The Fundamental Theorem of Calculus

Transcription

1 Chpter The Fundmentl Theorem of Clculus In this chpter we will formulte one of the most importnt results of clculus, the Fundmentl Theorem. This result will link together the notions of n integrl nd derivtive. Using this result will llow us to replce the technicl clcultions of Chpter by much simpler procedures involving ntiderivtives of function.. The definite integrl In Chpter, we defined the definite integrl, I, of function f() > on n intervl [, b] s the re under the grph of the function over the given intervl b. We used the nottion I = b f()d to represent tht quntity. We lso set up technique for computing res: the procedure for clculting the vlue of I is to write down sum of res of rectngulr strips nd to compute limit s the number of strips increses: I = b f()d = lim N k= N f( k ), (.) where N is the number of strips used to pproimte the region, k is n inde ssocited with the k th strip, nd = k+ k is the width of the rectngle. As the number of strips increses (N ), nd their width decreses ( ), the sum becomes better nd better pproimtion of the true re, nd hence, of the definite integrl, I. Emple of such clcultions (tedious s they were) formed the min theme of Chpter. We cn generlize the definite integrl to include functions tht re not strictly positive, s shown in Figure.. To do so, note wht hppens s we incorporte strips corresponding to regions of the grph below the is: These re ssocited with negtive vlues of the function, so tht the quntity f( k ) in the bove sum would be negtive for ech rectngle in the negtive portions of the function. This mens tht regions of the grph below the is will contribute negtively to the net vlue of I. 6

2 64 Chpter. The Fundmentl Theorem of Clculus If we refer to A s the re corresponding to regions of the grph of f() bove the is, nd A s the totl re of regions of the grph under the is, then we will find tht the vlue of the definite integrl I shown bove will be I = A A. Thus the notion of re under the grph of function must be interpreted little crefully when the function dips below the is. y y=f() y y=f() y () y (b) y=f() y=f() (c) (d) b c Figure.. () If f() is negtive in some regions, there re terms in the sum (.) tht crry negtive signs: this hppens for ll rectngles in prts of the grph tht dip below the is. (b) This mens tht the definite integrl I = b f()d will correspond to the difference of two res, A A where A is the totl re (drk) of positive regions minus the totl re (light) of negtive portions of the grph. Properties of the definite integrl: (c) illustrtes Property. (d) illustrtes Property.. Properties of the definite integrl The following properties of definite integrl stem from its definition, nd the procedure for clculting it discussed so fr. For emple, the fct tht summtion stisfies the distributive

3 .. The re s function 65 property mens tht n integrl will stisfy the sme the sme property. We illustrte some of these in Fig c b b b f()d =, f()d = b Cf()d = C f()d + b (f() + g())d = f()d = b c b f()d, b f()d. f() + f()d, b g()d, Property sttes tht the re of region with no width is zero. Property shows how region cn be broken up into two pieces whose totl re is just the sum of the individul res. Properties nd 4 reflect the fct tht the integrl is ctully just sum, nd so stisfies properties of simple ddition. Property 5 is obtined by noting tht if we perform the summtion in the opposite direction, then we must replce the previous rectngle width given by = k+ k by the new width which is of opposite sign: k k+. This ccounts for the sign chnge shown in Property 5.. The re s function In Chpter, we investigted how the re under the grph of function chnges s one of the endpoints of the intervl moves. We defined function tht represents the re under the grph of function f, from some fied strting point, to n endpoint. A() = f(t) dt. This endpoint is considered s vrible 4, i.e. we will be interested in the wy tht this re chnges s the endpoint vries (Figure.()). We will now investigte the interesting connection between A() nd the originl function, f(). We would like to study how A() chnges s is incresed ever so slightly. Let = h represent some (very smll) increment in. (Cution: do not confuse h with height here. It is ctully step size long the is.) Then, ccording to our definition, A( + h) = +h f(t) dt. 4 Recll tht the dummy vrible t inside the integrl is just plce holder, nd is used to void confusion with the endpoint of the integrl ( in this cse). Also note tht the vlue of A() does not depend in ny wy on t, so ny letter or symbol in its plce would do just s well.

4 66 Chpter. The Fundmentl Theorem of Clculus y y y=f() y=f() A() A(+h) y () y (b) +h y=f() y=f() A(+h) A() f() (c) +h (d) h Figure.. When the right endpoint of the intervl moves by distnce h, the re of the region increses from A() to A( + h). This leds to the importnt Fundmentl Theorem of Clculus, given in Eqn. (.). In Figure.()(b), we illustrte the res represented by A() nd by A( + h), respectively. The difference between the two res is thin sliver (shown in Figure.(c)) tht looks much like rectngulr strip (Figure.(d)). (Indeed, if h is smll, then the pproimtion of this sliver by rectngle will be good.) The height of this sliver is specified by the function f evluted t the point, i.e. by f(), so tht the re of the sliver is pproimtely f() h. Thus, A( + h) A() f()h or A( + h) A() f(). h As h gets smll, i.e. h, we get better nd better pproimtion, so tht, in the limit, A( + h) A() lim = f(). h h The rtio bove should be recognizble. It is simply the derivtive of the re function, i.e. f() = da d = lim h A( + h) A(). (.) h

5 .4. The Fundmentl Theorem of Clculus 67 We hve just given simple rgument in support of n importnt result, clled the Fundmentl Theorem of Clculus, which is restted below...4 The Fundmentl Theorem of Clculus.4. Fundmentl theorem of clculus: Prt I Let f() be bounded nd continuous function on n intervl [, b]. Let A() = f(t) dt. Then for < < b, da d = f(). In other words, this result sys tht A() is n ntiderivtive (or nti-derivtive ) of the originl function, f(). Proof See bove rgument. nd Figure...4. Emple: n ntiderivtive Recll the connection between functions nd their derivtives. Consider the following two functions: g () =, g = +. Clerly, both functions hve the sme derivtive: g () = g () =. We would sy tht / is n ntiderivtive of nd tht ( /) + is lso n ntiderivtive of. In fct, ny function of the form g() = + C where C is ny constnt is lso n ntiderivtive of. This emple illustrtes tht dding constnt to given function will not ffect the vlue of its derivtive, or, stted nother wy, ntiderivtives of given function re defined only up to some constnt. We will use this fct shortly: if A() nd F () re both ntiderivtives of some function f(), then A() = F () + C..4. Fundmentl theorem of clculus: Prt II Let f() be continuous function on [, b]. Suppose F () is ny ntiderivtive of f(). Then for b, A() = f(t) dt = F () F ().

6 68 Chpter. The Fundmentl Theorem of Clculus Proof From comments bove, we know tht function f() could hve mny different ntiderivtives tht differ from one nother by some dditive constnt. We re told tht F () is n ntiderivtive of f(). But from Prt I of the Fundmentl Theorem, we know tht A() is lso n ntiderivtive of f(). It follows tht A() = However, by property of definite integrls, Thus, f(t) dt = F () + C, where C is some constnt. (.) A() = f(t) = F () + C =. C = F (). Replcing C by F () in eqution. leds to the desired result. Thus Remrk : Implictions A() = f(t) dt = F () F (). This theorem hs tremendous implictions, becuse it llows us to use powerful new tool in determining res under curves. Insted of the drudgery of summtions in order to compute res, we will be ble to use shortcut: find n ntiderivtive, evlute it t the two endpoints, b of the intervl of interest, nd subtrct the results to get the re. In the cse of elementry functions, this will be very esy nd convenient. Remrk : Nottion We will often use the nottion F (t) = F () F () to denote the difference in the vlues of function t two endpoints..5 Review of derivtives (nd ntiderivtives) By remrks bove, we see tht integrtion is relted to nti-differentition. This motivtes review of derivtives of common functions. Tble. lists functions f() nd their derivtives f () (in the first two columns) nd functions f() nd their ntiderivtives F () in the subsequent two columns. These will prove very helpful in our clcultions of bsic integrls. As n emple, consider the polynomil p() =

7 .5. Review of derivtives (nd ntiderivtives) 69 function derivtive function ntiderivtive f() f () f() F () C C C C n n n m m+ m+ sin() cos() cos(b) (/b) sin(b) cos() sin() sin(b) (/b) cos(b) tn() sec () sec (b) (/b) tn(b) e k ke k e k e k /k ln() rctn() rcsin() ln() + + rctn() rcsin() Tble.. Common functions nd their derivtives (on the left two columns) lso result in corresponding reltionships between functions nd their ntiderivtives (right two columns). In this tble, we ssume tht m, b, k. Also, when using ln() s ntiderivtive for /, we ssume tht >. This polynomil could hve mny other terms (or even n infinite number of such terms, s we discuss much lter, in Chpter ). Its ntiderivtive cn be found esily using the power rule together with the properties of ddition of terms. Indeed, the ntiderivtive is F () = C This cn be checked esily by differentition 5. 5 In fct, it is very good prctice to perform such checks.

8 7 Chpter. The Fundmentl Theorem of Clculus.6 Emples: Computing res with the Fundmentl Theorem of Clculus.6. Emple : The re under polynomil Consider the polynomil p() = (Here we hve tken the first few terms from the emple of the lst section with coefficients ll set to.) Then, computing I = p() d leds to I = ( ) d = ( ).6. Emple : Simple res = Determine the vlues of the following definite integrls by finding ntiderivtives nd using the Fundmentl Theorem of Clculus:. I =. I =. I = 4. I = Solutions π π d, ( ) d, e d, ( ) sin d,. An ntiderivtive of f() = is F () = ( /), thus I = d = F () = (/)( ) = ( ) =.. An ntiderivtive of f() = ( ) is F () = ( /), thus ) I = ( ) d = F () = ( ) ) = ( (( ) ( ) = 4/ See comment below for simpler wy to compute this integrl.

9 .6. Emples: Computing res with the Fundmentl Theorem of Clculus 7 y= Figure.. We cn eploit the symmetry of the function f() = in the second integrl of Emples.6.. We cn integrte over nd double the result.. An ntiderivtive of e is F () = ( /)e. Thus, I = e d = F () = ( /)(e ) = ( /)(e e ). 4. An ntiderivtive of sin(/) is F () = cos(/)/(/) = cos(/). Thus π ( ) ( ) π I = sin d = cos = (cos (π) cos ( π)) π π = ( ( )) =. See comment below for simpler wy to compute this integrl. Comment : The evlution of Integrl in the emples bove is tricky only in tht signs cn esily get grbled when we plug in the endpoint t -. However, we cn simplify our work by noting the symmetry of the function f() = on the given intervl. As shown in Fig., the res to the right nd to the left of = re the sme for the intervl. This stems directly from the fct tht the function considered is even 6. Thus, we cn immeditely write I = ( ) d = ( ) d = ) ( ) = ( = 4/. Note tht this clcultion is simpler since the endpoint t = is trivil to plug in. We stte the generl result we hve obtined, which holds true for ny function with even symmetry integrted on symmetric intervl bout = : If f() is n even function, then f() d = f() d. (.4) 6 Recll tht function f() is even if f() = f( ) for ll. A function is odd if f() = f( ).

10 7 Chpter. The Fundmentl Theorem of Clculus Cution: The function f() must be integrble, i.e. f() must eist nd be defined over the entire intervl [, ]. For emple, the integrnd in d is even but requires more creful considertions becuse it does not eist t = (see Chpter 7). Similrly, if f() is n odd function then the symmetry yields n even simpler result on symmetric intervl bout = : If f() is n odd function, then f() d = for ny. (.5) Cution: Agin, f() must be integrble over the entire intervl [, ]. For emple, even though the integrnd in d is odd, the integrl requires more creful considertions becuse the integrnd does not eist t = (see Chpter 7). Comment : We cn eploit symmetries for simpler evlution of Integrl 4 bove: by relizing tht sin( ) is n odd function nd lso noticing tht the integrtion bounds re symmetric bout =, we immeditely conclude tht the integrl evlutes to zero without requiring ny ctul clcultions..6. Emple : The re between two curves The definite integrl is n re of somewht specil type of region, i.e., n is, two verticl lines ( = nd = b) nd the grph of function. However, using dditive (or subtrctive) properties of res, we cn generlize to computing res of other regions, including those bounded by the grphs of two functions. () Find the re enclosed between the grphs of the functions y = nd y = / in the first qudrnt. (b) Find the re enclosed between the grphs of the functions y = nd y = in the first qudrnt. (c) Wht is the reltionship of these two res? Wht is the reltionship of the functions y = nd y = / tht leds to this reltionship between the two res? Solution () The two curves, y = nd y = /, intersect t = nd t = in the first qudrnt. Thus the intervl tht we will be concerned with is < <. On this intervl, / >, so tht the re we wnt to find cn be epressed s: A = ( / ) d.

11 .6. Emples: Computing res with the Fundmentl Theorem of Clculus y y= y= / y= A A Figure.4. In Emple, we compute the res A nd A shown bove. Thus, A = 4/ 4/ 4 4 = 4 4 =. (b) The two curves y = nd y = lso intersect t = nd t = in the first qudrnt, nd on the intervl < < we hve >. The re cn be represented s ( A = ) d. A = 4 4 = 4 = 4. (c) The re clculted in () is twice the re clculted in (b). The reson for this is tht / is the inverse of the function, which mens geometriclly tht the grph of / is the mirror imge of the grph of reflected bout the line y =. Therefore, the re A between y = / nd y = is twice s lrge s the re A between y = nd y = clculted in prt (b): A = A (see Figure.4)..6.4 Emple 4: Are of lnd Find the ect re of the piece of lnd which is bounded by the y is on the west, the is in the south, the lke described by the function y = f() = + (/) in the north nd the line = in the est. Solution The re is A = ( ( ) ) + d. = ( ( ) ) + d.

12 74 Chpter. The Fundmentl Theorem of Clculus Note tht the multiplictive constnt (/) is not ffected by integrtion. The result is A =.7 Qulittive ides + ( ) = 4 5. In some cses, we re given sketch of the grph of function, f(), from which we would like to construct sketch of the ssocited function A(). This sketching skill is illustrted in the figures shown in this section. Suppose we re given function s shown in the top left hnd pnel of Figure.5. We would like to ssemble sketch of A() = f(t)dt which corresponds to the re ssocited with the grph of the function f. As moves from left to right, we show how the re ccumulted long the grph grdully chnges. (See A() in bottom pnels of Figure.5): We strt with no re, t the point = (since, by definition A() = ) nd grdully build up to some net positive mount, but then we encounter portion of the grph of f below the is, nd this subtrcts from the mount ccrued. (Hence the grph of A() hs little pek tht corresponds to the point t which f =.) Every time the function f() crosses the is, we see tht A() hs either mimum or minimum vlue. This fits well with our ide of A() s the ntiderivtive of f(): Plces where A() hs criticl point coincide with plces where da/d = f() =. Sketching the function A() is thus nlogous to sketching function g() when we re given sketch of its derivtive g (). Recll tht this ws one of the skills we built up in lerning the connection between functions nd their derivtives in first semester clculus course. Remrks The following remrks my be helpful in gining confidence with sketching the re function A() = f(t) dt, from the originl function f():. The endpoint of the intervl, on the is indictes the plce t which A() =. This follows from Property of the definite integrl, i.e. from the fct tht A() = f(t) dt =.. Whenever f() is positive, A() is n incresing function - this follows from the fct tht the re continues to ccumulte s we sweep cross positive regions of f().. Wherever f(), chnges sign, the function A() hs locl minimum or mimum. This mens tht either the re stops incresing (if the trnsition is from positive to negtive vlues of f), or else the re strts to increse (if f crosses from negtive to positive vlues).

13 .7. Qulittive ides 75 f() f() A() A() () f() (b) f() A() A() (c) (d) Figure.5. Given function f(), we here show how to sketch the corresponding re function A(). (The reltionship is tht f() is the derivtive of A() 4. Since da/d = f() by the Fundmentl Theorem of Clculus, it follows tht (tking derivtive of both sides) d A/d = f (). Thus, when f() hs locl mimum or minimum, (i.e. f () = ), it follows tht A () =. This mens tht t such points, the function A() would hve n inflection point. Given function f(), Figure.6 shows in detil how to sketch the corresponding function g() =.7. Emple: sketching A() f(t)dt. Consider the f() whose grph is shown in the top prt of Figure.7. Sketch the corresponding function g() = f()d.

14 76 Chpter. The Fundmentl Theorem of Clculus f() g() Figure.6. Given function f() (top, solid line), we ssemble plot of the corresponding function g() = f(t)dt (bottom, solid line). g() is n ntiderivtive of f(). Whether f() is positive (+) or negtive (-) in portions of its grph, determines whether g() is incresing or decresing over the given intervls. Plces where f() chnges sign correspond to mim nd minim of the function g() (Two such plces re indicted by dotted verticl lines). The bo in the middle of the sketch shows configurtions of tngent lines to g() bsed on the sign of f(). Where f() =, those tngent lines re horizontl. The function g() is drwn s smooth curve whose direction is prllel to the tngent lines shown in the bo. While the function f() hs mny ntiderivtives (e.g., dshed curve prllel to g()), only one of these stisfies g() = s required by Property of the definite integrl. (See dshed verticl line t = ). This determines the height of the desired function g(). Solution See Figure.7

15 .8. Prelude to improper integrls 77 f() + + g() Figure.7. The originl functions, f() is shown bove. The corresponding functions g() is drwn below..8 Prelude to improper integrls The Fundmentl Theorem hs number of restrictions tht must be stisfied before its results cn be pplied. In this section we look t some emples in which cre must be used. The emples of this section should be compred with our chpter on Improper Integrls. In prticulr one must be creful in pplying the Fundmentl theorem when the integrnd hs discontinuity within the region of integrtion..8. Function unbounded I Consider the definite integrl d. The function f() = is discontinuous t =, nd unbounded on ny intervl tht contins the point =, i.e. lim ± = ±. Hence we cnnot directly evlute this integrl using the Fundmentl theorem. However we cn show tht lim ɛ + ɛ d =. Indeed the integrnd is continuous on the intervl (ɛ, ) for ll ɛ >. The Fundmentl theorem therefore tells us d = log() log(ɛ). ɛ

16 78 Chpter. The Fundmentl Theorem of Clculus But lim ɛ + log(ɛ) =, nd therefore lim ɛ + the integrl d diverges..8. Function unbounded II Consider the definite integrl d. ɛ d =, s climed. We sy tht As in the previous emple the integrnd is discontinuous nd unbounded t =. To blindly pply the Fundmentl theorem (which in this cse is unjustified) yields =. = However the integrl d certinly cnnot be negtive, since we re integrting positive function over finite domin (nd the grph bounds some positive possibly infinite re). Splitting the integrl s sum d = d + d, we cn see by n rgument entirely similr to our previous emple tht sepertely ech improper integrl d, d diverges, i.e. both ɛ lim ɛ d, lim ɛ ɛ d diverge to. Consequently our integrl the sum of two integrls diverging to + diverges..8. Emple: Function discontinuous or with distinct prts Suppose we re given the integrl I = d. This function is ctully mde up of two distinct prts, nmely { if > f() = if <. The integrl I must therefore be split up into two prts, nmely I = d = ( ) d + d.

17 .8. Prelude to improper integrls 79 We find tht I = + [ = ] [ ] 4 + =.5 y y= Figure.8. In this emple, to compute the integrl over the intervl, we must split up the region into two distinct prts..8.4 Function undefined Now let us emine the integrl / d. We see tht there is problem here. Recll tht / =. Hence, the function is not defined for < nd the intervl of integrtion is inpproprite. Hence, this integrl does not mke sense..8.5 Integrting over n infinite domin Consider the integrl I = b e r d, where r >, nd b > re constnts. Simple integrtion using the ntiderivtive in Tble. (for k = r) leds to the result I = e r r b = ( e rb e ) = ( e rb ). r r This is the re under the eponentil curve between = nd = b. Now consider wht hppens when b, the upper endpoint of the integrl increses, so tht b. Then the vlue of the integrl becomes I = lim b b e r ( d = lim e rb ) = b r r ( ) = r.

18 8 Chpter. The Fundmentl Theorem of Clculus (We used the fct tht e rb s b.) We hve, in essence, found tht I = e r d = r. (.6) An integrl of the form (.6) is clled n improper integrl. Even though the domin of integrtion of this integrl is infinite, (, ), observe tht the vlue we computed is finite, so long s r. Not ll such integrls hve bounded finite vlue. Lerning to distinguish between those tht do nd those tht do not will form n importnt theme in Chpter Regions tht need specil tretment So fr, we hve lerned how to compute res of regions in the plne tht re bounded by one or more curves. In ll our emples so fr, the bsis for these clcultions rests on imgining rectngles whose heights re specified by one or nother function. Up to now, ll the rectngulr strips we considered hd bses (of width ) on the is. In Figure.9 we observe n emple in which it would not be possible to use this technique. We re y y y =g(y) Figure.9. The re in the region shown here is best computed by integrting in the y direction. If we do so, we cn use the curved boundry s single function tht defines the region. (Note tht the curve cnnot be epressed in the form of function in the usul sense, y = f(), but it cn be epressed in the form of function = f(y).) sked to find the re between the curve y y + = nd the y is. However, one nd the sme curve, y y + = forms the boundry from both the top nd the bottom of the region. We re unble to set up series of rectngles with bses long the is whose heights re described by this curve. This mens tht our definite integrl (which is relly just convenient wy of crrying out the process of re computtion) hs to be hndled with cre. Let us consider this problem from new ngle, i.e. with rectngles bsed on the y is, we cn chieve the desired result. To do so, let us epress our curve in the form = g(y) = y y. Then, plcing our rectngles long the intervl < y < on the y is (ech hving bse of width y) leds to the integrl ( ) y I = g(y) dy = (y y )dy = y = = 6.

19 .9. Summry 8.9 Summry In this chpter we first recpped the definition of the definite integrl in Section., reclled its connection to n re in the plne under the grph of some function f(), nd emined its bsic properties. If one of the endpoints, of the integrl is llowed to vry, the re it represents, A(), becomes function of. Our construction in Figure. showed tht there is connection between the derivtive A () of the re nd the function f(). Indeed, we showed tht A () = f() nd rgued tht this mkes A() n ntiderivtive of the function f(). This importnt connection between integrls nd ntiderivtives is the cru of Integrl Clculus, forming the Fundmentl Theorem of Clculus. Its significnce is tht finding res need not be s tedious nd lbored s the clcultion of Riemnn sums tht formed the bulk of Chpter. Rther, we cn tke shortcut using ntidifferentition. Motivted by this very importnt result, we reviewed some common functions nd derivtives, nd used this to relte functions nd their ntiderivtives in Tble.. We used these ntiderivtives to clculte res in severl emples. Finlly, we etended the tretment to include qulittive sketches of functions nd their ntiderivtives. As we will see in upcoming chpters, the ides presented here hve much wider rnge of pplicbility thn simple re clcultions. Indeed, we will shortly show tht the sme concepts cn be used to clculte net chnges in continully vrying processes, to compute volumes of vrious shpes, to determine displcement from velocity, mss from densities, s well s host of other quntities tht involve process of ccumultion. These ides will be investigted in Chpters 4, nd 5.

20 8 Chpter. The Fundmentl Theorem of Clculus. Eercises Eercise. () Give concise sttement of the Fundmentl Theorem of Clculus. (b) Why is it useful prcticl tool? Eercise. Consider the function y = f() = e on the intervl [, ]. Find the re under the grph of this function over this intervl using the Fundmentl Theorem of Clculus. Eercise. Determine the vlues of the integrls shown below, using the Fundmentl Theorem of Clculus (i.e. find the nti-derivtive of ech of the functions nd evlute t the two endpoints.) () (b) - d ( )d Eercise.4 Use the Fundmentl Theorem of Clculus to compute ech of the following integrls. The lst few re little more chllenging, nd will require specil cre. Some of these integrls my not eist. Eplin why. () (d) (g) (j) (m) (p) (s) π π 4 π 4 sin() d (b) cos() d (e) ( + ) d (h) / d d / d d (k) (n) (q) (t) π sin() d (c) ( ) d (f) ( + ) d (i) ( + ) d (l) e d d (o) (r) π 4 cos() d ( ) d / d d / d d

21 .. Eercises 8 Eercise.5 Find the following integrls using the Fundmentl theorem of Clculus. () (d) (g) (j) b b e kt dt q dq s ds dt (b) (e) (h) T c T A cos(ks) ds sec (5) d d (i) / (c) (f) b Ct m dt + t dt sin(y) dy Eercise.6 () Use n integrl to estimte the sums N k= k. (b) For N = 4 drw sketch in which the vlue computed by summing the four terms is compred to the vlue found by the integrtion. (Your sketch should show the grph of the pproprite function nd set of steps tht represent the bove sum.) Eercise.7 Find the re under the grphs of these functions: () f() = / between = nd =. (b) v(t) = t between t = nd t = T, where, T re fied constnts nd >. (c) h(u) = u between u = / nd u =. Eercise.8 Find the re S between the two curves y = nd y = for >. Eplin the reltionship of your nswer to the two integrls I = ( )d nd I = ( )d. Eercise.9 Find the re S between the grphs of y = f() = nd y =. Eercise. () Find the re enclosed between the grphs of the functions y = f() = n nd the stright line y = in the first qudrnt. (Note tht we re considering positive vlues of n nd tht for n = the re is zero.) (b) Use your nswer in prt () to find the re between the grphs of the functions y = f() = n nd y = g() = /n in the first qudrnt. (Hint: wht is the reltionship between these two functions nd wht sort of symmetry do their grphs stisfy?) Eercise. A piece of tin shped like lef blde is to be cut from squre m m sheet of tin. The shpe of one of the sides is given by the function y = nd the shpe is to be symmetric bout the line y =, see Figure..

22 84 Chpter. The Fundmentl Theorem of Clculus () How much tin goes into mking the shpe? (b) How much is left over? Assume tht the thickness of the sheet is such tht ech squre cm weighs one grm (g), nd tht y nd re in meters..4. y = ^ y Figure.. For problem. Eercise. Find the re S between the grphs of the functions y = f() =, y = g() = + nd the y is. Eercise. Find the re A of the finite plne region bounded by the prbol y = 6 nd the prbol y = 4. [from the April 97 Finl Em]. Eercise.4 Find the re A of the shpe shown in Figure.. Eercise.5 Find the re A under the function shown in Figure. between = nd = 5. To do so, determine the equtions of the line segments mking up this grph nd use integrtion methods. Eercise.6 Let g() = () Evlute g(), g(), g(), g() nd g(6). (b) On wht intervls is g() incresing? (c) Where does g() hve mimum vlue? (d) Sketch rough grph of g(). f(t)dt, where f(t) is the function shown in Figure..

23 .. Eercises y = ^ y Figure.. For problem.4 4 (,) y (4,) y=f() (,) Figure.. For problem.5 Eercise.7 Consider the functions shown in Figure.4 () nd (b). In ech cse, use the sketch of this function y = f() to drw sketch of the grph of the relted function F () = f(t) dt. (Assume tht F ( ) = 4 in (), nd F () =.5 t the left end of the intervl in (b).) Eercise.8 Consider the functions shown in Figure.5 () nd (b). In ech cse, use the sketch of this function y = f() to drw sketch of the grph of the relted function F () = f(t)dt. Eercise.9 Consider the functions shown in Figure.6 () nd (b). In ech cse, use the sketch of this function y = f() to drw sketch of the grph of the relted function F () = f(t) dt. (Assume tht F () = in both () nd (b).)

24 86 Chpter. The Fundmentl Theorem of Clculus 5 4 y f () Figure.. For problem Figure.4. For problem Figure.5. For problem.8

25 .. Eercises Figure.6. For problem.9 Eercise. Leves revisited Use the Fundmentl Theorem of Clculus (i.e. integrtion techniques) to find the res of the leves shown in Figure.7. These leves re generted by the functions () y = ( ) (b) y = ( ) (c) y = ( ) () y = ( )

26 88 Chpter. The Fundmentl Theorem of Clculus.4.4 y=f()=^ (-) y=f()= (-)^ Tpered lef Tpered lef y=f()= (-).4 y=f()= (-^) -.4 Symmetric lef Brod lef.. Figure.7. The shpe of leves for problem..

27 .. Solutions 5. Solutions Solution to. () The Fundmentl Theorem of Clculus sttes tht if F () is ny nti-derivtive of f() then b f()d = F (b) F () (b) the Fundmentl Theorem of Clculus provides shortcut for clculting integrls. Without it, we would hve to use tedious summtion nd limits to compute res, volumes, etc. of irregulr regions. Solution to. e Solution to. () (b) Solution to.4 () (b) (c) (d) (e) 8 4 (f) (g) (h) (i) (j) (k) (l) ln (m) ln() (n) (e ) (o) ( / /) 4 (q) DNE (q) DNE (r) (s) DNE (t) DNE DNE: does not eist. Solution to.5 () ( e k e k) (b) k 6 A k sin(k) (c) C ( m+ b m+), m m + (c) C(ln() ln(b)), m =, b (d) DNE (e) (tn(5t ) tn(5c)) 5 (f) rctn() π ( (g) b ) (h) ( T ), T >, > (i) ( cos()) (j) b DNE: does not eist. Solution to.6 () N / (b) see Figure.

28 6 Chpter. The Fundmentl Theorem of Clculus Figure.. Solution to problem.6 for N = 4 nd N = 6. Solution to.7 () ln (b) T (c) Solution to.8 S = 7 6 Solution to.9 S = 8 Solution to. () n (n + ) (b) twice the re of (). Solution to. () kg (b) kg Solution to. S = Solution to. S = 64 Solution to.4 A = Solution to.5 A = 8.5

29 .. Solutions 7 Solution to.6 (), 5, 7, (b) [, ] (c) Solution to.7 For solution, see Figure Figure.4. Solution to problem.7 Solution to.8 For solution, see Figure Figure.5. Solution to problem.8 Solution to.9 For solution, see Figure.6. Solution to. () 6 (b) 6 (c) (d)

30 8 Chpter. The Fundmentl Theorem of Clculus Figure.6. Solution to problem.9