!0 f(x)dx + lim!0 f(x)dx. The latter is sometimes also referred to as improper integrals of the. k=1 k p converges for p>1 and diverges otherwise.

Size: px
Start display at page:

Download "!0 f(x)dx + lim!0 f(x)dx. The latter is sometimes also referred to as improper integrals of the. k=1 k p converges for p>1 and diverges otherwise."

Transcription

1 Chpter 7 Improper integrls 7. Introduction The gol of this chpter is to meningfully extend our theory of integrls to improper integrls. There re two types of so-clled improper integrls: the first involves integrting function over n infinite domin nd the second involves integrnds tht re undefined t points within the domin of integrtion. In order to integrte over n infinite domin, we R b consider its of the form b! f(x)dx. If the integrnd is not defined t c ( <c< b) then we split the integrl nd consider the its R b f(x)dx = R c! f(x)dx + R b! f(x)dx. The ltter is sometimes lso referred to s improper integrls of the c+ second kind. Such situtions occur, for exmple, for rtionl functions f(x) =p(x)/q(x) whenever q(x) hs zeroes in the domin of integrtion. The notions of convergence nd divergence s discussed in Chpter & in the context of sequences nd series will be very importnt to determine these its. Improper integrls (of both types) rise frequently in pplictions nd in probbility. By relting improper integrls to infinite series we derive the lst convergence test: the Integrl Comprison test. As n ppliction we finlly prove tht the p-series P k= k p converges for p> nd diverges otherwise. 7.2 Integrtion over n infinite domin We will see tht there is close connection between certin infinite series nd improper integrls, which involve integrls over n infinite domin. We hve lredy encountered exmples of improper integrls in Section 3.8 nd in the context of rdioctive decy in Section 8.4. Recll the following definition: Definition (Improper integrl (first kind)). An improper integrl of the first kind is n integrl performed over n infinite domin, e.g. Z f(x) dx. 89

2 9 Chpter 7. Improper integrls The vlue of such n integrl is understood to be it, s given in the following definition: Z f(x) dx = b! f(x) dx. We evlute n improper integrl by first computing definite integrl over finite domin pple x pple b, nd then tking it s the endpoint b moves off to lrger nd lrger vlues. The definite integrl cn be interpreted s n re under the grph of the function. The essentil question being ddressed here is whether tht re remins bounded when we include the infinite til of the function (i.e. s the endpoint b moves to lrger vlues.) For some functions (whose vlues get smll enough, fst enough) the nswer is yes. Definition 2 (Convergence). If the it b! f(x) dx exists, we sy tht the improper integrl converges. Otherwise we sy tht the improper integrl diverges. With these definitions in mind, we cn compute number of clssic integrls Exmple: Decying exponentil Recll tht the improper integrl of decying exponentil converges we hve seen this erlier, in Section 3.8.5, nd gin in pplictions in Sections 7.3 nd Here we recpitulte this importnt result in the context of improper integrls. Suppose tht r> nd let Then b! b r e rt Z e rt dt b! e rt dt. = r (e rb e )= b! r ( b! e rb {z } ) = r, where we hve used the fct tht b! e rb =for r>. Thus the it exists (is finite) nd hence the integrl converges. More precisely, it converges to the vlue /r Exmple: The improper integrl of /x diverges We now consider clssic nd counter-intuitive result, nd one of the most importnt results in this chpter. Consider the function y = f(x) = x.

3 7.2. Integrtion over n infinite domin 9 Exmining the grph of this function for positive x, e.g. for the intervl (, ), we know tht vlues decrese to zero s x increses 26. The function is not only bounded, but lso flls to rbitrrily smll vlues s x increses (see Figure 7.). Nevertheless, this is insufficient y y= /x y=/x 2 x Figure 7.. In Sections nd 7.2.3, we consider two functions whose vlues decrese long the x xis, f(x) =/x nd f(x) =/x 2. We show tht one, but not the other encloses finite (bounded) re over the intervl (, ). To do so, we compute n improper integrl for ech one. The hevy rrow is ment to remind us tht we re considering res over n unbounded domin. to gurntee tht the enclosed re remins finite! We mde similr observtion in the context of series in Section.3.. We show this in the following clcultion. Z Z b b dx = dx = x b! x ln(x) = (ln(b) ln()) b! b! = ln(b) = b! The fct tht we get n infinite vlue for this integrl follows from the observtion tht ln(b) increses without bound s b increses, tht is the it does not exist (is not finite). Thus, the re under the curve f(x) =/x over the intervl pple x ppleis infinite. We sy tht the improper integrl of /x diverges (or does not converge). We will use this result gin in Section Exmple: The improper integrl of /x 2 converges Now consider the relted function y = f(x) = x 2 nd the corresponding integrl Z x 2 dx. 26 We do not chose the intervl (, ) becuse this function is undefined t x =. Here we wnt to emphsize the behviour t infinity, not the blow up tht occurs close to x =.

4 92 Chpter 7. Improper integrls Then b! b x 2 dx = ( x ) = b! b! b =. Thus, the it exists, nd is. In contrst to the exmple in Section 7.2.2, this integrl converges. We observe tht the behviours of the improper integrls of the functions /x nd /x 2 re very different. The former diverges, while the ltter converges. The only difference between these functions is the power of x. As shown in Figure 7., tht power ffects how rpidly the grph flls off to zero s x increses. The function /x 2 decreses much fster thn /x. Consequently /x 2 hs sufficiently s infinite til, such tht the re under its grph does not become infinite - not n esy concept to digest! This observtions leds us to wonder wht power p is needed to mke the improper integrl of function /x p converge. We nswer this question below When does the integrl of /x p converge? Here we consider n rbitrry power, p, tht cn be ny rel number. We sk when the corresponding improper integrl converges or diverges. Let Z x p dx. For p =we hve lredy estblished tht this integrl diverges (see Section 7.2.2), nd for p =2we hve seen tht it is convergent (see Section 7.2.3). By similr clcultion, we find tht b! x p ( p) b = b! p b p. Thus, this integrl converges provided tht the term b p does not blow up s b increses. For this to be true, we require tht the exponent ( p) should be negtive, i.e. p< or p>. In this cse, we hve p. To summrize our result, Z dx converges if p>, nd diverges if p pple. xp Exmples: (i) The integrl Z p x dx, diverges. We see this from the following rgument: p x = x 2, so p = 2 generl result, this integrl diverges. <. Thus, by the

5 7.3. Appliction: Present vlue of continuous income strem 93 (ii) The integrl Z x. dx, converges. Here p =. >, so the result implies convergence of the integrl. 7.3 Appliction: Present vlue of continuous income strem Here we discuss the vlue of n nnuity, which is kind of svings ccount tht gurntees continuous strem of income. You would like to py P dollrs to purchse n nnuity tht will py you n income f(t) every yer from now on, for t>. In some cses, we might wnt constnt income every yer, in which cse f(t) would be constnt. More generlly, we cn consider the cse tht t ech future yer t, we sk for income f(t) tht could vry from yer to yer. If the bnk interest rte is r, how much should you py now? Solution If we invest P dollrs (the principl i.e., the mount deposited) in the bnk with interest r then the mount A(t) in the ccount t time t (in yers), will grow s follows: A(t) =P + r n nt, where r is the nnul interest rte (e.g. 5%) nd n is the number of times per yer tht interest is compound (e.g. n = 2mens interest compounded twice per yer, n = 2 mens monthly compounded interest, etc.). Define h = r n. Then t time t, we hve tht A(t) =P ( + h) h rt h = P ( + h) h i rt Pe rt for lrge n or smll h. Here we hve used the fct tht when h is smll (i.e. frequent intervls of compounding) the expression in squre brckets bove cn be pproximted by e, the bse of the nturl logrithms. Recll tht h i e = ( + h) h. h! This result ws obtined in first semester clculus course by selecting the bse of exponentils such tht the derivtive of e x is just e x itself. Thus, we hve found tht the mount in the bnk t time t will grow s A(t) =Pe rt, continully compounded interest. (7.) Hving estblished the exponentil growth of n investment, we return to the question of how to set up n nnuity for continuous strem of income in the future. Rewriting

6 94 Chpter 7. Improper integrls Eqn. (7.), the principle mount tht we should invest in order to hve A(t) to spend t time t is P = A(t)e rt. Suppose we wnt to hve f(t) spending money for ech yer t. We refer to the present vlue of yer t s the quntity P = f(t)e rt, i.e. we must py P now, in the present, to get f(t) in future yer t. Summing over ll the yers, we find tht the present vlue of the continuous income strem is P = LX t= f(t)e rt {z} t Z L f(t)e rt dt, where L is the expected number of yers left in the lifespn of the individul to whom this nnuity will be pid, nd where we hve pproximted sum of pyments by n integrl (of continuous income strem). One problem is tht we do not know in dvnce how long the lifespn L will be. As crude pproximtion, we could ssume tht this income strem continues forever, i.e. tht L. In such n pproximtion, we hve to compute the integrl: P = Z f(t)e rt dt. (7.2) The integrl in Eqn. (7.2) is n improper integrl (i.e. integrl over n unbounded domin), s we hve lredy encountered in Section We shll hve more to sy bout the properties of such integrls, nd bout their technicl definition, existence, nd properties in Chpter 7. We refer to the quntity P = Z s the present vlue of continuous income strem f(t). Exmple: Setting up n nnuity f(t)e rt dt, (7.3) Suppose we wnt n nnuity tht provides us with n nnul pyment of, from the bnk, i.e. in this cse f(t) = $, is function tht hs constnt vlue for every yer. Then from Eqn (7.3), P = Z e rt dt = By previous clcultion in Section 3.8.5, we find tht P = r, Z e rt dt. e.g. if interest rte is 5% (nd ssumed constnt over future yers), then P =.5 = $2,. Therefore, we need to py $2, tody to get, nnully for every future yer.

7 7.4. Integrl comprison test Integrl comprison test The integrls discussed bove cn be used to mke comprisons tht help us to identify when other improper integrls converge or diverge 27. The following importnt result estblishes how these comprisons work: Suppose we re given two functions, f(x) nd g(x), both continuous on some infinite intervl [, ). Suppose, moreover, tht t ll points on this intervl the first function is smller thn the second, i.e. pple f(x) pple g(x). Then the following conclusions cn be mde: (i) (ii) If (iii) If Z f(x) dx pple Z g(x) dx. The re under f(x) is smller thn the re under g(x). Z g(x) dx converges, then Z f(x) dx converges. If the lrger re is finite, so is the smller one. Z f(x) dx diverges, then Z g(x) dx diverges. If the smller re is infinite, so is the lrger one. These sttements hve to be crefully noted. Wht is ssumed nd wht is concluded works one wy. Tht is the order if...then is importnt. Reversing tht order leds to common error. Exmple: Determine whether the following integrl converges: Z x +x 3 dx. Solution: by noting tht for ll x> Note tht for x> pple x +x 3 pple x x 3 = x 2. pple x +x 3 pple x x 3 = x 2. Thus, Z Z x dx pple +x3 x 2 dx. Since the lrger integrl on the right is known to converge, so does the smller integrl on the left. } 27 Similr ides will be employed for the comprison of infinite series in Chpter. A recurring theme in this course is the close connection between series nd integrls, for exmple, recll the Riemnn sums in Chpter 2.

8 96 Chpter 7. Improper integrls 7.5 Integrtion of n unbounded integrnd The second kind of improper integrls refers to integrnds tht re undefined t one (or more) points of the domin of integrtion [, b]. Suppose f(x) is continuous on the open intervl (, b) but becomes infinite t the lower bound, x =. Then the integrl of f(x) over the domin [ +, b] for >hs definite vlue regrdless of how smll is chosen. Therefore, we cn consider the it f(x)dx,! + + where! + mens tht pproches from bove, i.e. > lwys holds. If this it exists nd is equl to L, then we define f(x)dx = L. If n nti-derivtive of f(x), sy F (x) is known, then the fundmentl theorem of clculus permits us to compute + f(x)dx = F (b) F ( + ). We re thus led to determine the existence (or nonexistence) of the it F ( + ).! + Exmple : Clculte the following integrl for p 6= : Solution: dx (x ) p. We interpret the integrl s the following it:! + + pple dx (x ) p =! + p (b ) p p. Thus, for p>the term p becomes rbitrrily lrge s! + nd hence the re diverges nd the integrl does not exist. Conversely, for p<the term p converges to s! + nd hence the improper integrl exists nd is p (b ). p Finlly, note tht for p =the nti-derivtive is undefined nd the integrl does not exist. Alterntively, for p =we directly see tht + dx b x =ln

9 7.5. Integrtion of n unbounded integrnd 97 5 y e e -e 2 -e e e 2 x 5 ln x 2 4e 2. Z e 2 Figure 7.2. Consider the improper integrl ln x 2 dx. The integrnd e 2 does not exist for x =. Nevertheless, the definite integrl exists nd equls diverges s! +. } Note tht for =the bove exmple recovers the integrnd /x p tht ws discussed in Section In prticulr, we find tht the improper integrl of the second kind Z x p dx exists for p < nd equls /( p) but does not exist for p. Conversely, in Section we observed tht the improper integrl of the first kind Z x p dx exists for p> nd equls /(p ) but does not exist for p pple. Note tht for p = neither of the integrls exists. Exmple 2: Clculte the following integrl: Z e 2 e 2 ln x 2 dx. Solution: A grph of the integrnd is shown in Figure 7.2. First, we note tht the integrnd is not defined t x =. Therefore, we split the integrl into two prts such tht

10 98 Chpter 7. Improper integrls the undefined point mrks once the upper nd once the lower bound nd we write the two integrls s it: Z e 2 e 2 ln x 2 dx =! Z e 2 ln x 2 Z e 2 dx + ln x 2! + dx, where! mens pproches from below, i.e. < lwys holds. Second, the integrnd is n even function nd the integrl runs over symmetric domin. Hence we get Z e 2 2 ln x 2! + dx. Check it! The integrl cn be solved using the substitution u = x 2 followed by n integrtion by prts. This yields 2x ln x2 2! + e 2 =4 2 e x (ln(x) )! + =4e 2 (ln( ) ).! + The fct tht the it exists nd converges to cn be seen by setting =/k nd considering the it k!: (ln( ) ) =! + k! k (ln( ) ) = (ln(k) + ). k k! k Now we cn use either de l Hôpitl s rule or simply recognize tht k grows much fster thn ln k nd hence the it converges to. Thus, we find tht the improper integrl exists nd is 4e L Hôpitl s rule This section introduces powerful method to evlute tricky its of the form x! f(x) g(x). The rule is nmed fter the French mthemticin Guillume de l Hôpitl, who published it in the 7 th century. For our purposes, the rule is often prticulrly useful to evlute the its tht rise in improper integrls of the first (unbounded domin) nd second kind (unbounded integrnd). Consider two functions, f(x) nd g(x), nd suppose tht the following four prerequisites re stisfied: () f(x) nd g(x) re differentible ner x =, but not necessrily t x =. (b) g (x) 6= for x ner but x 6=. (c) x! f (x) g (x) exists. }

11 7.6. L Hôpitl s rule 99 (d) nd either (i) f(x) = g(x) =, or x! x! (ii) f(x) =±, g(x) =±. x! x! Then, l Hôpitl s rule sttes tht f(x) x! g(x) = f (x) x! g (x). (7.4) Note: vlues of the it cn include ±. In loose terms, l Hôpitl s rule cn be used if f(x) x! g(x) is of type or ±. Let us now explore the power of l Hôpitl s rule through severl exmples. Exmple : Clculte the following it: x! sin x e x. Solution: x! (e x In this exmple l Hôpitl s rule cn be used becuse x! sin x =nd ) =. Thus, x! sin x e x = x! (sin x) (e x ) = cos x x! e x cos x x! = (becuse of non-zero denomintor) x! ex = cos =. } Exmple 2: Clculte the following it: t! t ln t t 2 +.

12 2 Chpter 7. Improper integrls Solution: In this exmple, we cn use l Hôpitl s rule becuse both the numertor nd the denomintor diverge: t! (t ln t) = nd t! (t 2 + ) =. Thus, t! t ln t t 2 + = t! (t ln t) (t 2 + ) = t! ln t +. 2t Unfortuntely, the new it tht results from pplying l Hôpitl s rule remins tricky. However, we cn simply pply l Hôpitl s rule gin becuse the prerequisites re still stisfied. In this cse, the numertor nd denomintor still diverge: t! (ln t + ) = nd t! 2t =. Using l Hôpitl s rule gin, we obtin ln t + (ln t + ) = t! 2t t! (2t) = t! t 2 = t 2 t! =. Note: As long s ll prerequisites of l Hôpitl s rule remin stisfied, the rule cn be pplied repetedly. } Exmple 3: Clculte the following it: sin x x!+ x 2. Note: plus-sign (+) dded to the it mens tht x pproches the it from bove (or from the right). In the present cse, x>lwys holds s x pproches zero. Lter in this exmple we will see why this subtle point is importnt. In nlogy, minus-sign ( ) dded to the it mens tht the it is pproched from below (or from the left). If no sign is dded, it does not mtter whether the it is pproched from bove, below or even in n lternting mnner. Solution: Agin, l Hôpitl s rule cn be used becuse x! sin x =nd x! x 2 =. Thus, sin x (sin x) cos x x!+ x 2 = x!+ (x 2 ) = x!+ 2x. Check it! Check it! In order to evlute this new it, we might be tempted to pply l Hôpitl s rule gin. Is this permissible? Stop for moment nd think bout why or why not. Of course, it is not permissible becuse the numertor x!+ cos x = nd hence violtes the prerequisites for pplying l Hôpitl s rule. Ignoring the prerequisites nd blindly pplying l Hôpitl s rule gin would yield n incorrect it of zero. Insted, we get cos x x!+ 2x = cos x x!+ x!+ 2x =+. Hence, the it does not exist, it diverges to +. In order to see why it ws importnt to pproch the it from bove, x! +, clculte the it s x pproches zero from below:

13 7.7. Summry 2 x! sin x x 2. In contrst to the bove result, the it is now pproches zero. becuse x<lwys holds s x } Exmple 4: Clculte the following it: ln.!+ Note: The it of zero cn only be pproched from bove,! +, becuse ln is undefined for <. Solution: At first, this it does not seem to hve the correct form to pply l Hôpitl s rule. However, we cn use net little trick nd rewrite the it s quotient of two functions: ln =!+!+ ln. Once rewritten, it becomes pprent tht l Hôpitl s rule cn indeed be pplied becuse gin both the numertor nd the denomintor diverge:!+ ln = nd!+ / =. Thus, ln (ln )!+ =!+ ( = )!+ 2 =!+ ( ) =. } 7.7 Summry The min points of this chpter cn be summrized s follows:. We reviewed the definition of n improper integrl (type one) over n infinite domin: Z f(x) dx = b! f(x) dx. 2. We computed some exmples of improper integrls nd discussed their convergence or divergence. We reclled (from erlier chpters) tht wheres Z Z e rt dt x dx converges, diverges.

14 22 Chpter 7. Improper integrls 3. More generlly, we showed tht Z dx converges if p>, diverges if p pple. xp 4. We reviewed the definition of improper integrls (type two) for integrnds tht re unbounded t either end of the domin of integrtion, sy x = : f(x) dx = f(x) dx.! + 5. If the integrnd is not defined t one (or more) point(s) in the interior of the domin of integrtion, then the integrl is split into two (or more) prts nd we proceed s bove. 6. In prticulr, we showed tht Z dx converges if p<, diverges if p. xp 7. L Hôpitl s rule is powerful tool to evlute tricky its tht my rise for improper integrls of both kinds. It sttes tht f(x) x! g(x) = f (x) x! g (x) if f() =g() =or x! f(x) = x! g(x) =± s well s some, more subtle prerequisites.

15 7.8. Exercises Exercises Exercise 7. Consider the integrl A = Z D x p dx () Sketch region in the plne whose re represents this if (i) p> nd (ii) p<. (b) Evlute the integrl for p 6=. (c) How does the re A depend on the vlue of D in ech of the cses (i) nd (ii). Does the re increse without bound s D increses? Or does the re pproch some constnt? (d) With this in mind, how might we try to understnd n integrl of the form Exercise 7.2 cse. () (d) Z x p dx Which of the following improper integrls converge? Give reson in ech Z Z dx x. (b) e x dx (e) Z Z xdx e 2x dx (c) (f) Z Z x 3 dx xe x dx Exercise 7.3 The grvittionl force between two objects of mss m nd m 2 is F = Gm m 2 /r 2 where r is the distnce of seprtion. Initilly the objects re distnce D prt. The work done in moving n object from position D to position x ginst force F is defined s W = Z x D F (r) dr. Find the totl work needed to move one of these objects infinitely fr wy. Exercise 7.4 Gbriel s Horn is the surfce of revolution formed by rotting the grph of the function y = f(x) =/x bout the x xis for pple x<. () Find the volume of ir inside this shpe nd show tht it is finite. (b) When we cut cross-section of this horn long the x y-plne, we see flt re which is wedged between the curves y =/x nd y = /x. Show tht this cross-sectionl re is infinite. (c) The surfce re of surfce of revolution generted by revolving the function y = f(x) for pple x pple b bout the x-xis is given by S = 2 f(x) p +(f (x)) 2 dx Write down n integrl tht would represent the surfce re of Gbriel s Horn.

16 24 Chpter 7. Improper integrls (d) The integrl in prt (c) is not esy to evlute explicitly i.e. we cnnot find n ntiderivtive. However, we cn show tht it diverges. Set up comprison tht shows tht the surfce re of Gbriel s horn is infinite. Exercise 7.5 Z sin (x) Does the integrl x 4 + x 2 dx converge or diverge? + Exercise 7.6 Suppose tht n irborne disese is introduced into lrge popultion t time t =. At time t >, the rte t which this disese is spreding is r (t) = 4te 4t new infections per dy. () After how long is this disese most infectious? (b) How mny people in totl cquire the disese? (c) Wht eventully hppens to the rte of new infections? Exercise 7.7 Z 5 x Does the integrl x 2 + x dx converge or diverge? 2 Exercise 7.8 Suppose tht you plce $,, in n ccount tht erns 5% nnul interest, nd you wish to withdrw z dollrs from this ccount one yer from now, 2z dollrs two yers from now, 3z dollrs three yers from now, nd so on. Wht is the mximum vlue of z so tht you never run out of money? Exercise 7.9 For which vlues of p does the integrl Exercise 7. Evlute the following its: () x! xe4x (b) x sin x! x Z 2 x (ln x) p dx converge? (c) cos ( )! 2 2 Exercise 7. Determine the convergence of the following integrls. If they converge, find their vlues: () Z xe x dx (b) Z Z 2x p dx (c) x 2 x 3 dx Exercise 7.2 Convergence of n integrl of the form splitting the integrl up into two prts: Z f (x) dx = Z f (x) dx + Z Z f (x) dx is determined by f (x) dx.

17 7.8. Exercises 25 Z The integrl f (x) dx is sid to converge if both of these two integrls converge. Cn you come up with n exmple of function f (x) for which Z but the integrl f (x) dx diverges? f (x) dx < b! b Exercise 7.3 Exercise 7.4 Evlute the integrl Evlute the integrl Z Z cos p sin 3 d. e x sin xdx.

18 7.9. Solutions Solutions Solution to 7. () See Figure 7.3 (b) A = D p p (c) p>: A pproches constnt s D increses p<: A increses with D. DNE: does not exist. y y (d) p>: A = p p<: A DNE. D x D x Figure 7.3. Solution for problem?? Solution to 7.2 () convergent, p> (b) divergent, p< (c) convergent, p> (d) divergent,! (e) convergent, 2 (f) convergent, Solution to 7.3 W = Gm m 2 D Solution to 7.4 () V = (b) A! Z p p x4 + x4 + (c) S =2 x 3 dx (d) x 3 > for x> ) S! x Solution to 7.5 The integrl converges. Solution to 7.6 () fter 6 hours (b) 25 people (c) the rte tends to

19 336 Chpter 7. Improper integrls Solution to 7.7 It converges: R 5 Solution to 7.8 z = 25 dollrs x x 2 +x 2 dx = R 5 Solution to 7.9 It converges if p> nd diverges otherwise. Solution to 7. x+2 dx = ln (7) ln (2). () (b) (c) Solution to 7. () it converges to (b) it converges to 2 (c) it diverges Solution to 7.2 One possible exmple is f (x) =xe x2 ; ny odd function will do. Solution to 7.3 Solution to 7.4 Z Z cos p d =. sin 3 e x sin xdx= 2.

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph

More information

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

More information

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

More information

5.7 Improper Integrals

5.7 Improper Integrals 458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the

More information

Overview of Calculus I

Overview of Calculus I Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,

More information

The Regulated and Riemann Integrals

The Regulated and Riemann Integrals Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue

More information

Math 116 Calculus II

Math 116 Calculus II Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................

More information

Improper Integrals, and Differential Equations

Improper Integrals, and Differential Equations Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted

More information

The Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.

The Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve. Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F

More information

Review of Calculus, cont d

Review of Calculus, cont d Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

More information

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite

More information

APPROXIMATE INTEGRATION

APPROXIMATE INTEGRATION APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose nti-derivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be

More information

4.4 Areas, Integrals and Antiderivatives

4.4 Areas, Integrals and Antiderivatives . res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order

More information

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

More information

Math 1B, lecture 4: Error bounds for numerical methods

Math 1B, lecture 4: Error bounds for numerical methods Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the

More information

MATH 144: Business Calculus Final Review

MATH 144: Business Calculus Final Review MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives

More information

7.2 The Definite Integral

7.2 The Definite Integral 7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where

More information

Math 8 Winter 2015 Applications of Integration

Math 8 Winter 2015 Applications of Integration Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl

More information

Section 6.1 INTRO to LAPLACE TRANSFORMS

Section 6.1 INTRO to LAPLACE TRANSFORMS Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform

More information

P 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)

P 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0) 1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this

More information

1 The Riemann Integral

1 The Riemann Integral The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re

More information

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl

More information

Definite integral. Mathematics FRDIS MENDELU

Definite integral. Mathematics FRDIS MENDELU Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the

More information

Lecture 1. Functional series. Pointwise and uniform convergence.

Lecture 1. Functional series. Pointwise and uniform convergence. 1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is

More information

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1 The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the

More information

MA 124 January 18, Derivatives are. Integrals are.

MA 124 January 18, Derivatives are. Integrals are. MA 124 Jnury 18, 2018 Prof PB s one-minute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,

More information

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a). The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples

More information

Section 4.8. D v(t j 1 ) t. (4.8.1) j=1

Section 4.8. D v(t j 1 ) t. (4.8.1) j=1 Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions

More information

practice How would you find: e x + e x e 2x e x 1 dx 1 e today: improper integrals

practice How would you find: e x + e x e 2x e x 1 dx 1 e today: improper integrals prctice How would you find: dx e x + e x e 2x e x 1 dx e 2x 1 e x dx 1. Let u=e^x. Then dx=du/u. Ans = rctn ( e^x ) + C 2. Let u=e^x. Becomes u du / (u-1), divide to get u/(u-1)=1+1/(u-1) Ans = e^x + ln

More information

Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30

Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30 Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function

More information

F (x) dx = F (x)+c = u + C = du,

F (x) dx = F (x)+c = u + C = du, 35. The Substitution Rule An indefinite integrl of the derivtive F (x) is the function F (x) itself. Let u = F (x), where u is new vrible defined s differentible function of x. Consider the differentil

More information

INTRODUCTION TO INTEGRATION

INTRODUCTION TO INTEGRATION INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide

More information

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O 1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the x-xis etween nd is denoted y f(x) dx nd clled the

More information

different methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).

different methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s). Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different

More information

Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1

Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1 Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH-115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixed-point itertion to converge when solving the eqution

More information

Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim

Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)

More information

11.1 Exponential Functions

11.1 Exponential Functions . Eponentil Functions In this chpter we wnt to look t specific type of function tht hs mny very useful pplictions, the eponentil function. Definition: Eponentil Function An eponentil function is function

More information

Riemann Sums and Riemann Integrals

Riemann Sums and Riemann Integrals Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties

More information

Section 6.1 Definite Integral

Section 6.1 Definite Integral Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined

More information

SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus

SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is

More information

Chapters 4 & 5 Integrals & Applications

Chapters 4 & 5 Integrals & Applications Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO - Ares Under Functions............................................ 3.2 VIDEO - Applictions

More information

MAT187H1F Lec0101 Burbulla

MAT187H1F Lec0101 Burbulla Chpter 6 Lecture Notes Review nd Two New Sections Sprint 17 Net Distnce nd Totl Distnce Trvelled Suppose s is the position of prticle t time t for t [, b]. Then v dt = s (t) dt = s(b) s(). s(b) s() is

More information

1 Probability Density Functions

1 Probability Density Functions Lis Yn CS 9 Continuous Distributions Lecture Notes #9 July 6, 28 Bsed on chpter by Chris Piech So fr, ll rndom vribles we hve seen hve been discrete. In ll the cses we hve seen in CS 9, this ment tht our

More information

Riemann Sums and Riemann Integrals

Riemann Sums and Riemann Integrals Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct

More information

Math& 152 Section Integration by Parts

Math& 152 Section Integration by Parts Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

More information

Section 7.1 Integration by Substitution

Section 7.1 Integration by Substitution Section 7. Integrtion by Substitution Evlute ech of the following integrls. Keep in mind tht using substitution my not work on some problems. For one of the definite integrls, it is not possible to find

More information

Lecture 14: Quadrature

Lecture 14: Quadrature Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be rel-vlues nd smooth The pproximtion of n integrl by numericl

More information

x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b

x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick

More information

How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?

How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function? Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those

More information

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,

More information

Section 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40

Section 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40 Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since

More information

Suppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = -2.

Suppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = -2. Mth 43 Section 6. Section 6.: Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot

More information

MATH , Calculus 2, Fall 2018

MATH , Calculus 2, Fall 2018 MATH 36-2, 36-3 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly

More information

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral. Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:

More information

Math 360: A primitive integral and elementary functions

Math 360: A primitive integral and elementary functions Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:

More information

a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1

a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1 Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the

More information

Main topics for the First Midterm

Main topics for the First Midterm Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the

More information

f(a+h) f(a) x a h 0. This is the rate at which

f(a+h) f(a) x a h 0. This is the rate at which M408S Concept Inventory smple nswers These questions re open-ended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnk-out-n-nswer problems! (There re plenty of those in the

More information

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1 3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =

More information

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp. MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

More information

Indefinite Integral. Chapter Integration - reverse of differentiation

Indefinite Integral. Chapter Integration - reverse of differentiation Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the

More information

Math 113 Exam 2 Practice

Math 113 Exam 2 Practice Mth 3 Exm Prctice Februry 8, 03 Exm will cover 7.4, 7.5, 7.7, 7.8, 8.-3 nd 8.5. Plese note tht integrtion skills lerned in erlier sections will still be needed for the mteril in 7.5, 7.8 nd chpter 8. This

More information

Week 10: Line Integrals

Week 10: Line Integrals Week 10: Line Integrls Introduction In this finl week we return to prmetrised curves nd consider integrtion long such curves. We lredy sw this in Week 2 when we integrted long curve to find its length.

More information

MATH SS124 Sec 39 Concepts summary with examples

MATH SS124 Sec 39 Concepts summary with examples This note is mde for students in MTH124 Section 39 to review most(not ll) topics I think we covered in this semester, nd there s exmples fter these concepts, go over this note nd try to solve those exmples

More information

We know that if f is a continuous nonnegative function on the interval [a, b], then b

We know that if f is a continuous nonnegative function on the interval [a, b], then b 1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going

More information

Numerical Analysis: Trapezoidal and Simpson s Rule

Numerical Analysis: Trapezoidal and Simpson s Rule nd Simpson s Mthemticl question we re interested in numericlly nswering How to we evlute I = f (x) dx? Clculus tells us tht if F(x) is the ntiderivtive of function f (x) on the intervl [, b], then I =

More information

1 The fundamental theorems of calculus.

1 The fundamental theorems of calculus. The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl- new nme for nti-derivtive. Differentiting integrls. Tody we provide the connection

More information

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019 ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil

More information

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS. THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem

More information

Review of Riemann Integral

Review of Riemann Integral 1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.

More information

Interpreting Integrals and the Fundamental Theorem

Interpreting Integrals and the Fundamental Theorem Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of

More information

Math 116 Final Exam April 26, 2013

Math 116 Final Exam April 26, 2013 Mth 6 Finl Exm April 26, 23 Nme: EXAM SOLUTIONS Instructor: Section:. Do not open this exm until you re told to do so. 2. This exm hs 5 pges including this cover. There re problems. Note tht the problems

More information

Chapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...

Chapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx... Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting

More information

Section 6.1 INTRO to LAPLACE TRANSFORMS

Section 6.1 INTRO to LAPLACE TRANSFORMS Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform

More information

The practical version

The practical version Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht

More information

Best Approximation. Chapter The General Case

Best Approximation. Chapter The General Case Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given

More information

Math 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8

Math 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8 Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite

More information

Math 113 Exam 1-Review

Math 113 Exam 1-Review Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between

More information

W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying

W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)

More information

Integrals - Motivation

Integrals - Motivation Integrls - Motivtion When we looked t function s rte of chnge If f(x) is liner, the nswer is esy slope If f(x) is non-liner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but

More information

7 Improper Integrals, Exp, Log, Arcsin, and the Integral Test for Series

7 Improper Integrals, Exp, Log, Arcsin, and the Integral Test for Series 7 Improper Integrls, Exp, Log, Arcsin, nd the Integrl Test for Series We hve now ttined good level of understnding of integrtion of nice functions f over closed intervls [, b]. In prctice one often wnts

More information

Stuff You Need to Know From Calculus

Stuff You Need to Know From Calculus Stuff You Need to Know From Clculus For the first time in the semester, the stuff we re doing is finlly going to look like clculus (with vector slnt, of course). This mens tht in order to succeed, you

More information

Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004

Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004 Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when

More information

1 The fundamental theorems of calculus.

1 The fundamental theorems of calculus. The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl- new nme for nti-derivtive. Differentiting integrls. Theorem Suppose f is continuous

More information

Review of basic calculus

Review of basic calculus Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below

More information

Lecture 1: Introduction to integration theory and bounded variation

Lecture 1: Introduction to integration theory and bounded variation Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You

More information

Test , 8.2, 8.4 (density only), 8.5 (work only), 9.1, 9.2 and 9.3 related test 1 material and material from prior classes

Test , 8.2, 8.4 (density only), 8.5 (work only), 9.1, 9.2 and 9.3 related test 1 material and material from prior classes Test 2 8., 8.2, 8.4 (density only), 8.5 (work only), 9., 9.2 nd 9.3 relted test mteril nd mteril from prior clsses Locl to Globl Perspectives Anlyze smll pieces to understnd the big picture. Exmples: numericl

More information

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence

More information

2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).

2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ). AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following

More information

AP Calculus Multiple Choice: BC Edition Solutions

AP Calculus Multiple Choice: BC Edition Solutions AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this

More information

Chapter 0. What is the Lebesgue integral about?

Chapter 0. What is the Lebesgue integral about? Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous

More information

Math 113 Exam 2 Practice

Math 113 Exam 2 Practice Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.-7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number

More information

13.4. Integration by Parts. Introduction. Prerequisites. Learning Outcomes

13.4. Integration by Parts. Introduction. Prerequisites. Learning Outcomes Integrtion by Prts 13.4 Introduction Integrtion by Prts is technique for integrting products of functions. In this Section you will lern to recognise when it is pproprite to use the technique nd hve the

More information

and that at t = 0 the object is at position 5. Find the position of the object at t = 2.

and that at t = 0 the object is at position 5. Find the position of the object at t = 2. 7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we

More information

x = b a N. (13-1) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is

x = b a N. (13-1) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is Jnury 28, 2002 13. The Integrl The concept of integrtion, nd the motivtion for developing this concept, were described in the previous chpter. Now we must define the integrl, crefully nd completely. According

More information

MAA 4212 Improper Integrals

MAA 4212 Improper Integrals Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly well-defined, is too restrictive for mny purposes; there re functions which

More information

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows: Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl

More information

Week 10: Riemann integral and its properties

Week 10: Riemann integral and its properties Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the

More information

Section 6: Area, Volume, and Average Value

Section 6: Area, Volume, and Average Value Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find

More information

We divide the interval [a, b] into subintervals of equal length x = b a n

We divide the interval [a, b] into subintervals of equal length x = b a n Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:

More information