1 . 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 to focus on these connections nd their geometric mening, ll of the functions in this section re nonnegtive, but in the next section we will generlize (nd prove the results for ll continuous functions. This section lso introduces exmples showing how you cn use the reltionships between res, integrls nd ntiderivtives in vrious pplictions. When f is continuous, nonnegtive function, the re function A(x = f (t dt represents the re of the region bounded by the grph of f, the t-xis, nd verticl lines t t = nd t = x (see mrgin figure, nd the derivtive of A(x represents the rte of chnge (growth of A(x s the verticl line t = x moves rightwrd. Exmples nd 3 of Section.3 showed tht for certin functions f, A (x = f (x so tht A(x ws n ntiderivtive of f (x. The next theorem sys the result is true for every continuous, nonnegtive function f. The Are Function Is n Antiderivtive If then so f is continuous, nonnegtive function nd A(x = f (t dt for x ( d x f (t dt = A (x = f (x dx A(x is n ntiderivtive of f (x. This result relting integrls nd ntiderivtives is specil cse (for nonnegtive functions f of the first prt of the Fundmentl Theorem of Clculus (FTC, which we will prove in Section.5. This result is importnt for two resons: It sys tht lrge collection of functions hve ntiderivtives. It leds to n esy wy to exctly evlute definite integrls. Exmple. Define A(x = f (t dt for the function f (t shown in the mrgin. Estimte the vlues of A(x nd A (x for x =, 3, nd 5 nd use these vlues to sketch grph of y = A(x. Solution. Dividing the region into squres nd tringles, it is esy to see tht A( =, A(3 =.5, A( = 7 nd A(5 = 8.5. Becuse A (x = f (x, we know tht A ( = f ( =, A (3 = f (3 = 3, A ( = f ( = nd A (5 = f (5 =. A grph of y = A(x ppers in the mrgin t the top of the next pge.
2 33 the integrl It is importnt to recognize tht f is not differentible t x = or x = 3 but tht the vlues of A chnge smoothly ner x = nd x = 3, nd the function A is differentible t those points nd t every other point between x = nd x = 5. Also note tht f ( = ( f is clerly decresing ner x = but tht A ( = f ( = is positive (the re A is growing even though f is getting smller. Prctice. Let B(x be the re bounded by the horizontl xis, verticl lines t t = nd t = x, nd the grph of f (t shown in the mrgin. Estimte the vlues of B(x nd B (x for x =,, 3, nd 5. Exmple. Let G(x = d ( sin(t dt. Evlute G(x for x = π, π nd 3π. dx Solution. The middle mrgin figure shows A(x = sin(t dt grphiclly. By the theorem, A (x = sin(x, so: ( π ( G = A π ( π = sin ( π ( G = A π ( π = sin = ( ( ( 3π G = A 3π 3π = sin =.77 =.77 The penultimte mrgin figure shows grph of y = A(x nd the bottom mrgin figure shows the grph of y = A (x = G(x. Using Antiderivtives to Evlute f (x dx Now we combine the ides of res nd ntiderivtives to devise technique for evluting definite integrls tht is exct nd often esy. If A(x = A(b = f (t dt, then we know tht A( = f (t dt =, f (t dt nd tht A(x is n ntiderivtive of f, so A (x = f (x. We lso know tht if F(x is ny ntiderivtive of f, then F(x nd A(x hve the sme derivtive so F(x nd A(x re prllel functions nd differ by constnt: F(x = A(x + C for ll x nd some constnt C. As consequence: F(b F( = [A(b + C] [A( + C] = A(b A( = f (t dt f (t dt = f (t dt This result sys tht, to evlute definite integrl A(b = f (t dt, we cn find ny ntiderivtive F of f nd simply evlute F(b F(.
3 . res, integrls nd ntiderivtives 335 This result is specil cse of the second prt of the Fundmentl Theorem of Clculus (FTC, stted nd proved in Section.5, which you will use hundreds of times over the next severl chpters. Antiderivtives nd Definite Integrls If then f is continuous, nonnegtive function nd F is ny ntiderivtive of f (so tht F (x = f (x on [, b] f (t dt = F(b F( The problem of finding the exct vlue of definite integrl hs been reduced to finding some (ny ntiderivtive F of the integrnd nd then evluting F(b F(. Even finding one ntiderivtive cn be difficult, so for now we will restrict our ttention to functions tht hve esy ntiderivtives. Lter we will explore some methods for finding ntiderivtives of more difficult functions. Becuse n evlution of the form F(b F( will occur quite often, we represent it symboliclly s F(x b ] b [F(x or Exmple 3. Evlute x dx in two wys: ( by sketching grph of y = x nd finding the re represented by the definite integrl. (b by finding n ntiderivtive F(x of f (x = x nd evluting F(3 F(. Solution. ( A grph of y = x ppers in the mrgin; the re of the trpezoidl region in question hs re. (b One ntiderivtive of x is F(x = ( x x (you should check for yourself tht D = x, so: 3 F(x = F(3 F( = (3 ( = 9 = which grees with the re from prt (. If someone chose nother ntiderivtive of x, sy F(x = x + 7 ( x (you should check for yourself tht D + 7 = x, then: 3 [ ] [ ] F(x = F(3 F( = (3 + 7 ( + 7 = 3 5 = No mtter which ntiderivtive F we choose, F(3 F( =.. Prctice. Evlute previous Exmple. (x dx in the two wys specified in the
4 336 the integrl This ntiderivtive method provides n extremely powerful wy to evlute some definite integrls, nd we will use it often. Exmple. Find the re of the region in the first qudrnt bounded by the grph of y = cos(x, the horizontl xis, nd the line x =. Solution. The re we wnt (see mrgin is π cos(x dx so we need n ntiderivtive of f (x = cos(x. F(x = sin(x is one such ntiderivtive (you should check tht D (sin(x = cos(x, so π cos(x dx = sin(x is the re of the region in question. π ( π = sin sin( = = Prctice 3. Find the re of the region bounded by the grph of y = 3x, the horizontl xis nd the verticl lines x = nd x =. Integrls, Antiderivtives nd Applictions The ntiderivtive method for evluting definite integrls cn lso be used when we need to find more generl re, so it is often useful for solving pplied problems. Exmple 5. A robot hs been progrmmed so tht when it strts to move, its velocity fter t seconds will be 3t feet per second. ( How fr will the robot trvel during its first four seconds of movement? (b How fr will the robot trvel during its next four seconds of movement? (c How long will it tke for the robot to move 79 feet from its strting plce? Solution. ( The distnce during the first four seconds will be the re under the grph of the velocity function (see mrgin figure from t = to t =, n re we cn compute with the definite integrl 3t dt. One ntiderivtive of 3t is t 3 so: 3t dt = [t 3] = 3 3 = 6 nd we cn conclude tht the robot will be 6 feet wy from its strting position fter four seconds. (b Proceeding similrly: 8 3t dt = [t 3] 8 = 83 3 = 5 6 = 8 feet
5 . res, integrls nd ntiderivtives 337 (c This question is different from the first two. Here we know the lower integrtion endpoint, t =, nd the totl distnce, 79 feet, nd need to find the upper integrtion endpoint (the time when the robot is 79 feet wy from its strting position. Clling this upper endpoint T, we know tht: 79 = 3t dt = [t 3] T = T3 3 = T 3 so T = 3 79 = 9. The robot is 79 feet wy fter 9 seconds. Prctice. Refer to the robot from the previous Exmple. ( How fr will the robot trvel between t = nd t = 5 seconds? (b How long will it tke for the robot to move 33 feet from its strting plce? Exmple 6. Suppose tht t minutes fter plcing, bcteri on Petri plte the rte of growth of the bcteri popultion is 6t bcteri per minute. ( How mny new bcteri re dded to the popultion during the first seven minutes? (b Wht is the totl popultion fter seven minutes? (c When will the totl popultion rech, bcteri? Solution. ( The number of new bcteri is represented by the re under the rte-of-growth grph (see mrgin nd one ntiderivtive of 6t is 3t (check tht D ( 3t = 6t so: new bcteri = 7 6t dt = [3t ] 7 = 3(7 3( = 7 (b [old popultion] + [new bcteri] = + 7 = 7 bcteri. (c When the totl popultion reches, bcteri, then there re = new bcteri, hence we need to find the time T required for tht mny new bcteri to grow: = 6t dt = [3t ] T = 3(T 3( = 3T so T = T =. After minutes, the totl bcteri popultion will be + =. Prctice 5. Refer to the bcteri popultion from the previous Exmple. ( How mny new bcteri will be dded to the popultion between t = nd t = 8 minutes? (b When will the totl popultion rech,875 bcteri?
6 338 the integrl. Problems In Problems 8, A(x = ( Grph y = A(x for x 5. f (t dt with f (t given. (b Estimte the vlues of A(, A(, A(3 nd A(. (c Estimte A (, A (, A (3 nd A (. In 9, use the given velocity of cr (in feet per second fter t seconds to find: ( how fr the cr trvels during the first seconds. (b how mny seconds it tkes the cr to trvel hlf the distnce in prt (. 9. v(t = t. v(t = 3t. v(t = t Problems 3 give the velocity of cr (in feet per second fter t seconds. ( How mny seconds does it tke for the cr to come to stop (velocity =? (b How fr does the cr trvel before coming to stop? (c How mny seconds does it tke the cr to trvel hlf the distnce in prt (b?. v(t = t 3. v(t = 75 3t 5. f (t = 6. f (t = + t 7. f (t = 6 t 8. f (t = + t In Problems 9 8, use the Antiderivtives nd Definite Integrls Theorem to evlute ech integrl. 9. (. (. (. ( 3. (. ( 5. ( 6. ( 7. ( 8. ( 5 3 x dx 6x dx x dx 3x dx (b (b (b (b (b (b (b 5 dx (b 3x dx (b 3 x dx 6x dx x dx 3x dx (c (c (c (c (c (c (c 5 dx (c 3x dx [ 3x ] dx (b (c 5 3 x dx 6x dx x dx 3x dx 5 dx 3x dx [ 3x ] dx. Find the exct re under hlf of one rch of the sine curve: π sin(x dx. 5. An rtist you know wnts to mke figure consisting of the region between the curve y = x nd the x-xis for x 3. ( Where should the rtist divide the region with verticl line (see figure below so tht ech piece hs the sme re? (b Where should she divide the region with verticl lines to get three pieces with equl res?
7 . res, integrls nd ntiderivtives 339. Prctice Answers. B( =.5, B( = 5, B(3 = 8.5, B( =, B(5 =.5 B(x = f (t dt B (x = d ( f (t dt = f (x dx (by the Are Function Is n Antiderivtive Theorem, hence: B ( = f ( =, B ( = f ( = 3, B (3 =, B ( = 3 nd B (5 =.. ( (x dx gives the re of the tringulr region between the grph of y = x nd the x-xis for x 3: re = (bse (height = (( = (b F(x = x x is n ntiderivtive of f (x = x so: 3. Are = (x dx = F(3 F( =. ( distnce = 3x dx = x 3 = 3 3 = 8 = 7 5 [ ] [ ] = 3t dt = t 3 5 = 5 = feet. (b We know the strting point is x = nd the totl distnce ( re under the velocity curve is 33 feet. We need to find the time T (see mrgin figure so tht 33 feet = 33 = hence T = 3 33 = 7 seconds. 5. ( new bcteri = 8 3t dt: T 3t dt = t 3 = T 3 = T 3 8 6t dt = 3t = = bcteri. (b We know the totl new popultion ( re under the rte-ofchnge grph is 875 = 875 so: 875 = T 6t dt = 3t = 3T = 3T T = 65 hence T = 65 = 5 minutes.