicl slice, its shpe is box with dimensions 3 (/2) z, so its volume is (3)(/2) z = 3 2 z. If we let z i be vlue of z somewhere in the slice, then we hve so tht ( ) 3 Weight of slice (5 z i ) 2 z = 3 2 (5 z i) z, 4 3 Weight of window = (5 z)dz 2 = 3 (5z z2 2 2 Here s yet nother vrint: ) z=4 z= = 3 [(6 8) ( )] = 888 pounds 2 Exmple 6.4. Suppose tht the popultion density in certin neighborhood in Brooklyn is ρ(r) = 4e 3r people/mi 2 where r is the distnce from the origin (the popultion is most dense right round crft brewery locted t the origin). How mny people live within 2.5 miles of the origin? We need to slice the neighborhood in such wy tht the density is pproximtely constnt on ech slice. Since the density depends on the distnce r from the origin, we need the slices to be t constnt r:
64 6.. DENSITY FUNCTIONS 258 ring with rdil thickness r. We need to compute the re of this typicl slice, in terms of r i, some vlue of r in the slice, nd r. If we let r out be the lrgest rdius in the slice nd r in the smllest rdius, then the re of the slice is πrout 2 πrin 2 (the re inside the outer circle minus the re inside the inner circle). So, if we choose r i to be the inner rdius, then Are of slice π(r i + r) 2 π(r i ) 2 = π[(r i ) 2 +2r i r+( r) 2 ] π(r i ) 2 = 2πr i r+π( r) 2. Thus, multiplying by the density: Popultion in the slice 4e 3r i ( 2πr i r + π( r) 2) We re left in bit of strnge sitution reltive to our usul pth to definite integrl. In ll previous exmples, those relted to geometry or those in this section, when we solved the problem on the slice, our nswer ws lwys some function of w i times single fctor of w (where w ws the slicing vrible in the problem). Now we hve n extr fctor of ( r) 2. Were we to follow our usul procedure, tht ( r) 2 term would become (dr) 2, which we would not be ble to interpret s definite integrl. But recll tht r is supposed to be smll quntity; indeed, when n s we tke more nd more slices, the quntity r. Thus, the ( r) 2 term in our expression for Popultion in the slice will be much smller thn the term involving single fctor of r. Therefore, we drop tht ( r) 2 term. Remember tht in every slicing problem, we re only solving the problem on the slice pproximtely, nd relying on the fct tht our pproximtion gets better nd better s n. This choice to drop ( r) 2 is just nother instnce of this ide. So, we hve Popultion in the slice 4e 3r i (2πr i r) = 8πr i e 3r i r, which mens tht Popultion within 2.5 miles = 2.5 8πre 3r dr.
65 6.2. PRESENT VALUE 259 This integrl cn be computing symboliclly by combintion of integrtion by prts (choosing u = πr nd v = e 3r ) nd substitution (using u = 3r to find the ntiderivtive of e 3r ). But since our focus here is on the ppliction, nd on the setup of the integrl, we will insted use numericl pproximtion, which is less messy. We will im to compute the nswer to within one. 4 We choose n = rbitrrily nd compute M nd T. Annoyingly, when we plot the integrnd 8πre 3r for r 2.5, we find it is sometimes concve down nd sometimes concve up, so we do not know for sure tht M nd T brcket the exct nswer. One pproch to this technicl sng would be to work out where the concvity chnges 5, nd then split the integrl into two prts t this point. We will tke the less precise pproch of using 2 M n T n s n estimte of the error in M n nd T n (we hve dded the fctor of 2 to ccount for the fct tht we no longer hve brcket). We compute (detils not shown) M = T = nd thus estimte the error in ech of those pproximtions to be 4 (twice the difference between them). The Rule of Thumb tells us tht if we multiply n by, we cn expect the error will go down by fctor of 4, which will tke us below our trget ccurcy. So we try n = next: M = , T = , which indeed re within of ech other, suggesting we hve reched our trget ccurcy. Thus, our numericl evlution of the integrl yields: Popultion within 2.5 miles = with error less thn 6.2. Present Vlue We next consider n ppliction to finnce tht incorportes techniques from our discussion of discrete dynmicl systems in Prt s well s from our more recent exmintions of definite integrls. The bsic question we seek to nswer is: given series of pyments t times in the future (either t discrete times or s continuous strem of income), how much re those pyments worth tody? The nswer to this question is clled the present vlue of the series of pyments, nd it cn be prticulrly useful in compring two such series Discrete Present Vlue. Let us begin with n esy version of the bsic question: Wht is more vluble: getting pid $ tody, or getting pid $ yer from now? The nswer to this question depends on severl prmeters: how lrge is infltion (or defltion)? How lrge return on investment do we get? To be concrete, let us ssume tht we cn sfely ern 3% on n investment (modeled s continuously 4 After ll, we re trying to mesure popultion, so getting n error less thn is bit silly. 5 Solving d 2 (8πre 3r ) = shows tht r = 2/3 is where the concvity chnges. dr 2
66 6.2. PRESENT VALUE 26 compounded interest s in Exmple 7.2) nd tht there is no infltion or, more relisticlly, tht the difference between the interest rte on sfe investment nd infltion is 3%. To begin to nswer this question, let us think bout wht would hppen in yer if you were pid $ tody. This is strightforwrd ppliction of our understnding of continuously compounded interest: in generl, we know from Exmple 7.2 tht if you cn sfely invest P dollrs in n ccount with n nnul continuously compounded interest rte r, then the vlue P of the ccount fter t yers is (6.) P (t) = P e rt. Thus, in this cse, the vlue of your pyment fter one yer is e We cn think of $3.5 s the future vlue of your pyment fter one yer. Thus, since this future vlue is lrger thn the $ pyment in one yer, you should prefer the $ pyment now. Another wy to think bout the comprison between the two pyments is to ynk the future pyment bck to the present rther thn pushing the present pyment to the future. In prticulr, we sk: how much is the future $ worth now? Tht is, how much would you hve to invest now to hve $ fter yer? To nswer this, we need to solve Eqution (6.) for P if we know tht r =.3 nd P () = $: = P e.3 = P = e We cll $97.4 the present vlue of pyment of $ yer from now. In generl, we mke the following definition: Definition 6.5. The present vlue P V of sum of money P pid t time t, ssuming previling rte of return on investment r, is P V = P e rt. The notion of present vlue is especilly useful when evluting the vlue of series of future pyments. Exmple 6.6. Suppose tht frm expects to sell $ worth of pumpkins ech fll nd tht the previling rte of return on investment is 4%. 6 First, let us sk for the present vlue of the first three yers of sles, ssuming tht the first sles re occurring right now. To figure this out, we dd up the present vlues of the sles from ech of the first three yers: P V 3 = + e.4 + e Wht is the present vlue of ll future business? This my not seem like sensible question t first glnce; fter ll, the frm will mke dditionl money 6 Going bck to our erlier discussion of models, this would be good time to tese out nd criticize the ssumptions underlying the model in this exmple. For exmple, you my sk why the frm expects constnt sles in the future? Why is the return on investment constnt?
67 6.2. PRESENT VALUE 26 every yer. Even so, the concept of ynking those future sles bck to present dollrs will mke sense of the question. We begin by figuring out the present vlue fter the first n yers: P V n = + e.4 + e e.4(n ) = ( + (e.4 ) + (e.4 ) (e.4 ) n ) We cn recognize the prt of the formul inside the prentheses s geometric sum of the form + k + k k n s in Section 4.2, where k = e.4. As we let the time n increse without bound, we seek the sum of geometric series, which we determined to be k in Section 4.2 so long s k <. Thus, the present vlue of the entire income strem is P V = e.4 A good wy to interpret this present vlue is tht $ would be fir price to py for the frm s business tody. Remrk. In Section 4.2, we introduced geometric series nd their sums in our exmintion of the longterm behvior of ffine discrete dynmicl systems, which hve the form x n+ = kx n + b. We cn see just such DDS if we look t the future vlue of the pumpkin frm s income. The future vlue of the frm s income in yer n + is simply the sum of $ in new income nd the future vlue in yer n fter it hs erned yer s worth of interest. Tht is, we hve the following ffine DDS: F V n+ = e.4 F V n +. Note tht we cn recover the present vlue of the frm s income up to time n from the future vlue using our usul ynking bck method: P V n = e.4n F V n. This formul mkes it seem tht the present nd future vlue re equivlent concepts. Since the future vlue is, dmittedly, bit esier to think bout, why did we bse our discussion on present vlue? There re two resons: () The present vlue of series of pyments hs the useful interprettion of being fir price to py now for the entire series of pyments; see the lst exmple. (2) Future vlue does not mke sense for n infinite series of pyments s t the end of the lst exmple, while the present vlue does. Exmple 6.7. Mny lotteries often the winner choice between lumpsum pyment up front nd n nnuity, which insted offers pyments ech yer for some set number of yers. For exmple, ccording to usmeg.com, the nnuity option for $2 million Powerbll victory is shown below.
68 6.2. PRESENT VALUE 262 The nnuity my offer vriety of benefits, such s gurnteed flow of income over severl decdes, or possibly some tx benefits. On the other hnd, becuse some of the pyments re in the future, their vlue to the winner is decresed. To quntify this effect, we could compute the present vlue of the nnuity. To do so, we need the vlue of the interest/infltion rte prmeter r. If we ssume r =.2 ( plusible vlue coming from the CPI over the pst few decdes), then Present Vlue of nnuity = 8224e e.2() e.2(2) e.2(29) = 85, 543, 25 On the other hnd, if we ssume r =.8 (coming perhps from sttement from our stock broker tht our investments will typiclly ern 8% per yer), then Present Vlue of nnuity = 8224e e.8() e.8(2) e.8(29) = 36, 3, 25 Clerly, the present vlue depends quite drmticlly on the rte prmeter r Continuous Present Vlue. In ll of our pplictions of the definite integrl thus fr, we hve solved problems tht involved obtining informtion bout geometric quntity (the rc length of curve, the re of region, the volume of solid) by slicing up the object to be mesured, estimting the mesurement on ech slice, dding the estimtes up, 7 nd finlly tking the limit s the slices get ever thinner to obtin the desired result. In this section, we continue to use the slicing ide, but insted of slicing sptilly, we will slice temporlly. In contrst to the pumpkin frm in the lst section, lrger compny s sles re better modeled s continuous function rther thn s series of discrete 7 This step is the Riemnn sum.
69 6.2. PRESENT VALUE 263 pyments. More explicitly, we now think of income s continuous function P (t), with units in dollrs per yer. The units of P (t) hint tht you cn think of it s sort of density function. We re interested in understnding the present vlue of continuous income strem over period of time, sy between nd T yers. The ide is to slice up the time intervl [, T ] into smll intervls of length t. On one such intervl [t i, t i+ ], the mount of income erned is pproximtely P (t i ) t. It helps to think in terms of density functions here: if the income rrives t constnt rte P dollrs per unit time, then the mount of income erned over time intervl of length t is simply P t. Thus, if P (t i ) is the instntneous mount of income per unit time t time t i, nd if the time intervl [t i, t i+ ] is short enough so tht so the mount of income erned over the intervl is pproximtely constnt, then we obtin the pproximte income of P (t i ) t. The present vlue of this pproximtion to income over the intervl [t i, t i+ ] cn be found by ynking bck the income erned to the present dy: P V [ti,t i+ ] = e rt i P (t i ) t. As in rguments from the previous section, we then dd up pproximtions of present vlue over ll of the subintervls of [, T ] nd let t to obtin the following definition: Definition 6.8. The present vlue P V of continuous income strem P (t) over time intervl [, T ], ssuming previling rte of return on investment r, is P V = T P (t)e rt dt. Exmple 6.9. An internet photogrphy compny sells two kinds of pictures: one of dorble mrmots, nd the other of eqully dorble owls. Owls re growing more populr, while mrmots re holding stedy. In prticulr, the compny gets constnt income of $.2 million per yer from mrmot pictures, but growing income of $.5t million per yer from owl pictures. If the previling return on investment is 5%, which type of picture will be more vluble over the next three yers? To nswer this question, we turn to the present vlue definition: P V mrmot = P V owl = 3 3.2e.5t dt $3.35 million.5te.5t dt $2.3 million Thus, the mrmots re more vluble over the next three yers. 8 8 Will this lwys be true? Will there be time when the owl pictures will become more vluble thn the mrmots?
70 6.3. WEIGHTED AVERAGE Weighted Averge Definite integrls cn lso be used to computed weighted verges. As wrmup, let s consider finite number of prticles on the xxis, with positions x, x 2,..., x n. If we wnted to compute the verge of the positions, we would use the formul: x = x + x x n, n where we dd up the quntities nd divide by the number of terms. Another wy of writing this formul is: x = ( ) x + n ( ) x n ( ) x n. n This version of the formul describes the verge x s weighted sum, where ech of the n terms x i gets the sme weighting /n (representing the proportionl contribution tht term mkes to the whole, since it is one term out of n). Now suppose tht the n prticles hve different msses m, m 2,... m n (respectively). We would like to generlize the previous notion of verge to mssweighted verge, in which the terms with lrger msses get to contribute more substntilly to the verge. A nturl wy to define such weighted verge is to tke the previous eqution nd chnge the equl weightings /n to weightings bsed on the msses: ( m x = m + m m n ) ( x + m 2 m + m m n ) ( x m n m + m m n As before, ech weighting reflects the proportionl contribution of tht prticle to the whole, but this time using mss to compute tht contribution: the weighting is the mss of tht prticle divided by the totl mss of ll the prticles. If we tke this formul nd get common denomintor, we will get this formul: ) x n. Definition 6. (Mssweighted verge for finite number of prticles). Given prticles t positions x, x 2,..., x n, with msses m, m 2,..., m n (respectively), the weighted verge of the positions x i is x = m x + m 2 x m n x n m + m m n. Notice tht in the specil cse tht ll the msses re the sme: m = m 2 = = m n = m for some constnt m, the formul reduces to our initil formul for (unweighted) verge: x = mx + mx mx n m + m + + m = m(x + x x n ) nm = x + x x n. n To generlize this mssweighted verge to continuous objects (like rods in D, pltes in 2D, or solids in 3D) requires definite integrl. For ny such continuous object, if we crete slices perpendiculr to the xxis, nd let x i be n xvlue in the ith slice, nd m i be the mss of the ith slice, then n pproximtion of the
71 6.3. WEIGHTED AVERAGE 265 mssweighted verge of x will be exctly the formul in the Definition bove, nd when we tke the limit of the number of slices n, the numertor nd the denomintor of tht formul will ech become definite integrl. Exmple 6. (A mssweighted verge in D). Suppose tht rod extending from x = to x = 4 meters hs mssdensity ρ(x) = 5 + x 2 kg/m Find the mssweighted verge of x. We slice the rod into n equl pieces of width x (we know this equls 4/n, but s usul in slicing derivtions we write it in the form x, prepring for the moment when it becomes dx in definite integrl), nd let x i be vlue of x in the ith slice. Then, the mss of the ith slice is ( m i = ρ(x i ) x = 5 + x ) i x 2 Looking t the formul for mssweighted verge x, its denomintor is the totl mss of the rod, nd we know from the previous section how to compute tht once we hve the mss of the slice: Denom of x = Totl mss of rod = x= x=4 x= ( 5 + x ) dx. 2 Looking gin t the formul for x, its numertor is ( x m +x 2 m 2 + +x n m n = x 5 + x ) ( x+x x ) ( 2 x+ +x n 5 + x ) n x, nd s n, this lso pproches definite integrl, but with n extr fctor of x in the integrnd (coming from the fctor of x i in ech term of the sum): x=4 ( Num of x = x 5 + x ) dx. 2 Overll, we hve: x = 4 x ( 5 + x ) 2 dx ( ). 5 + x 2 dx 4 We will see this pttern repetedly in the exmples in this section. The computtion of the denomintor of weighted verge x is totlmss tht we compute by slicing construction exctly like wht we did in the lst section. Then the numertor of x is the sme definite integrl with n extr fctor of x in the integrnd.
72 6.3. WEIGHTED AVERAGE 266 To finish this exmple, we compute the two integrls: x = ( ) 5x 2 x=4 2 + x3 6 x= ( ) x=4 5x + x2 4 x= = ( ) ( + ) (2 + 4) ( + ) = 34/6 = Wht does this result men? In setting like this one, the number x is often clled the center of mss of the rod. It would be the point on the rod where you could put your finger nd hve the rod exctly blnce (not tip over one wy or the other). 9 The ide of center of mss is often exploited in physics, e.g., in some situtions you cn tret the rod s point prticle concentrted t position x. Note tht in the lst exmple, the center of the rod visully (ignoring the mssdensity ρ) would of course be t x = 2, hlfwy between x = nd x = 4. When we tke density into ccount, the verge shifts to the right to 2., which fits with ρ(x) being lrger for lrger vlues of x, so tht there is bit more mss in the right portion of the rod thn in the left portion. Our next exmple will involve 2dimensionl object. In tht setting we cn compute ȳ in ddition to x. Exmple 6.2 (A mssweighted verge in 2D). A steel plte with uniform density 3 kg/m 2 is bounded by the yxis nd the prbol x = 4 y 2. 9 Note tht this is not quite the sme thing s sying tht there is 5% of the mss to the left of your finger nd 5% to the right. Tht 55 notion of verge is distinct from wht we re computing with x. We will revisit this distinction when we tlk bout probbility, where x is often clled the men of x, wheres the 55 brekpoint of mss would be clled the medin of x.
73 6.3. WEIGHTED AVERAGE 267 Find the mssweighted verges of x nd of y. Let us begin with the computtion of x. We proceed s outlined erlier in this section, by slicing perpendiculr to the xxis: Then, the re of typicl slice is pproximtely the height of the rectngle times its width x. The rectngle extends from the bottom hlf of the prbol to the top hlf, so we solve for y in the eqution of the prbol to find y 2 = 4 x, which implies y = ± 4 x with the + sign for the top hlf nd the sign for the bottom. Thus, Are of slice (in m 2 ) [ 4 x i ( 4 x i )] x = 2 4 xi x
74 6.3. WEIGHTED AVERAGE 268 To find the mss m i of the ith slice, we multiply by the constnt density 3: This mens tht: Mss of slice (in kg) 6 4 x i x. Totl mss (in kg) = Denom of x = x=4 x= 6 4 x dx. So s to not distrct from the setup of the problem, let s postpone computtion of this integrl, nd insted consider the form of the numertor of x: x m +x 2 m 2 + +x n m n = x ( 6 4 x ) x+x2 ( 6 4 x2 ) x+ +xn ( 6 4 xn ) x, This fits the form of Riemnn sum, so s the number of slices n, we find Num of x = x=4 x= x(6 4 x)dx. (Its units re kg m, since it cme from sum of terms x i m i.) Therefore, x (in m) = x=4 x= x(6 4 x)dx x=4 x= 6 4 x dx. As in the previous exmple, we find tht x is the rtio of two integrls, with the top integrl hving n extr fctor of x in the integrnd. To get numericl vlue for x, we compute the two integrls. In ech cse, cued by the presence of the composition 4 x = (4 x) /2, we use the substitution u = 4 x (the inner function). For this substitution, we hve Thus Denom of x = Totl mss = du = = du = dx. dx u= = 6 = 6 u=4 u= ( 6 u( du) u=4 u 3/2 3/2 u /2 du ) u= u=4 = 6 ( 43/2 3/2 ) = 6(4) 3/2 2 = 32 kg. 3
75 Similrly, Therefore, Num of x = u= = 6 = 6 = WEIGHTED AVERAGE 269 u=4 4 ( [ (4 u)(6 u)( du) (4u /2 u 3/2 )du 4 u3/2 3/2 u5/2 5/2 ( ( ) ) u= 4 43/2 ( = /2 32 ) = 6 5/2 u=4 )] 3/2 45/2 5/2 ( ) 5 = kg m x = 256/5 = m, which fits with the picture, since the fct tht the plte is tller for smller x vlues shifts the verge left from x = 2. Now we shift to computing ȳ. Though we hve only written out formuls thus fr for x, the process of constructing ȳ would be exctly nlogous, just replcing ech x i in the weightedverge formul by y i. In the context of this problem, we need to reslice the steel plte perpendiculr to the yxis: Then, the re of typicl slice is pproximtely the width of the rectngle times its height y. The rectngle extends from the yxis (which hs eqution x = ) to the prbol x = 4 y 2, so Are of slice (in m 2 ) (4 (y i ) 2 ) y.
76 6.3. WEIGHTED AVERAGE 27 To find the mss m i of the ith slice, we multiply by the constnt density 3: This mens tht: Mss of slice (in kg) 3(4 (y i ) 2 ) y. Totl mss (in kg) = Denom of ȳ = y=2 y= 2 3(4 y 2 )dy. So s to not distrct from the setup of the problem, let s postpone computtion of this integrl, nd insted consider the form of the numertor of ȳ: y m +y 2 m 2 + +y n m n = y ( 3(4 (yi ) 2 ) ) +y 2 ( 3(4 (yi ) 2 ) ) + +y n ( 3(4 (yi ) 2 ) ), This fits the form of Riemnn sum, so s the number of slices n, we find Num of ȳ (in kg m) = y=2 y= 2 y(3)(4 y 2 )dy. Therefore, y=2 y= 2 ȳ = y(3)(4 y2 )dy y=2 y= 2 3(4 y2 )dy. Once gin, we find tht ȳ is the rtio of two integrls, with now the top integrl hving n extr fctor of y in the integrnd. With both integrnds being polynomils, the computtion of ȳ is reltively strightforwrd: 2 2 ȳ = (2y 3y3 )dy 2 2 (2 3y2 )dy ( 6y 2 3y 4 /4 ) y=2 y= 2 = (2y y 3 ) y=2 y= 2 (24 2) (24 2) = (24 8) ( ) = 32 = A few comments bout this exmple re in order: The computtion of the denomintor of ȳ ws, in sense, redundnt, since it equls the totl mss, lredy computed s 32 kg in the x computtion. However, in order to determine the numertor of ȳ, we hd to t lest set up the integrl for the denomintor of ȳ, since it involves the computtion of m i for slice i, key step for working out the definite integrl in the numertor of ȳ. If we were clever, we could hve skipped the computtion of ȳ entirely nd seen by symmetry tht it must be zero, since the plte nd the density function re symmetric with respect to the xxis. (We opted to work through the derivtion s demonstrtion of how to compute ȳ using slices perpendiculr to the yxis.) These results cn gin be interpreted s showing tht the point (8/5, ) is the center of mss of this plte. Thus, if you put your finger directly below tht point, the plte would blnce exctly. This intereprettion presumes tht your finger cn support 32 kg, the mss of the plte...
77 6.3. WEIGHTED AVERAGE 27 Exmple 6.3 (A mssweighted verge in 3D). An ornge (uniform density 3 g/in 3, rdius 2 in) is sliced down the middle. Wht is the center of mss of the top hlf? In principle, we hve three quntities to compute x, ȳ, nd z nd, s we sw in the previous exmple, ech would involve new slicing of the solid (perpendiculr to the xxis, then to the yxis, then to the zxis). However, reclling comment from the lst exmple, we cn use symmetry to see tht x =, since the solid nd the density function re symmetric in x (i.e., when we reflect through the yzplne), nd similrly ȳ =. Thus, we only need to compute z. To do so, we slice the solid perpendiculr to the zxis. We hve solid of revolution (with the zxis s the xis of revolution), so we begin by drwing the 2D region tht is rotted to form the solid, nd its slices:
78 6.3. WEIGHTED AVERAGE 272 When rotted bout the xis of revolution, ech slice will be pproximtely disk of some rdius r nd height z, with volume πr 2 z: Compring the 2D nd 3D figures, we cn see tht the rdius of the disk in 3D is equl to the width of the rectngle in 2D, which we cn figure out by solving the
79 6.3. WEIGHTED AVERAGE 273 eqution of the circle for x: Thus, x 2 + z 2 = 4 = x 2 = 4 z 2 = x = 4 z 2 Volume of slice π ( 4 (zi ) 2 ) 2 z = π(4 (zi ) 2 ) z. To get the mss of the slice, we mulitply by the constnt density: which mens Mss of slice 3π(4 (z i ) 2 ) z, Totl mss = Denom of z = z=2 z= 3π(4 z 2 )dz. Next we refer bck to the weightedverge formul to see tht Num of z z m +z 2 m 2 + +z n m n = 3πz (4 (z ) 2 ) z+3πz 2 (4 (z 2 ) 2 ) z+ +3πz n (4 (z n ) 2 ) z, which mens tht when we tke n, Therefore, Num of z = z = z=2 z= 2 3πz(4 z2 )dz 2 3π(4 z2 )dz = 3π 2 (4z z3 )dz 3π 2 (4 z2 )dz 2 = (4z z3 )dz 2 (4 z2 )dz ( ) z=2 2z 2 z4 = = ( 4z z3 3 4 z= ) z=2 z= 3πz(4 z 2 )dz. (8 4) ( ) 4 (8 8 = 3 ) ( ) 6/3 = 3 4, so tht the center of mss is t position (,, 3 4 ). It is sensible tht z is bit below z =, since the hemisphere is wider ner z = thn ner z = 2. Presumbly you cn interpret this 3D centerofmss s the loction where 4dimensionl being could blnce this 3D solid, nlogous to our interprettions of the D nd 2D centerofmss. For interprettions more useful to 3D beings like us, you might sk your locl physicist. For exmple, if you were to throw this hlfornge cross room, physicist would likely choose coordintes centered t the center of mss to describe its motion most tidily.
80 6.4. PROBABILITY DENSITY AND AVERAGES Probbility Density nd Averges In the pst two sections we hve considered density functions tht weight geometric object (in D or 2D or 3D) by something firly tngible, like mss or weight or popultion. Now we will explore weighting tht is more bstrct, but rgubly the most prevlent type of density used. It reltes to the mthemticl formultion of rndom numbers. The brnch of mthemtics tht involves rndom numbers is often clled probbility theory, nd our gol will be to understnd the mening of probbility density function A wrmup: discrete probbility. To set the stge, we first explore reltively simple kind of rndomness: the selection of rndom number from finite number of possible choices. The explortion of this problem is often clled discrete probbility theory. In order to hve concrete exmple to work through s we go, let s suppose we wnt to choose number from to (ech with equl likelihood). We cn implement this kind of rndomness quite esily. To choose from mong N possible rndom numbers (with equl likelihood), tke N slips of pper, write one of the numbers on ech slip, put the slips in n opque bg, nd then fish one out without looking. Ech number clerly hs inn chnce of being selected; in mthemticl lnguge, we sy its probbility of being chosen is /N. We could summrize the sitution using tble listing the possible outcomes nd the probbility of ech, e.g., in our exmple we would hve: x Probbility Given this tble, it is pretty strightforwrd to determine the probbility tht ny prticulr event of interest will occur: you locte the columns in which x chieves the event nd dd up the corresponding probbilities. For exmple, suppose tht in our to exmple, we wnt to know the chnces/probbility tht our number will be perfect squre. We determine tht the columns with x =, 4, 9 chieve the event, nd ech hs probbility / so: Probbility of perfect squre = + + = 3, or 3% chnce. A useful extension of this ide is to llow the rndom numbers to not ll be eqully likely this would be discrete probbility weighting. So, we would hve N possible numbers x, x 2,..., x N, nd ech would hve some likelihood p, p 2,..., p N (respectively). In order for the p j to qulify s likelihoods, we would hve to hve p + p p N = (representing the fct tht our rndom process is % gurnteed to produce single rndom outcome from our list). This fits perfectly with the setup of our tble, in tht we merely replce the /N entries with the p i. For instnce, here is n unequl weighting for our to exmple:
81 6.4. PROBABILITY DENSITY AND AVERAGES 275 x Probbility Notice tht the sum of the probbilities is one, s we stipulted it must be. This weighting gives more likelihood to the lrger numbers. In order to determine the probbility of ny prticulr event, the procedure remins the sme: locte the x vlues of interest nd dd up their probbilities. For exmple, now the probbility of picking perfect squre comes from summing the probbilities corresponding to, 4, nd 9: Probbility of perfect squre = = 27, or 27% chnce Continuous Probbility. We now seek to generlize the ide of rndom number to setting in which we re selecting from n infinite number of possibilities. Specificlly, we will seek to describe the process of choosing rel number t rndom from the intervl [, b] (for < b). For exmple, if we plce pencil on the xyplne nd spin it until it comes to rest, its ngle with the positive xxis would be rndom rel number in the intervl [, 2π]. Further, we would like to llow for nonequl weighting, s we sw t the end of the section on discrete probbility; for exmple, we my wnt to choose number t rndom from [, ] with the numbers closer to more likely to be chosen thn those closer to. This kind of weighting of n intervl [, b] is exctly nlogous to how mssdensity function ρ(x) weights different segments of D rod. Seeing this connection leds to the following: Definition 6.4 (Probbility Density Function). Given n intervl [, b], we sy tht function p(x) is probbility density function if p(x) for ll x in [, b] nd b p(x)dx =. Given probbility density function p(x) on the intervl [, b], we define rndom number governed by p(x) s follows: given ny intervl [c, d] inside [, b], Probbility the rndom number is between c nd d = d c p(x)dx. Consistent with our pproch elsewhere in these notes, we omit some technicl restrictions on p(x) tht re necessry in order for the definite integrls in the definition to exist. As long s p(x) is continuous on [, b], or even if it hs finite number of jump discontinuities, those technicl requirements will be stisfied. Let s compre this definition with the setup of discrete probbility. The requirements tht p(x) nd b p(x)dx re nlogous to our requirements for the tble of discrete probbilities: the probbilities in the tble must ech be nd must sum to one (s we hve seen in erlier pplictions, it is common for sum
82 6.4. PROBABILITY DENSITY AND AVERAGES 276 in discrete problem to become definite integrl in the continuous cse, just s rectngle sum becme n reundercurve definite integrl.). Similrly the lst line in the definition is nlogous to the computtion in the discrete cse in which we dd up probbilities over ll x vlues tht stisfy the event of interest (in this cse, the event being c x d). Let us puse to highlight feture of this definition tht some find disturbing. If you wnt to pick vlue of c nd sk for the probbility tht the rndom number equls c, the definition sys tht the probbility is c c p(x)dx, which we know to equl zero. Thus, for ny prticulr vlue of c, there is zero probbility tht the rndom number will equl c. And yet, of course, the rndom number will equl some number, which my pper to contrdict the fct tht for ech individul point, there is zero probbility of lnding on tht point. In terms of our definition of definite integrls, these two fcts re not in contrdiction: if we hve curve y = f(x) bove the xxis, we know tht b f(x)dx >, becuse it is the re under the curve, but we lso hve c c f(x)dx = for ech individul c between nd b, becuse the re from c to c is zero, s tht region hs zero width. In terms of the intuition of rndom numbers, consider the cse in which x is the height of person chosen t rndom from some lrge popultion. If you pick ny prticulr height, sy c = 72 inches, it is infinitesimlly unlikely tht the person you choose t rndom will hve height exctly to n infinite number of deciml plces. Wht then is the mening of the function p(x)? One nswer comes stright from the definition: it s function tht you integrte to get probbility (just like mssdensity is function you integrte to get mss). Here s nother nswer. For concreteness ske, let s return to the height exmple nd focus gin on the height of 72 inches. In tht cse, you cn think of p(72) s (pproximtely) the probbility tht person chosen t rndom will hve height 72 inches to the nerest inch. Why is tht? Well, by the mening of to the nerest inch, height will be 72 to the nerest inch if it is between 7.5 nd 72.5, nd then using the definition of probbility density function, we hve: Probbility the height is 72 to the nerest inch = p(x)dx. Now we ll pproximte the definite integrl by rectngle sum. Since the region of integrtion is quite nrrow (7.5 to 72.5), we should hve decent ccurcy using single rectngle, nd we my s well use the midpoint rectngle sum s one of our best pproximtions. Thus, p(x)dx p(72)( ) = p(72), where p(72) is the height of the rectngle (using the midpoint 72) nd is the width of the rectngle. Thus, we hve p(72) Probbility the height is 72 to the nerest inch.
83 6.4. PROBABILITY DENSITY AND AVERAGES 277 Exmple 6.5. Consider the function p(x) = +.25x 2.5 =.4 +.x. Let s verify tht p(x) is probbility density function for the intervl [, 2], nd then compute the probbility tht rndom number governed by p(x) is betweeen x = nd x =. From the formul for p(x), we cn see tht p(x) for ll x in [, 2]. All tht remins to verify tht p(x) is probbility density function is to confirm tht p(x)dx =, which we cn do by computing the integrl symboliclly: 2 2 (.4 +.x)dx = = ) x=2 (.4x +. x2 2 x= [( ] ) ( + ) =. To find the probbility tht the rndom number is between x = nd x =, we compute (.4 +.x)dx = = ) x= (.4x +. x2 2 x= [( ] ) ( + ) =.45 Thus, there is 45% chnce (or probbility of.45) tht the rndom number is between x = nd x =. We cn mke sense of this result be referring to grph of p(x): Since the probbility density function slopes up gently, it gives more likelihood to the lrger numbers in the intervl [, 2], which fits with the chnces of lnding between nd being bit less thn 5%, since those numbers re given somewht less weight thn the numbers in the right hlf of the intervl.
84 6.4. PROBABILITY DENSITY AND AVERAGES Constructing probbility density function. When one studies the mthemticl discipline clled probbility theory (or sttistics), one lerns bout wide vriety of probbility density functions. On the other hnd, it is lso not very hrd to construct such function oneself. Let s see this vi n exmple. The Frminghm Offspring study (sixth exmintion, ) collected dt on BMI (body mss index) for 348 prticipnts; these results re summrized in the histogrm below, where the height of ech br equls the number of prticipnts with BMI in the rnge covered by the br 2 : This dt suggests tht BMI obeys probbility density function on [6, 54] tht rises from zero t x = 6 to pek t x = 26 nd tils off to nerly zero t x = 54 (it looks to hve height of bout 2 % of the height of the pek). Let s construct probbility density function p(x) with these key fetures. There is not just one wy to construct resonble p(x), nd it will require some cretivity nd guesswork. First, let s imgine shifting the dt to the left by 6, so our desired curve strts t the origin rther thn the point (6, ); this should mke it esier to drem up symbolic formul for the curve. Our shifted curve should strt t (, ), rise up to pek t x = (= 26 6) nd decy to lmost zero by x = 38 (= 54 6). We know tht the functions e kx (with k > ) hve tils like we hve in the righthlf of the intervl. Since we wnt the til to be lmost gone by x = 38, we choose k so tht e k(38) is lrgely, but not entirely, decyed to zero, which suggests 38k 4, since e 4.2, consistent with til decying to 2% of the pek height. How cn we mintin tht til but lso crete pek in the left hlf of the intervl? One ide is to subtrct two exponentillydecying functions e kx e x (for, k > ). For lrger vlues of x, this difference will be dominted by the slowerdecying term, so if > k, then the first term should dominte the second term, so tht e kx e x e kx for x lrge. 2 Figure nd description tken from M.G. Lrson, Descriptive Sttistics nd Grphicl Displys, Circultion (26) 4:768.
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