Chpter 4 EVALUATING DEFINITE INTEGRALS If the defiite itegrl represets re betwee curve d the x-xis, d if you c fid the re by recogizig the shpe of the regio, the you c evlute the defiite itegrl. Those re big ifs. Most of the time the defiite itegrl will ot be set up to fid re, let loe simple re. This chpter will cosider much more efficiet wy: the Fudmetl Theorem of Clculus. First, however, is exmple illustrtig other techique. This pproch my be doe i preclculus course s fid the re betwee the grph d the xis problem without metioig Riem Sum or defiite itegrl. Evlute the defiite itegrl Ares Without Clculus Riem Sum d fidig tht limit. x 2 + x + ) dx by writig it s the limit of The itervl is [, d the fuctio is fx) =x 2 + x +. The regulrly spced prtitio poits re t x =,, 2,, k,,,. Formig the limit of the righthd Riem Sum gives: x 2 + x + ) ) ) 2 k dx = lim + k ) + Next expd d simplify k is the vrible, is costt): [ k ) 2 ) k ) ) ) [ lim + + ) = lim x 3 k 2 + 2 k + There re simple formuls for these three sums usully proved by mthemticl iductio i preclculus). This results i the limit of the sum of three rtiol expressios whose umertors re of the sme degree s their deomitors. The limits re esily foud. [ lim x 3 k 2 + 2 k + = lim x = 2 6 + 2 + = 6 [ + )2 + ) + ) + + 6 3 2 2 6
So x 2 + x + ) dx = 6. Eve this method is time cosumig, cumbersome, d oly pplicble to polyomils. The ext discussio will led to very importt result. Recogizig the Shpe Some Riem sums my be foud by usig the shpe of the regio, for exmple 3 9 x2 dx = 9 π sice the regio is semicircle with rdius of 3. There re 3 2 very ofte itegrtio questios o the AP Exms tht give studets the grph of fuctio with o equtio) d sk studets to evlute its defiite itegrl. Studets re expected to fid the re of the simple shpes bsed o the grph. See for exmple 999 AB 5 / BC 5 or 23 AB 4/BC 4 mog others.) My of the properties of itegrls see Chpter 6) c be see bsed o the loctio d shpe of the regio betwee the grph d the x-xis d the loctio of d b. So fr we ve cosidered fuctios tht were o-egtive d mootoic over the itervl. Neither of these restrictios is ecessry. If f is egtive over the etire itervl the the Riem sum gives the egtive of the re of the regio betwee the x-xis o top) d the curve. This is becuse i ech term the fuctio vlues re egtive. If f is both positive d egtive o the itervl, the Riem sum gives the lgebric sum of the positive re d the egtive re. The sum my be positive, egtive or zero. Exmple with o kowledge of tiderivtives ecessry): grph. 2π six)dx = by the symmetry of the Itegrtio is doe from left to right. However, if >b, the x = b < d fx)dx = b fx)dx. Eve this ide my be tested without equtios. I 22 AB 4/ BC 4 the grph of y = ft) ws give without equtio d fuctio g ws defied s gx) = x ft)dt. Studets were sked for the vlue of g ) requirig itegrtio from right to left. The umericl vlue ws foud by fidig the opposite of the re of the regio trigle). Fidig the Net Chge Suppose there is cotiuous fuctio tht rus from the poit, f) ) o the left to the poit b, fb) ) o the right. Wht is the et chge i the y-vlues of this fuctio from oe ed to the other? The swer is esy eough: fb) f). Keep this i mid s we cosider other wy to fid the et chge i y. E V A L U A T I N G D E F I N I T E I N T E G R A L S 7
Prtitio the itervl [, b i the usul wy, clculte the chge i y o ech subitervl d sum them up the chges my be positive, egtive, or zero). The et chge is pproximtely y k. Ufortutely, this is ot Riem Sum. But evertheless we would expect its limit s pproches ifiity to be the et chge i y. The et chge i y = lim y k. y = fx) C T y A θ x = dx dy B Figure 4. A close look t the fuctio o oe subitervl. Cosider the prt of the grph betwee y two prtitio poits See Figure 4.). Let AB = x = dx. AT is tget to the grph t A d mkes gle of θ with AB. The ctul chge i y, y = CB. The chge log the tget lie for the sme chge i x) is dy = TB. The s x pproches zero, dy pproches y. By trigoometry, dy = t θ)dx). Sice AT is tget t A, t θ = slope AT = f x) = dy dx, d dy = f x)dx. 2 So lim y k = lim dy = lim f x k ) x Actully it is Riem sum for the fuctio fy) =o the itervl with edpoits of f) d fb). This regio is rectgle with verticl side of fb) f) d horizotl side of. Its vlue, the re of rectgle, is fb) f). But we lredy kew tht. 2 Previously you my hve defied dy = f x)dx d discussed differetils dy d dx). Their itroductio t this poit seems turl. 8 C H A P T E R 4
This lst limit is Riem Sum so the et chge i y is lim f x k ) x = f x)dx Ad sice we lredy kow the et chge f x)dx = fb) f) This very importt result is kow s the Fudmetl Theorem of Clculus 3. It sys tht if we kow the fuctio of which the itegrd is the derivtive clled its tiderivtive or its idefiite itegrl), the the defiite itegrl is the differece of the tiderivtive evluted t the upper limit of itegrtio mius the vlue t the lower limit. Notice lso tht this equtio sys tht the defiite itegrl of the rte of chge is the et mout of chge. I fct, this pproch strted out by fidig the mout of chge of fuctio o itervl, without ssumig it ws the ccumulted rte of chge. As exmple, retur to the problem t the begiig of this chpter. Note tht if fx) = 3 x3 + 2 x2 + x the f x) =x 2 + x +. Tht is, 3 x3 + 2 x2 + x is tiderivtive of x 2 + x +. f) = 3 + 2 += 6 d f) =. Usig the Fudmetl Theorem of Clculus x 2 + x + ) dx = f) f) = 6 = 6 Studets ow hve very good reso to wt to kow how to fid tiderivtives. Oce tiderivtive is kow, evlutig the defiite itegrl ivolves oly evlutig tiderivtive twice d subtrctig. The fct tht there is more th oe tiderivtive of give fuctio turs out ot to be problem. Tht so my of the importt ides d cocepts reltig to itegrtio c be tught d lered without kowig how to fid tiderivtives is sigifict. Begiig clculus studets get bogged dow i the detils of fidig tiderivtives d ofte lose the big picture. This is good reso for doig the big ides first d seprtely from the techiques of tidifferetitio. Of course, they will hve to ler some of these techiques Chpter 6) d will use them with the pplictios Chpter 7). But first, i the ext chpter, we cosider more o the cocept of ccumultio d fuctios defied by itegrls. 3 Most, but ot ll, texts cll this the Secod Fudmetl Theorem of Clculus d the form i Chpter 5, the First. E V A L U A T I N G D E F I N I T E I N T E G R A L S 9
AP Exm Questios reltig to the Fudmetl Theorem of Clculus. The questios tht give the grph of the derivtive d sk questios bout the fuctio icludig itegrls re very commo o AP Clculus exms d studets should be prepred for them.. 995 AB 6; BC 6 2. 997 AB multiple-choice 84; BC multiple-choice 82, 89 3. 998 AB multiple-choice 5, 88; BC multiple-choice 28, 88 4. 999 AB 5 / BC 5 5. 2 AB 4; 6. 2 AB 3 / BC 3 7. 22 AB 2 - BC 2 8 23 Multiple choice: AB 23, 92; BC 8, 27; Free-respose AB 4/ BC 4; Form B AB 5 9. 24 AB 5. 25 AB 3/BC 3 C H A P T E R 4