5.5 The Substitution Rule


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1 5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n ntiderivtive is not esily recognizble, then we re in difficulty. So how, for instnce, do we find n integrl such s sin( ) d? If it cn be done, there re two possible pproches: Guess nd differentite Mke guess, differentitte it to check, nd modify your guess until you get it right! Use rule Reclling differentil clculus, we might try to formulte some helpful rules bsed on the chin nd product rules. The product rule will be resurrected lter (s integrtion by prts). This section, on the substitution rule, eplins how the chin rule my be pplied to integrl clculus. Returning to our emple, suppose tht we re unble to mke sensible guess. The net pproch is to try to subtitute wy the ugliest epression. In this cse, sin( ) is ugly so we define u =. The epression sin u is certinly less dunting thn sin( ). Wht s more, we know how to ntidifferentite sin u. But how do we del with the in the originl integrl? And wht bout the differentil d? These problems re delt with simultneously by differentiting our substitution du d = = d = du We cn now replce ll the pieces in the originl integrl, integrte nd substitute bck t the end: sin( ) d = sin u du = cos u + c = cos( ) + c This process is merely the chin rule written in different wy. Indeed we hve the following Theorem: Theorem (Substitution Rule). If u = g() is differentible function whose rnge is n intervl I, nd f is continuous on I, then f ( g() ) g () d = f (u) du Proof. Let u = g(), nd let F be n ntiderivtive of f on the intervl I, so tht F (u) = f (u). The chin rule sys tht d d F(g()) = F ( g() ) g () = f ( g() ) g () Any epression for derivtive my instntly be rephrsed using indefinite integrls. Thus, s required. f ( g() ) g () d = F(g()) = F(u) = f (u) du
2 You should eplicitly write out your substitutions nd their derivtives until you re comfortble with the method (nd lwys when the clcultion is long!). Obviously, there is no need to use substitution if you re cpble of mking sensible guess. The first emple tht follows is in the mster tble of ntiderivtives so you should be ble to stte the nswer without resorting to substitution.. To compute cos d we let u =. Then du d = = d = du. Therefore cos d = cos u du = sin u + c = sin + c. In the integrl d 4, the ugliest epression is in the squre root. Therefore we let u = 4, from which du = 4 d. It follows tht d 4 = du 4 u = 4 u / du = u/ + c = 4 + c Mking the wrong susbtitution It is very esy to mke n unhelpful substitution. With some prctice, you will quickly recognize when substitution is going wrong, nd try nother! For emple, suppose you hd forgotten tht d = sin + c nd you wnted to compute the integrl. Perhps your first ttempt is s follows: Try the substitution u =. Then = u, nd du d = = d = du = u du. Applying the substituion rule, we obtin d = u u du = u u du This looks even worse thn the intergl we strted with! Insted, different substitution previls: We hope to remove the squre root in the integrnd by using the identity sin θ = cos θ. We therefore try the substitution = sin θ. Then d dθ = cos θ = d = cos θ dθ. Applying the substituion rule, we obtin d = cos θ dθ = sin θ cos θ cos θ dθ = dθ = θ + c = sin + c s epected. Note tht the squreroot gives + cos θ. This is since the rnge of θ = sin is the intervl π < θ < π, on which cos θ >. Substitutions cn use ny letter, not just u!
3 Substitutions in Definite Integrls The substitution rule cn be pplied directly to definite integrls. The importnt point is tht you must chnge the limits! Theorem. If g is continuous on [, b] nd f is continuous on the rnge of u = g(), then b f ( g() ) g () d = g(b) g() f (u) du Emple To evlute 4 + d we substitute u = +. Then du = d, u() =, u(4) = 9 It follows tht d = u du (substitute nd chnge limits) = 9 u/ (find ntiderivtive) = ( 9 / /) = (evlute nd simplify) Notice tht once we substitute nd chnge the limits, we never see gin. An lterntive to chnging the limits is to first compute the indefinite integrl then substitute bck. For emple: Therefore + d = 4 + d = ( + )/ 4 u du = u/ + c = ( + )/ + c = ( 9 / /) = Both of these methods re cceptble. Wht is incorrect is to mi them. In wht follows, the errors re in red d = u du (substitute without chnging limits) = 9 u/ (find ntiderivtive) = ( + 9 )/ (substitute bck) = ( 9 / /) = (evlute nd simplify) Since you obtined the correct solution (), you d likely ssume you did the clcultion correctly. Don t mke this mistke!
4 Odd nd Even Functions When functions hve symmetry, we cn often use the symmetry to compute integrls very quickly. Functions which re odd or even re prticulr strightforwrd, if integrted over symmetric intervl [, ]. Theorem. Suppose tht f is continuous on [, ].. If f is even ( f ( ) = f ()) then. If f is odd ( f ( ) = f ()) then y f () d = f () d = f () d y An Even Function An Odd Function Proof. Let f be odd nd substitute u =. Then du = d, u() = nd u() =. Therefore f () d = = = = f ( u)( du) = f ( u) du f (u) du f () d f ( u)( du) Therefore f () d = = f () d =. The proof for f even is similr: substitute u = for the integrl. ( ) is even, hence ( ) d = (swp limits) (cncel negtive signs) (since f is odd) (since u is dummy vrible) f () d. ( ( ) d = 4 + d = 5 ) + = π π sin( 4 + sin ) d = π/ π/ 4 sin + cos d = π/ cos d = sin π/ = 4
5 Suggested Problems. Evlute the following integrls: () s cos(ln s) ds. (b) cos( ) + ( ) d.. () Evlute the integrl t sin(t ) cos 8 (t ) dt. (b) Use the chnge of vribles u = to evlute the integrl. () Evlute the integrl d + + (strt with u = +... ). (b) Let c be nonzero constnt. For ny integrble function f, prove tht b f (c) d = c bc c f (u) du. d. 5
6 Advnced: misusing the substitution rule Suppose we wnted to evlute ( ) d = 4 du u ( ) d. We substitute u = to get du = (6 ) d: = 4 u = + = There seems to be nothing wrong with this. Now try the sme thing with different limits: ( ) d = 4 du u = u = + 4 = This second clcultion is incorrect: why? One understnd this immeditely by thinking bout the grph of the integrnd ( ). ( ) 5 5 The integrnd is discontinuous (indeed does not eist!) t both = nd =, both of which re in the intervl of integrtion [, ]. We cnnot pply the fundmentl theorem of clculus to evlute the integrl. The re under the curve looks confusingly like, which doesn t mke sense. Indeed ( ) d = DNE. But why then did the subtitution rule trnsform noneistnt integrl into something tht we cn evlute correctly? u is certinly continuous on the intervl [, 4] so everything fter the first equlity is correct. The nswer comes from creful reding of the Substitution Rule Theorem. We wish to compute b f ( g() ) g () d with [, b] = [, ], g() = nd f (u) = u. Given the domin [, ], the rnge of u = g() is the intervl [, 4]. f is not continuous on this intervl, so the hypotheses of the substitution rule re not stisfied. 6
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