The Definite Integral

Chpter 4 The Definite Integrl 4. Determining distnce trveled from velocity Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: If we know the velocity of moving body t every point in given intervl, cn we determine the distnce the object hs trveled on the time intervl? How is the problem of finding distnce trveled relted to finding the re under certin curve? Wht does it men to ntidifferentite function nd why is this process relevnt to finding distnce trveled? If velocity is negtive, how does this impct the problem of finding distnce trveled? Introduction In the very first section of the text, we considered sitution where moving object hd known position t time t. In prticulr, we stipulted tht tennis bll tossed into the ir hd its height s (in feet) t time t (in seconds) given by s(t) = 64 6(t ) 2. From this strting point, we investigted the verge velocity of the bll on given intervl [, b], computed by the difference quotient s(b) s() b, nd eventully found tht we could determine the exct instntneous velocity of the bll t time t by tking the derivtive of 207

208 4.. DETERMINING DISTANCE TRAVELED FROM VELOCITY the position function, s s(t + h) s(t) (t) = lim. h 0 h Thus, given differentible position function, we re ble to know the exct velocity of the moving object t ny point in time. Moreover, from this foundtionl problem involving position nd velocity we hve lerned gret del. Given differentible function f, we re now ble to find its derivtive nd use this derivtive to determine the function s instntneous rte of chnge t ny point in the domin, s well s to find where the function is incresing or decresing, is concve up or concve down, nd hs reltive extremes. The vst mjority of the problems nd pplictions we hve considered hve involved the sitution where prticulr function is known nd we seek informtion tht relies on knowing the function s instntneous rte of chnge. Tht is, we hve typiclly proceeded from function f to its derivtive, f, nd then used the mening of the derivtive to help us nswer importnt questions. In much smller number of situtions so fr, we hve encountered the reverse sitution where we insted know the derivtive, f, nd hve tried to deduce informtion bout f. It is this prticulr problem tht will be the focus of our ttention in most of Chpter 4: if we know the instntneous rte of chnge of function, re we ble to determine the function itself? To begin, we strt with more focused question: if we know the instntneous velocity of n object moving long stright line pth, cn we determine its corresponding position function? Preview Activity 4.. Suppose tht person is tking wlk long long stright pth nd wlks t constnt rte of 3 miles per hour. () On the left-hnd xes provided in Figure 4., sketch lbeled grph of the velocity function v(t) = 3. Note tht while the scle on the two sets of xes is the sme, 8 mph 8 miles 4 4 hrs hrs 2 2 Figure 4.: At left, xes for plotting y = v(t); t right, for plotting y = s(t). the units on the right-hnd xes differ from those on the left. The right-hnd xes

4.. DETERMINING DISTANCE TRAVELED FROM VELOCITY 209 will be used in question (d). (b) How fr did the person trvel during the two hours? How is this distnce relted to the re of certin region under the grph of y = v(t)? (c) Find n lgebric formul, s(t), for the position of the person t time t, ssuming tht s(0) = 0. Explin your thinking. (d) On the right-hnd xes provided in Figure 4., sketch lbeled grph of the position function y = s(t). (e) For wht vlues of t is the position function s incresing? Explin why this is the cse using relevnt informtion bout the velocity function v. Are under the grph of the velocity function In Preview Activity 4., we encountered fundmentl fct: when moving object s velocity is constnt (nd positive), the re under the velocity curve over given intervl tells us the distnce the object trveled. As seen t left in Figure 4.2, if we consider n object 3 mph 3 mph y = v(t) v(t) = 2 A A 2 hrs hrs 2 3 2 3 Figure 4.2: At left, constnt velocity function; t right, non-constnt velocity function. moving t 2 miles per hour over the time intervl [,.5], then the re A of the shded region under y = v(t) on [,.5] is A = 2 miles hour hours = mile. 2 This principle holds in generl simply due to the fct tht distnce equls rte times time, provided the rte is constnt. Thus, if v(t) is constnt on the intervl [, b], then the

20 4.. DETERMINING DISTANCE TRAVELED FROM VELOCITY distnce trveled on [, b] is the re A tht is given by A = v()(b ) = v() t, where t is the chnge in t over the intervl. Note, too, tht we could use ny vlue of v(t) on the intervl [, b], since the velocity is constnt; we simply chose v(), the vlue t the intervl s left endpoint. For severl exmples where the velocity function is piecewise constnt, see http://gvsu.edu/s/9t. The sitution is obviously more complicted when the velocity function is not constnt. At the sme time, on reltively smll intervls on which v(t) does not vry much, the re principle llows us to estimte the distnce the moving object trvels on tht time intervl. For instnce, for the non-constnt velocity function shown t right in Figure 4.2, we see tht on the intervl [,.5], velocity vries from v() = 2.5 down to v(.5) 2.. Hence, one estimte for distnce trveled is the re of the pictured rectngle, A 2 = v() t = 2.5 miles hour hours =.25 miles. 2 Becuse v is decresing on [,.5] nd the rectngle lies bove the curve, clerly A 2 =.25 is n over-estimte of the ctul distnce trveled. If we wnt to estimte the re under the non-constnt velocity function on wider intervl, sy [0, 3], it becomes pprent tht one rectngle probbly will not give good pproximtion. Insted, we could use the six rectngles pictured in Figure 4.3, find the 3 mph y = v(t) 2 3 hrs Figure 4.3: Using six rectngles to estimte the re under y = v(t) on [0, 3]. re of ech rectngle, nd dd up the totl. Obviously there re choices to mke nd issues to understnd: how mny rectngles should we use? where should we evlute the function to decide the rectngle s height? wht hppens if velocity is sometimes negtive? Mrc Renult, clculus pplets.

4.. DETERMINING DISTANCE TRAVELED FROM VELOCITY 2 cn we ttin the exct re under ny non-constnt curve? These questions nd more re ones we will study in wht follows; for now it suffices to relize tht the simple ide of the re of rectngle gives us powerful tool for estimting both distnce trveled from velocity function s well s the re under n rbitrry curve. To explore the setting of multiple rectngles to pproximte re under non-constnt velocity function, see the pplet found t http://gvsu.edu/s/9u. 2 Activity 4.. Suppose tht person is wlking in such wy tht her velocity vries slightly ccording to the informtion given in the tble below nd grph given in Figure 4.4. t 0.00 0.25 0.50 0.75.00.25.50.75 2.00 v(t).500.789.938.992 2.000 2.008 2.063 2.2 2.500 3 mph y = v(t) 2 2 hrs Figure 4.4: The grph of y = v(t). () Using the grid, grph, nd given dt ppropritely, estimte the distnce trveled by the wlker during the two hour intervl from t = 0 to t = 2. You should use time intervls of width t = 0.5, choosing wy to use the function consistently to determine the height of ech rectngle in order to pproximte distnce trveled. (b) How could you get better pproximtion of the distnce trveled on [0, 2]? Explin, nd then find this new estimte. (c) Now suppose tht you know tht v is given by v(t) = 0.5t 3.5t 2 +.5t +.5. Remember tht v is the derivtive of the wlker s position function, s. Find formul for s so tht s = v. (d) Bsed on your work in (c), wht is the vlue of s(2) s(0)? Wht is the mening of this quntity? 2 Mrc Renult, clculus pplets.

22 4.. DETERMINING DISTANCE TRAVELED FROM VELOCITY Two pproches: re nd ntidifferentition When the velocity of moving object is positive, the object s position is lwys incresing. While we will soon consider situtions where velocity is negtive nd think bout the rmifictions of this condition on distnce trveled, for now we continue to ssume tht we re working with positive velocity function. In tht setting, we hve estblished tht whenever v is ctully constnt, the exct distnce trveled on n intervl is the re under the velocity curve; furthermore, we hve observed tht when v is not constnt, we cn estimte the totl distnce trveled by finding the res of rectngles tht help to pproximte the re under the velocity curve on the given intervl. Hence, we see the importnce of the problem of finding the re between curve nd the horizontl xis: besides being n interesting geometric question, in the setting of the curve being the (positive) velocity of moving object, the re under the curve over n intervl tells us the exct distnce trveled on the intervl. We cn estimte this re ny time we hve grph of the velocity function or tble of dt tht tells us some relevnt vlues of the function. In Activity 4., we lso encountered n lternte pproch to finding the distnce trveled. In prticulr, if we know formul for the instntneous velocity, y = v(t), of the moving body t time t, then we relize tht v must be the derivtive of some corresponding position function s. If we cn find formul for s from the formul for v, it follows tht we know the position of the object t time t. In ddition, under the ssumption tht velocity is positive, the chnge in position over given intervl then tells us the distnce trveled on tht intervl. For simple exmple, consider the sitution from Preview Activity 4., where person is wlking long stright line nd hs velocity function v(t) = 3 mph. As pictured in 8 mph 8 miles s(t) = 3t 4 v(t) = 3 4 A = 3.25 = 3.75 2 hrs s(.5) = 4.5 s(0.25) = 0.75 hrs 2 Figure 4.5: The velocity function v(t) = 3 nd corresponding position function s(t) = 3t. Figure 4.5, we see the lredy noted reltionship between re nd distnce trveled on the left-hnd grph of the velocity function. In ddition, becuse the velocity is constnt

4.. DETERMINING DISTANCE TRAVELED FROM VELOCITY 23 t 3, we know tht if 3 s(t) = 3t, then s (t) = 3, so s(t) = 3t is function whose derivtive is v(t). Furthermore, we now observe tht s(.5) = 4.5 nd s(0.25) = 0.75, which re the respective loctions of the person t times t = 0.25 nd t =.5, nd therefore s(.5) s(0.25) = 4.5 0.75 = 3.75 miles. This is not only the chnge in position on [0.25,.5], but lso precisely the distnce trveled on [0.25,.5], which cn lso be computed by finding the re under the velocity curve over the sme intervl. There re profound ides nd connections present in this exmple tht we will spend much of the reminder of Chpter 4 studying nd exploring. For now, it is most importnt to observe tht if we re given formul for velocity function v, it cn be very helpful to find function s tht stisfies s = v. In this context, we sy tht s is n ntiderivtive of v. More generlly, just s we sy tht f is the derivtive of f for given function f, if we re given function g nd G is function such tht G = g, we sy tht G is n ntiderivtive of g. For exmple, if g(x) = 3x 2 + 2x, n ntiderivtive of g is G(x) = x 3 + x 2, since G (x) = g(x). Note tht we sy n ntiderivtive of g rther thn the ntiderivtive of g becuse H(x) = x 3 + x 2 + 5 is lso function whose derivtive is g, nd thus H is nother ntiderivtive of g. Activity 4.2. A bll is tossed verticlly in such wy tht its velocity function is given by v(t) = 32 32t, where t is mesured in seconds nd v in feet per second. Assume tht this function is vlid for 0 t 2. () For wht vlues of t is the velocity of the bll positive? Wht does this tell you bout the motion of the bll on this intervl of time vlues? (b) Find n ntiderivtive, s, of v tht stisfies s(0) = 0. (c) Compute the vlue of s() s( 2 ). Wht is the mening of the vlue you find? (d) Using the grph of y = v(t) provided in Figure 4.6, find the exct re of the region under the velocity curve between t = 2 nd t =. Wht is the mening of the vlue you find? (e) Answer the sme questions s in (c) nd (d) but insted using the intervl [0, ]. (f) Wht is the vlue of s(2) s(0)? Wht does this result tell you bout the flight of the bll? How is this vlue connected to the provided grph of y = v(t)? Explin. 3 Here we re mking the implicit ssumption tht s(0) = 0; we will further discuss the different possibilities for vlues of s(0) in subsequent study.

24 4.. DETERMINING DISTANCE TRAVELED FROM VELOCITY ft/sec 24 2-2 v(t) = 32 32t 2 sec -24 Figure 4.6: The grph of y = v(t). When velocity is negtive Most of our work in this section hs occurred under the ssumption tht velocity is positive. This hypothesis gurntees tht the movement of the object under considertion is lwys in single direction, nd hence ensures tht the moving body s chnge in position is the sme s the distnce it trvels on given intervl. As we sw in Activity 4.2, there re nturl settings in which moving object s velocity is negtive; we would like to understnd this scenrio fully s well. Consider simple exmple where person goes for wlk on bech long stretch of very stright shoreline tht runs est-west. We cn nturlly ssume tht their initil position is s(0) = 0, nd further stipulte tht their position function increses s they move est from their strting loction. For instnce, position of s = mile represents being one mile est of the strt loction, while s = tells us the person is one mile west of where they begn wlking on the bech. Now suppose the person wlks in the following mnner. From the outset t t = 0, the person wlks due est t constnt rte of 3 mph for.5 hours. After.5 hours, the person stops bruptly nd begins wlking due west t the constnt rte of 4 mph nd does so for 0.5 hours. Then, fter nother brupt stop nd strt, the person resumes wlking t constnt rte of 3 mph to the est for one more hour. Wht is the totl distnce the person trveled on the time intervl t = 0 to t = 3? Wht is the person s totl chnge in position over tht time? On one hnd, these re elementry questions to nswer becuse the velocity involved is constnt on ech intervl. From t = 0 to t =.5, the person trveled D [0,.5] = 3 miles per hour.5 hours = 4.5 miles. Similrly, on t =.5 to t = 2, hving different rte, the distnce trveled is D [.5,2] = 4 miles per hour 0.5 hours = 2 miles.

4.. DETERMINING DISTANCE TRAVELED FROM VELOCITY 25 Finlly, similr clcultions revel tht in the finl hour, the person wlked so the totl distnce trveled is D [2,3] = 3 miles per hour hours = 3 miles, D = D [0,.5] + D [.5,2] + D [2,3] = 4.5 + 2 + 3 = 9.5 miles. Since the velocity on.5 < t < 2 is ctully v = 4, being negtive to indicte motion in the westwrd direction, this tells us tht the person first wlked 4.5 miles est, then 2 miles west, followed by 3 more miles est. Thus, the wlker s totl chnge in position is chnge in position = 4.5 2 + 3 = 5.5 miles. While we hve been ble to nswer these questions firly esily, it is lso importnt to think bout this problem grphiclly in order tht we cn generlize our solution to the more complicted setting when velocity is not constnt, s well s to note the prticulr impct tht negtive velocity hs. In Figure 4.7, we see how the distnces we computed 4.5 3.0.5 -.5-3.0-4.5 mph y = v(t) A = 4.5 A 3 = 3 hrs 3 A 2 = 2 (3,5.5) 4.5 miles (.5,4.5) 3.0 y = s(t).5 (2,2.5) hrs 3 -.5-3.0-4.5 Figure 4.7: At left, the velocity function of the person wlking; t right, the corresponding position function. bove cn be viewed s res: A = 4.5 comes from tking rte times time (3.5), s do A 2 nd A 3 for the second nd third rectngles. The big new issue is tht while A 2 is n re (nd is therefore positive), becuse this re involves n intervl on which the velocity function is negtive, its re hs negtive sign ssocited with it. This helps us to distinguish between distnce trveled nd chnge in position. The distnce trveled is the sum of the res, D = A + A 2 + A 3 = 4.5 + 2 + 3 = 9.5 miles.

26 4.. DETERMINING DISTANCE TRAVELED FROM VELOCITY But the chnge in position hs to ccount for the sign ssocited with the re, where those bove the t-xis re considered positive while those below the t-xis re viewed s negtive, so tht s(3) s(0) = (+4.5) + ( 2) + (+3) = 5.5 miles, ssigning the 2 to the re in the intervl [.5, 2] becuse there velocity is negtive nd the person is wlking in the negtive direction. In other words, the person wlks 4.5 miles in the positive direction, followed by two miles in the negtive direction, nd then 3 more miles in the positive direction. This ffect of velocity being negtive is lso seen in the grph of the function y = s(t), which hs negtive slope (specificlly, its slope is 4) on the intervl.5 < t < 2 since the velocity is 4 on tht intervl, which shows the person s position function is decresing due to the fct tht she is wlking est, rther thn west. On the intervls where she is wlking west, the velocity function is positive nd the slope of the position function s is therefore lso positive. To summrize, we see tht if velocity is sometimes negtive, this mkes the moving object s chnge in position different from its distnce trveled. By viewing the intervls on which velocity is positive nd negtive seprtely, we my compute the distnce trveled on ech such intervl, nd then depending on whether we desire totl distnce trveled or totl chnge in position, we my ccount for negtive velocities tht ccount for negtive chnge in position, while still contributing positively to totl distnce trveled. We close this section with one dditionl ctivity tht further explores the effects of negtive velocity on the problem of finding chnge in position nd totl distnce trveled. Activity 4.3. Suppose tht n object moving long stright line pth hs its velocity v (in meters per second) t time t (in seconds) given by the piecewise liner function whose grph is pictured in Figure 4.8. We view movement to the right s being in the positive direction (with positive velocity), while movement to the left is in the negtive direction. Suppose 4 m/sec 2 y = v(t) 8 4 2 4 6 8 sec 2 4 6 8-2 -4-4 -8 Figure 4.8: The velocity function of moving object.

4.. DETERMINING DISTANCE TRAVELED FROM VELOCITY 27 further tht the object s initil position t time t = 0 is s(0) =. () Determine the totl distnce trveled nd the totl chnge in position on the time intervl 0 t 2. Wht is the object s position t t = 2? (b) On wht time intervls is the moving object s position function incresing? Why? On wht intervls is the object s position decresing? Why? (c) Wht is the object s position t t = 8? How mny totl meters hs it trveled to get to this point (including distnce in both directions)? Is this different from the object s totl chnge in position on t = 0 to t = 8? (d) Find the exct position of the object t t =, 2, 3,..., 8 nd use this dt to sketch n ccurte grph of y = s(t) on the xes provided t right. How cn you use the provided informtion bout y = v(t) to determine the concvity of s on ech relevnt intervl? Summry In this section, we encountered the following importnt ides: If we know the velocity of moving body t every point in given intervl nd the velocity is positive throughout, we cn estimte the object s distnce trveled nd in some circumstnces determine this vlue exctly. In prticulr, when velocity is positive on n intervl, we cn find the totl distnce trveled by finding the re under the velocity curve nd bove the t-xis on the given time intervl. We my only be ble to estimte this re, depending on the shpe of the velocity curve. An ntiderivtive of function f is new function F whose derivtive is f. Tht is, F is n ntiderivtive of f provided tht F = f. In the context of velocity nd position, if we know velocity function v, n ntiderivtive of v is position function s tht stisfies s = v. If v is positive on given intervl, sy [, b], then the chnge in position, s(b) s(), mesures the distnce the moving object trveled on [, b]. In the setting where velocity is sometimes negtive, this mens tht the object is sometimes trveling in the opposite direction (depending on whether velocity is positive or negtive), nd thus involves the object bcktrcking. To determine distnce trveled, we hve to think bout the problem seprtely on intervls where velocity is positive nd negtive nd ccount for the chnge in position on ech such intervl.

28 4.. DETERMINING DISTANCE TRAVELED FROM VELOCITY Exercises. Along the estern shore of Lke Michign from Lke Mctw (ner Hollnd) to Grnd Hven, there is bike bth tht runs lmost directly north-south. For the purposes of this problem, ssume the rod is completely stright, nd tht the function s(t) trcks the position of the biker long this pth in miles north of Pigeon Lke, which lies roughly hlfwy between the ends of the bike pth. Suppose tht the biker s velocity function is given by the grph in Figure 4.9 on the time intervl 0 t 4 (where t is mesured in hours), nd tht s(0) =. 0 mph y = v(t) 0 miles 6 6 2-2 2 3 4 5 hrs 2-2 2 3 4 5 hrs -6-6 -0-0 Figure 4.9: The grph of the biker s velocity, y = v(t), t left. At right, xes to plot n pproximte sketch of y = s(t). () Approximtely how fr north of Pigeon Lke ws the cyclist when she ws the gretest distnce wy from Pigeon Lke? At wht time did this occur? (b) Wht is the cyclist s totl chnge in position on the time intervl 0 t 2? At t = 2, ws she north or south of Pigeon Lke? (c) Wht is the totl distnce the biker trveled on 0 t 4? At the end of the ride, how close ws she to the point t which she strted? (d) Sketch n pproximte grph of y = s(t), the position function of the cyclist, on the intervl 0 t 4. Lbel t lest four importnt points on the grph of s. 2. A toy rocket is lunched verticlly from the ground on dy with no wind. The rocket s verticl velocity t time t (in seconds) is given by v(t) = 500 32t feet/sec. () At wht time fter the rocket is lunched does the rocket s velocity equl zero? Cll this time vlue. Wht hppens to the rocket t t =? (b) Find the vlue of the totl re enclosed by y = v(t) nd the t-xis on the intervl 0 t. Wht does this re represent in terms of the physicl

4.. DETERMINING DISTANCE TRAVELED FROM VELOCITY 29 setting of the problem? (c) Find n ntiderivtive s of the function v. Tht is, find function s such tht s (t) = v(t). (d) Compute the vlue of s() s(0). Wht does this number represent in terms of the physicl setting of the problem? (e) Compute s(5) s(). Wht does this number tell you bout the rocket s flight? 3. An object moving long horizontl xis hs its instntneous velocity t time t in seconds given by the function v pictured in Figure 4.0, where v is mesured in feet/sec. Assume tht the curves tht mke up the prts of the grph of y = v(t) re either y = v(t) 2 3 4 5 6 7 - Figure 4.0: The grph of y = v(t), the velocity function of moving object. portions of stright lines or portions of circles. () Determine the exct totl distnce the object trveled on 0 t 2. (b) Wht is the vlue nd mening of s(5) s(2), where y = s(t) is the position function of the moving object? (c) On which time intervl did the object trvel the gretest distnce: [0, 2], [2, 4], or [5, 7]? (d) On which time intervl(s) is the position function s incresing? At which point(s) does s chieve reltive mximum? 4. Filters t wter tretment plnt become dirtier over time nd thus become less effective; they re replced every 30 dys. During one 30-dy period, the rte t which pollution psses through the filters into nerby lke (in units of prticulte mtter per dy) is mesured every 6 dys nd is given in the following tble. The time t is mesured in dys since the filters were replced. Dy, t 0 6 2 8 24 30 Rte of pollution in units per dy, p(t) 7 8 0 3 8 35

220 4.. DETERMINING DISTANCE TRAVELED FROM VELOCITY () Plot the given dt on set of xes with time on the horizontl xis nd the rte of pollution on the verticl xis. (b) Explin why the mount of pollution tht entered the lke during this 30-dy period would be given exctly by the re bounded by y = p(t) nd the t-xis on the time intervl [0, 30]. (c) Estimte the totl mount of pollution entering the lke during this 30-dy period. Crefully explin how you determined your estimte.

4.2. RIEMANN SUMS 22 4.2 Riemnn Sums Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: How cn we use Riemnn sum to estimte the re between given curve nd the horizontl xis over prticulr intervl? Wht re the differences mong left, right, middle, nd rndom Riemnn sums? How cn we write Riemnn sums in n bbrevited form?? Introduction In Section 4., we lerned tht if we hve moving object with velocity function v, whenever v(t) is positive, the re between y = v(t) nd the t-xis over given time intervl tells us the distnce trveled by the object over tht time period; in ddition, if v(t) is sometimes negtive nd we view the re of ny region below the t-xis s hving n ssocited negtive sign, then the sum of these signed res over given intervl tells us the moving object s chnge in position over the time intervl. For instnce, for the velocity function y = v(t) A A 3 A 2 b Figure 4.: A velocity function tht is sometimes negtive. given in Figure 4., if the res of shded regions re A, A 2, nd A 3 s lbeled, then the totl distnce D trveled by the moving object on [, b] is D = A + A 2 + A 3,

222 4.2. RIEMANN SUMS while the totl chnge in the object s position on [, b] is s(b) s() = A A 2 + A 3. Becuse the motion is in the negtive direction on the intervl where v(t) < 0, we subtrct A 2 when determining the object s totl chnge in position. Of course, finding D nd s(b) s() for the sitution given in Figure 4. presumes tht we cn ctully find the res represented by A, A 2, nd A 3. In most of our work in Section 4., such s in Activities 4.2 nd 4.3, we worked with velocity functions tht were either constnt or liner, so tht by finding the res of rectngles nd tringles, we could find the re bounded by the velocity function nd the horizontl xis exctly. But when the curve tht bounds region is not one for which we hve known formul for re, we re unble to find this re exctly. Indeed, this is one of our biggest gols in Chpter 4: to lern how to find the exct re bounded between curve nd the horizontl xis for s mny different types of functions s possible. To begin, we expnd on the ides in Activity 4., where we encountered nonliner velocity function nd pproximted the re under the curve using four nd eight rectngles, respectively. In the following preview ctivity, we focus on three different options for deciding how to find the heights of the rectngles we will use. Preview Activity 4.2. A person wlking long stright pth hs her velocity in miles per hour t time t given by the function v(t) = 0.25t 3.5t 2 + 3t + 0.25, for times in the intervl 0 t 2. The grph of this function is lso given in ech of the three digrms in Figure 4.2. Note tht in ech digrm, we use four rectngles to estimte the re under 3 mph y = v(t) 3 mph y = v(t) 3 mph y = v(t) 2 2 2 A 2 A 3 A 4 hrs B 4 B 3 B 2 B hrs C 4 C 3 C 2 C hrs A 2 2 2 Figure 4.2: Three pproches to estimting the re under y = v(t) on the intervl [0, 2]. y = v(t) on the intervl [0, 2], but the method by which the four rectngles respective heights re decided vries mong the three individul grphs.

4.2. RIEMANN SUMS 223 () How re the heights of rectngles in the left-most digrm being chosen? Explin, nd hence determine the vlue of S = A + A 2 + A 3 + A 4 by evluting the function y = v(t) t ppropritely chosen vlues nd observing the width of ech rectngle. Note, for exmple, tht A 3 = v() 2 = 2 2 =. (b) Explin how the heights of rectngles re being chosen in the middle digrm nd find the vlue of T = B + B 2 + B 3 + B 4. (c) Likewise, determine the pttern of how heights of rectngles re chosen in the right-most digrm nd determine U = C + C 2 + C 3 + C 4. (d) Of the estimtes S, T, nd U, which do you think is the best pproximtion of D, the totl distnce the person trveled on [0, 2]? Why? Sigm Nottion It is pprent from severl different problems we hve considered tht sums of res of rectngles is one of the min wys to pproximte the re under curve over given intervl. Intuitively, we expect tht using lrger number of thinner rectngles will provide wy to improve the estimtes we re computing. As such, we nticipte deling with sums with lrge number of terms. To do so, we introduce the use of so-clled sigm nottion, nmed for the Greek letter Σ, which is the cpitl letter S in the Greek lphbet. For exmple, sy we re interested in the sum + 2 + 3 + + 00, which is the sum of the first 00 nturl numbers. Sigm nottion provides shorthnd nottion tht recognizes the generl pttern in the terms of the sum. It is equivlent to write 00 k = + 2 + 3 + + 00. k=

224 4.2. RIEMANN SUMS 00 We red the symbol k s the sum from k equls to 00 of k. The vrible k is k= usully clled the index of summtion, nd the letter tht is used for this vrible is immteril. Ech sum in sigm nottion involves function of the index; for exmple, 0 k= nd more generlly, (k 2 + 2k) = ( 2 + 2 ) + (2 2 + 2 2) + (3 2 + 2 3) + + (0 2 + 2 0), n f (k) = f () + f (2) + + f (n). k= Sigm nottion llows us the flexibility to esily vry the function being used to trck the pttern in the sum, s well s to djust the number of terms in the sum simply by chnging the vlue of n. We test our understnding of this new nottion in the following ctivity. Activity 4.4. For ech sum written in sigm nottion, write the sum long-hnd nd evlute the sum to find its vlue. For ech sum written in expnded form, write the sum in sigm nottion. 5 () (k 2 + 2) (b) k= 6 (2i ) i=3 (c) 3 + 7 + + 5 + + 27 (d) 4 + 8 + 6 + 32 + + 256 (e) 6 i= 2 i Riemnn Sums When moving body hs positive velocity function y = v(t) on given intervl [, b], we know tht the re under the curve over the intervl is the totl distnce the body trvels on [, b]. While this is the fundmentl motivting force behind our interest in the re bounded by function, we re lso interested more generlly in being ble to find the exct re bounded by y = f (x) on n intervl [, b], regrdless of the mening or context of the function f. For now, we continue to focus on determining n ccurte estimte of this re through the use of sum of the res of rectngles, doing so in the

4.2. RIEMANN SUMS 225 setting where f (x) 0 on [, b]. Throughout, unless otherwise indicted, we lso ssume tht f is continuous on [, b]. The first choice we mke in ny such pproximtion is the number of rectngles. If we b x 0 x x 2 x i x i+ x x n x n Figure 4.3: Subdividing the intervl [, b] into n subintervls of equl length x. sy tht the totl number of rectngles is n, nd we desire n rectngles of equl width to subdivide the intervl [, b], then ech rectngle must hve width x = b n. We observe further tht x = x 0 + x, x 2 = x 0 + 2 x, nd thus in generl x i = + i x, s pictured in Figure 4.3. We use ech subintervl [x i, x i+ ] s the bse of rectngle, nd next must choose how to decide the height of the rectngle tht will be used to pproximte the re under y = f (x) on the subintervl. There re three stndrd choices: use the left endpoint of ech subintervl, the right endpoint of ech subintervl, or the midpoint of ech. These re precisely the options encountered in Preview Activity 4.2 nd seen in Figure 4.2. We next explore how these choices cn be reflected in sigm nottion. If we now consider n rbitrry positive function f on [, b] with the intervl subdivided s shown in Figure 4.3, nd choose to use left endpoints, then on ech intervl of the form [x i, x i+ ], the re of the rectngle formed is given by A i+ = f (x i ) x, s seen in Figure 4.4. If we let L n denote the sum of the res of rectngles whose heights re given by the function vlue t ech respective left endpoint, then we see tht L n = A + A 2 + + A i+ + + A n = f (x 0 ) x + f (x ) x + + f (x i ) x + + f (x n ) x. In the more compct sigm nottion, we hve L n = n i=0 f (x i ) x. Note prticulrly tht since the index of summtion begins t 0 nd ends t n, there re indeed n terms in this sum. We cll L n the left Riemnn sum for the function f on the intervl [, b].

226 4.2. RIEMANN SUMS y = f (x) A A 2 A i+ A n x 0 x x 2 x i x i+ x n x n Figure 4.4: Subdividing the intervl [, b] into n subintervls of equl length x nd pproximting the re under y = f (x) over [, b] using left rectngles. There re now two fundmentl issues to explore: the number of rectngles we choose to use nd the selection of the pttern by which we identify the height of ech rectngle. It is best to explore these choices dynmiclly, nd the pplet 4 found t http://gvsu.edu/s/9 is prticulrly useful one. There we see the imge shown in Figure 4.5: A snpshot of the pplet found t http://gvsu.edu/s/9. Figure 4.5, but with the opportunity to djust the slider brs for the left endpoint nd the number of subintervls. By moving the sliders, we cn see how the heights of the rectngles chnge s we consider left endpoints, midpoints, nd right endpoints, s well s the impct tht lrger number of nrrower rectngles hs on the pproximtion of the exct re bounded by the function nd the horizontl xis. To see how the Riemnn sums for right endpoints nd midpoints re constructed, 4 Mrc Renult, Geogebr Clculus Applets.

4.2. RIEMANN SUMS 227 we consider Figure 4.6. For the sum with right endpoints, we see tht the re of the y = f (x) y = f (x) B B 2 B i+ B n C C 2 C i+ C n x 0 x x 2 x i x i+ x n x n x 0 x x 2 x i x i+ x n x n Figure 4.6: Riemnn sums using right endpoints nd midpoints. rectngle on n rbitrry intervl [x i, x i+ ] is given by B i+ = f (x i+ ) x, so tht the sum of ll such res of rectngles is given by R n = B + B 2 + + B i+ + + B n = f (x ) x + f (x 2 ) x + + f (x i+ ) x + + f (x n ) x n = f (x i ) x. i= We cll R n the right Riemnn sum for the function f on the intervl [, b]. For the sum tht uses midpoints, we introduce the nottion x i+ = x i + x i+ 2 so tht x i+ is the midpoint of the intervl [x i, x i+ ]. For instnce, for the rectngle with re C in Figure 4.6, we now hve C = f (x ) x. Hence, the sum of ll the res of rectngles tht use midpoints is M n = C + C 2 + + C i+ + + C n = f (x ) x + f (x 2 ) x + + f (x i+ ) x + + f (x n ) x n = f (x i ) x, i= nd we sy tht M n is the middle Riemnn sum for f on [, b]. When f (x) 0 on [, b], ech of the Riemnn sums L n, R n, nd M n provides n

228 4.2. RIEMANN SUMS estimte of the re under the curve y = f (x) over the intervl [, b]; momentrily, we will discuss the mening of Riemnn sums in the setting when f is sometimes negtive. We lso recll tht in the context of nonnegtive velocity function y = v(t), the corresponding Riemnn sums re pproximting the distnce trveled on [, b] by the moving object with velocity function v. There is more generl wy to think of Riemnn sums, nd tht is to not restrict the choice of where the function is evluted to determine the respective rectngle heights. Tht is, rther thn sying we ll lwys choose left endpoints, or lwys choose midpoints, we simply sy tht point x i+ will be selected t rndom in the intervl [x i, x i+ ] (so tht x i x i+ x i+), which mkes the Riemnn sum given by f (x ) x + f (x 2 ) x + + f (x i+ ) x + + f (x n) x = n f (x i ) x. i= At http://gvsu.edu/s/9, the pplet noted erlier nd referenced in Figure 4.5, by unchecking the reltive box t the top left, nd insted checking rndom, we cn esily explore the effect of using rndom point loctions in subintervls on given Riemnn sum. In computtionl prctice, we most often use L n, R n, or M n, while the rndom Riemnn sum is useful in theoreticl discussions. In the following ctivity, we investigte severl different Riemnn sums for prticulr velocity function. Activity 4.5. Suppose tht n object moving long stright line pth hs its velocity in feet per second t time t in seconds given by v(t) = 2 9 (t 3)2 + 2. () Crefully sketch the region whose exct re will tell you the vlue of the distnce the object trveled on the time intervl 2 t 5. (b) Estimte the distnce trveled on [2, 5] by computing L 4, R 4, nd M 4. (c) Does verging L 4 nd R 4 result in the sme vlue s M 4? If not, wht do you think the verge of L 4 nd R 4 mesures? (d) For this question, think bout n rbitrry function f, rther thn the prticulr function v given bove. If f is positive nd incresing on [, b], will L n overestimte or under-estimte the exct re under f on [, b]? Will R n over- or under-estimte the exct re under f on [, b]? Explin. When the function is sometimes negtive For Riemnn sum such s L n = n i=0 f (x i ) x,

4.2. RIEMANN SUMS 229 we cn of course compute the sum even when f tkes on negtive vlues. We know tht when f is positive on [, b], the corresponding left Riemnn sum L n estimtes the re bounded by f nd the horizontl xis over the intervl. For function such s the y = f (x) y = f (x) y = f (x) A A 3 A 2 b c d b c d b c d Figure 4.7: At left nd center, two left Riemnn sums for function f tht is sometimes negtive; t right, the res bounded by f on the intervl [, d]. one pictured in Figure 4.7, where in the first figure left Riemnn sum is being tken with 2 subintervls over [, d], we observe tht the function is negtive on the intervl b x c, nd so for the four left endpoints tht fll in [b, c], the terms f (x i ) x hve negtive function vlues. This mens tht those four terms in the Riemnn sum produce n estimte of the opposite of the re bounded by y = f (x) nd the x-xis on [b, c]. In Figure 4.7, we lso see evidence tht by incresing the number of rectngles used in Riemnn sum, it ppers tht the pproximtion of the re (or the opposite of the re) bounded by curve ppers to improve. For instnce, in the middle grph, we use 24 left rectngles, nd from the shded res, it ppers tht we hve decresed the error from the pproximtion tht uses 2. When we proceed to Section 4.3, we will discuss the nturl ide of letting the number of rectngles in the sum increse without bound. For now, it is most importnt for us to observe tht, in generl, ny Riemnn sum of continuous function f on n intervl [, b] pproximtes the difference between the re tht lies bove the horizontl xis on [, b] nd under f nd the re tht lies below the horizontl xis on [, b] nd bove f. In the nottion of Figure 4.7, we my sy tht L 24 A A 2 + A 3, where L 24 is the left Riemnn sum using 24 subintervls shown in the middle grph, nd A nd A 3 re the res of the regions where f is positive on the intervl of interest, while A 2 is the re of the region where f is negtive. We will lso cll the quntity A A 2 + A 3 the net signed re bounded by f over the intervl [, d], where by the phrse signed re we indicte tht we re ttching minus sign to the res of regions tht fll below the horizontl xis.

230 4.2. RIEMANN SUMS Finlly, we recll from the introduction to this present section tht in the context where the function f represents the velocity of moving object, the totl sum of the res bounded by the curve tells us the totl distnce trveled over the relevnt time intervl, while the totl net signed re bounded by the curve computes the object s chnge in position on the intervl. Activity 4.6. Suppose tht n object moving long stright line pth hs its velocity v (in feet per second) t time t (in seconds) given by v(t) = 2 t2 3t + 7 2. () Compute M 5, the middle Riemnn sum, for v on the time intervl [, 5]. Be sure to clerly identify the vlue of t s well s the loctions of t 0, t,, t 5. In ddition, provide creful sketch of the function nd the corresponding rectngles tht re being used in the sum. (b) Building on your work in (), estimte the totl chnge in position of the object on the intervl [, 5]. (c) Building on your work in () nd (b), estimte the totl distnce trveled by the object on [, 5]. (d) Use pproprite computing technology 5 to compute M 0 nd M 20. Wht exct vlue do you think the middle sum eventully pproches s n increses without bound? Wht does tht number represent in the physicl context of the overll problem? Summry In this section, we encountered the following importnt ides: A Riemnn sum is simply sum of products of the form f (x i ) x tht estimtes the re between positive function nd the horizontl xis over given intervl. If the function is sometimes negtive on the intervl, the Riemnn sum estimtes the difference between the res tht lie bove the horizontl xis nd those tht lie below the xis. The three most common types of Riemnn sums re left, right, nd middle sums, plus we cn lso work with more generl, rndom Riemnn sum. The only difference 5 For instnce, consider the pplet t http://gvsu.edu/s/9 nd chnge the function nd djust the loctions of the blue points tht represent the intervl endpoints nd b.

4.2. RIEMANN SUMS 23 mong these sums is the loction of the point t which the function is evluted to determine the height of the rectngle whose re is being computed in the sum. For left Riemnn sum, we evlute the function t the left endpoint of ech subintervl, while for right nd middle sums, we use right endpoints nd midpoints, respectively. The left, right, nd middle Riemnn sums re denoted L n, R n, nd M n, with formuls L n = f (x 0 ) x + f (x ) x + + f (x n ) x = R n = f (x ) x + f (x 2 ) x + + f (x n ) x = M n = f (x ) x + f (x 2 ) x + + f (x n ) x = n i=0 f (x i ) x, n f (x i ) x, i= n f (x i ) x, where x 0 =, x i = + i x, nd x n = b, using x = b n. For the midpoint sum, x i = (x i + x i )/2. i= Exercises. Consider the function f (x) = 3x + 4. () Compute M 4 for y = f (x) on the intervl [2, 5]. Be sure to clerly identify the vlue of x, s well s the loctions of x 0, x,..., x 4. Include creful sketch of the function nd the corresponding rectngles being used in the sum. (b) Use fmilir geometric formul to determine the exct vlue of the re of the region bounded by y = f (x) nd the x-xis on [2, 5]. (c) Explin why the vlues you computed in () nd (b) turn out to be the sme. Will this be true if we use number different thn n = 4 nd compute M n? Will L 4 or R 4 hve the sme vlue s the exct re of the region found in (b)? (d) Describe the collection of functions g for which it will lwys be the cse tht M n, regrdless of the vlue of n, gives the exct net signed re bounded between the function g nd the x-xis on the intervl [, b]. 2. Let S be the sum given by S = ((.4) 2 +) 0.4+((.8) 2 +) 0.4+((2.2) 2 +) 0.4+((2.6) 2 +) 0.4+((3.0) 2 +) 0.4. () Assume tht S is right Riemnn sum. For wht function f nd wht intervl [, b] is S n pproximtion of the re under f nd bove the x-xis on [, b]? Why?

232 4.2. RIEMANN SUMS (b) How does your nswer to () chnge if S is left Riemnn sum? middle Riemnn sum? (c) Suppose tht S relly is right Riemnn sum. Wht is geometric quntity does S pproximte? (d) Use sigm nottion to write new sum R tht is the right Riemnn sum for the sme function, but tht uses twice s mny subintervls s S. 3. A cr trveling long stright rod is brking nd its velocity is mesured t severl different points in time, s given in the following tble. seconds, t 0 0.3 0.6 0.9.2.5.8 Velocity in ft/sec, v(t) 00 88 74 59 40 9 0 () Plot the given dt on set of xes with time on the horizontl xis nd the velocity on the verticl xis. (b) Estimte the totl distnce trveled during the cr the time brkes using middle Riemnn sum with 3 subintervls. (c) Estimte the totl distnce trveled on [0,.8] by computing L 6, R 6, nd 2 (L 6 + R 6 ). (d) Assuming tht v(t) is lwys decresing on [0,.8], wht is the mximum possible distnce the cr trveled before it stopped? Why? 4. The rte t which pollution escpes scrubbing process t mnufcturing plnt increses over time s filters nd other technologies become less effective. For this prticulr exmple, ssume tht the rte of pollution (in tons per week) is given by the function r tht is pictured in Figure 4.8. 4 3 tons/week y = r(t) 2 2 3 4 weeks Figure 4.8: The rte, r(t), of pollution in tons per week. () Use the grph to estimte the vlue of M 4 on the intervl [0, 4].

4.2. RIEMANN SUMS 233 (b) Wht is the mening of M 4 in terms of the pollution dischrged by the plnt? (c) Suppose tht r(t) = 0.5e 0.5t. Use this formul for r to compute L 5 on [0, 4]. (d) Determine n upper bound on the totl mount of pollution tht cn escpe the plnt during the pictured four week time period tht is ccurte within n error of t most one ton of pollution.

234 4.3. THE DEFINITE INTEGRAL 4.3 The Definite Integrl Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: How does incresing the number of subintervls ffect the ccurcy of the pproximtion generted by Riemnn sum? Wht is the definition of the definite integrl of function f over the intervl [, b]? Wht does the definite integrl mesure exctly, nd wht re some of the key properties of the definite integrl? Introduction In Figure 4.7, which is repeted below s Figure 4.9, we see visul evidence tht incresing the number of rectngles in Riemnn sum improves the ccurcy of the pproximtion of the net signed re tht is bounded by the given function on the intervl under considertion. We thus explore the nturl ide of llowing the number of rectngles to y = f (x) y = f (x) y = f (x) A A 3 A 2 b c d b c d b c d Figure 4.9: At left nd center, two left Riemnn sums for function f tht is sometimes negtive; t right, the exct res bounded by f on the intervl [, d]. increse without bound in n effort to compute the exct net signed re bounded by function on n intervl. In ddition, it is importnt to think bout the differences mong left, right, nd middle Riemnn sums nd the different results they generte s the vlue of n increses. As we hve done throughout our investigtions with re, we begin with functions tht re exclusively positive on the intervl under considertion.

4.3. THE DEFINITE INTEGRAL 235 Preview Activity 4.3. Consider the pplet found t http://gvsu.edu/s/w 6. There, you will initilly see the sitution shown in Figure 4.20. Note tht the vlue of the chosen Figure 4.20: A right Riemnn sum with 0 subintervls for the function f (x) = sin(2x) x 2 0 + 3 on the intervl [, 7]. The vlue of the sum is R 0 = 4.90595. Riemnn sum is displyed next to the word reltive, nd tht you cn chnge the type of Riemnn sum being computed by drgging the point on the slider br below the phrse smple point plcement. Explore to see how you cn chnge the window in which the function is viewed, s well s the function itself. You cn set the minimum nd mximum vlues of x by clicking nd drgging on the blue points tht set the endpoints; you cn chnge the function by typing new formul in the f(x) window t the bottom; nd you cn djust the overll window by pnning nd zooming by using the Shift key nd the scrolling feture of your mouse. More informtion on how to pn nd zoom cn be found t http://gvsu.edu/s/fl. Work ccordingly to djust the pplet so tht it uses left Riemnn sum with n = 5 subintervls for the function is f (x) = 2x +. You should see the updted figure shown in Figure 4.2. Then, nswer the following questions. () Updte the pplet (nd view window, s needed) so tht the function being considered is f (x) = 2x + on [, 4], s directed bove. For this function on this intervl, compute L n, M n, R n for n = 5, n = 25, nd n = 00. Wht ppers to be the exct re bounded by f (x) = 2x + nd the x-xis on [, 4]? (b) Use bsic geometry to determine the exct re bounded by f (x) = 2x + nd 6 Mrc Renult, Shippensburg University, Geogebr Applets for Clclulus, http://gvsu.edu/s/5p.

236 4.3. THE DEFINITE INTEGRAL Figure 4.2: A left Riemnn sum with 5 subintervls for the function f (x) = 2x + on the intervl [, 4]. The vlue of the sum is L 5 = 6.2. the x-xis on [, 4]. (c) Bsed on your work in () nd (b), wht do you observe occurs when we increse the number of subintervls used in the Riemnn sum? (d) Updte the pplet to consider the function f (x) = x 2 + on the intervl [, 4] (note tht you need to enter x 2 + for the function formul). Use the pplet to compute L n, M n, R n for n = 5, n = 25, nd n = 00. Wht do you conjecture is the exct re bounded by f (x) = x 2 + nd the x-xis on [, 4]? (e) Why cn we not compute the exct vlue of the re bounded by f (x) = x 2 + nd the x-xis on [, 4] using formul like we did in (b)? The definition of the definite integrl In both exmples in Preview Activity 4.3, we sw tht s the number of rectngles got lrger nd lrger, the vlues of L n, M n, nd R n ll grew closer nd closer to the sme vlue. It turns out tht this occurs for ny continuous function on n intervl [, b], nd even more generlly for Riemnn sum using ny point x i+ in the intervl [x i, x i+ ]. Sid differently, s we let n, it doesn t relly mtter where we choose to evlute the

4.3. THE DEFINITE INTEGRAL 237 function within given subintervl, becuse lim L n = lim R n = lim M n = lim n n n n n f (x i ) x. Tht these limits lwys exist (nd shre the sme vlue) for continuous 7 function f llows us to mke the following definition. Definition 4.. The definite integrl of continuous function f on the intervl [, b], denoted f (x) dx, is the rel number given by f (x) dx = lim n i= n f (x i ) x, where x = b n, x i = + i x (for i = 0,..., n), nd x i stisfies x i x i x i (for i =,..., n). We cll the symbol the integrl sign, the vlues nd b the limits of integrtion, nd the function f the integrnd. The process of determining the rel number f (x) dx is clled evluting the definite integrl. While we will come to understnd tht there re severl different interprettions of the vlue of the definite integrl, for now the most importnt is tht f (x) dx mesures the net signed re bounded by y = f (x) nd the x-xis on the intervl [, b]. For exmple, in the nottion of the definite integrl, if f is the function pictured in Figure 4.22 nd A, A 2, nd A 3 re the exct res bounded by f nd the x-xis on the respective intervls [, b], [b, c], nd [c, d], then nd f (x) dx = A, d c b i= f (x) dx = A 2, d f (x) dx = A A 2 + A 3. c f (x) dx = A 3, We cn lso use definite integrls to express the chnge in position nd distnce trveled by moving object. In the setting of velocity function v on n intervl [, b], it follows from our work bove nd in preceding sections tht the chnge in position, s(b) s(), is given by s(b) s() = v(t) dt. 7 It turns out tht function need not be continuous in order to hve definite integrl. For our purposes, we ssume tht the functions we consider re continuous on the intervl(s) of interest. It is strightforwrd to see tht ny function tht is piecewise continuous on n intervl of interest will lso hve well-defined definite integrl.

238 4.3. THE DEFINITE INTEGRAL y = f (x) A A 3 A 2 b c d Figure 4.22: A continuous function f on the intervl [, d]. If the velocity function is nonnegtive on [, b], then v(t) dt tells us the distnce the object trveled. When velocity is sometimes negtive on [, b], the res bounded by the function on intervls where v does not chnge sign cn be found using integrls, nd the sum of these vlues will tell us the distnce the object trveled. If we wish to compute the vlue of definite integrl using the definition, we hve to tke the limit of sum. While this is possible to do in select circumstnces, it is lso tedious nd time-consuming; moreover, computing these limits does not offer much dditionl insight into the mening or interprettion of the definite integrl. Insted, in Section 4.4, we will lern the Fundmentl Theorem of Clculus, result tht provides shortcut for evluting lrge clss of definite integrls. This will enble us to determine the exct net signed re bounded by continuous function nd the x-xis in mny circumstnces, including exmples such s 4 (x2 + ) dx, which we pproximted by Riemnn sums in Preview Activity 4.3. For now, our gol is to understnd the mening nd properties of the definite integrl, rther thn how to ctully compute its vlue using ides in clculus. Thus, we temporrily rely on the net signed re interprettion of the definite integrl nd observe tht if given curve produces regions whose res we cn compute exctly through known re formuls, we cn thus compute the exct vlue of the integrl. For instnce, if we wish to evlute the definite integrl 4 (2x + ) dx, we cn observe tht the region bounded by this function nd the x-xis is the trpezoid shown in Figure 4.23, nd by the known formul for the re of trpezoid, its re is A = 2 (3 + 9) 3 = 8, so 4 (2x + ) dx = 8.

4.3. THE DEFINITE INTEGRAL 239 9 f (x) = 2x + 3 4 (2x + )dx 4 Figure 4.23: The re bounded by f (x) = 2x + nd the x-xis on the intervl [, 4]. Activity 4.7. Use known geometric formuls nd the net signed re interprettion of the definite integrl to evlute ech of the definite integrls below. () (b) (c) (d) 0 4 4 3 3x dx (2 2x) dx x 2 dx g(x) dx, where g is the function pictured in Figure 4.24. Assume tht ech portion of g is either prt of line or prt of circle. y = g(x) -3-2 - 2 3 4 - Figure 4.24: A function g tht is piecewise defined; ech piece of the function is prt of circle or prt of line.

240 4.3. THE DEFINITE INTEGRAL Some properties of the definite integrl With the perspective tht the definite integrl of function f over n intervl [, b] mesures the net signed re bounded by f nd the x-xis over the intervl, we nturlly rrive t severl different stndrd properties of the definite integrl. In ddition, it is helpful to remember tht the definite integrl is defined in terms of Riemnn sums tht fundmentlly consist of the res of rectngles. If we consider the definite integrl f (x) dx for ny rel number, it is evident tht no re is being bounded becuse the intervl begins nd ends with the sme point. Hence, If f is continuous function nd is rel number, then f (x) dx = 0. y = f (x) A A 2 b c Figure 4.25: The re bounded by y = f (x) on the intervl [, c]. Next, we consider the results of subdividing given intervl. In Figure 4.25, we see tht f (x) dx = A, c b f (x) dx = A 2, nd c f (x) dx = A + A 2, which is indictive of the following generl rule. If f is continuous function nd, b, nd c re rel numbers, then c f (x) dx = c f (x) dx + b f (x) dx. While this rule is most pprent in the sitution where < b < c, it in fct holds in generl for ny vlues of, b, nd c. This result is connected to nother property of the definite integrl, which sttes tht if we reverse the order of the limits of integrtion, we

4.3. THE DEFINITE INTEGRAL 24 chnge the sign of the integrl s vlue. If f is continuous function nd nd b re rel numbers, then b f (x) dx = f (x) dx. This result mkes sense becuse if we integrte from to b, then in the defining Riemnn sum x = b b n, while if we integrte from b to, x = n = b n, nd this is the only chnge in the sum used to define the integrl. There re two dditionl properties of the definite integrl tht we need to understnd. Recll tht when we worked with derivtive rules in Chpter 2, we found tht both the Constnt Multiple Rule nd the Sum Rule held. The Constnt Multiple Rule tells us tht if f is differentible function nd k is constnt, then d dx [k f (x)] = k f (x), nd the Sum Rule sttes tht if f nd g re differentible functions, then d dx [ f (x) + g(x)] = f (x) + g (x). These rules re useful becuse they enble us to del individully with the simplest prts of certin functions nd tke dvntge of the elementry opertions of ddition nd multiplying by constnt. They lso tell us tht the process of tking the derivtive respects ddition nd multiplying by constnts in the simplest possible wy. It turns out tht similr rules hold for the definite integrl. First, let s consider the sitution pictured in Figure 4.26, where we exmine the effect of multiplying function by B = 2 f (x i ) x y = 2 f (x) A = f (x i ) x A y = f (x) B x i x i+ b x i x i+ b Figure 4.26: The res bounded by y = f (x) nd y = 2 f (x) on [, b].

242 4.3. THE DEFINITE INTEGRAL fctor of 2 on the re it bounds with the x-xis. Becuse multiplying the function by 2 doubles its height t every x-vlue, we see tht if we consider typicl rectngle from Riemnn sum, the difference in re comes from the chnged height of the rectngle: f (x i ) for the originl function, versus 2 f (x i ) in the doubled function, in the cse of left sum. Hence, in Figure 4.26, we see tht for the pictured rectngles with res A nd B, it follows B = 2A. As this will hppen in every such rectngle, regrdless of the vlue of n nd the type of sum we use, we see tht in the limit, the re of the red region bounded by y = 2 f (x) will be twice tht of the re of the blue region bounded by y = f (x). As there is nothing specil bout the vlue 2 compred to n rbitrry constnt k, it turns out tht the following generl principle holds. Constnt Multiple Rule: If f is continuous function nd k is ny rel number then k f (x) dx = k f (x) dx. Finlly, we see similr sitution geometriclly with the sum of two functions f nd g. In prticulr, s shown in Figure 4.27, if we tke the sum of two functions f nd g, t every C = ( f (x i ) + g(x i )) x A = f (x i ) x A f B = g(x i ) x B g C f + g x i x i+ b x i x i+ b x i x i+ b Figure 4.27: The res bounded by y = f (x) nd y = g(x) on [, b], s well s the re bounded by y = f (x) + g(x). point in the intervl, the height of the function f + g is given by ( f + g)(x i ) = f (x i ) + g(x i ), which is the sum of the individul function vlues of f nd g (tken t left endpoints). Hence, for the pictured rectngles with res A, B, nd C, it follows tht C = A + B, nd becuse this will occur for every such rectngle, in the limit the re of the gry region will be the sum of the res of the blue nd red regions. Stted in terms of definite integrls,

4.3. THE DEFINITE INTEGRAL 243 we hve the following generl rule. Sum Rule: If f nd g re continuous functions, then [ f (x) + g(x)] dx = f (x) dx + g(x) dx. More generlly, the Constnt Multiple nd Sum Rules cn be combined to mke the observtion tht for ny continuous functions f nd g nd ny constnts c nd k, Activity 4.8. [c f (x) ± kg(x)] dx = c f (x) dx ± k g(x) dx. Suppose tht the following informtion is known bout the functions f, g, x 2, nd x 3 : 2 0 f (x) dx = 3; 5 f (x) dx = 2 2 2 0 g(x) dx = 4; 5 g(x) dx = 2 2 0 x2 dx = 8 3 ; 5 2 x2 dx = 7 3 2 0 x3 dx = 4; 5 2 x3 dx = 609 4 Use the provided informtion nd the rules discussed in the preceding section to evlute ech of the following definite integrls. () 2 5 f (x) dx (b) 5 g(x) dx 0 (c) 5 ( f (x) + g(x)) dx 0 (d) 5 2 (3x2 4x 3 ) dx (e) 0 5 (2x3 7g(x)) dx How the definite integrl is connected to function s verge vlue One of the most vluble pplictions of the definite integrl is tht it provides wy to meningfully discuss the verge vlue of function, even for function tht tkes on infinitely mny vlues. Recll tht if we wish to tke the verge of n numbers y, y 2,..., y n, we do so by computing Avg = y + y 2 + + y n. n

244 4.3. THE DEFINITE INTEGRAL Since integrls rise from Riemnn sums in which we dd n vlues of function, it should not be surprising tht evluting n integrl is something like verging the output vlues of function. Consider, for instnce, the right Riemnn sum R n of function f, which is given by R n = f (x ) x + f (x 2 ) x + + f (x n ) x = ( f (x ) + f (x 2 ) + + f (x n )) x. Since x = b n, we cn thus write R n = ( f (x ) + f (x 2 ) + + f (x n )) b n = (b ) f (x ) + f (x 2 ) + + f (x n ). (4.) n Here, we see tht the right Riemnn sum with n subintervls is the length of the intervl (b ) times the verge of the n function vlues found t the right endpoints. And just s with our efforts to compute re, we see tht the lrger the vlue of n we use, the more ccurte our verge of the vlues of f will be. Indeed, we will define the verge vlue of f on [, b] to be f (x ) + f (x 2 ) + + f (x n ) f AVG[,b] = lim. n n But we lso know tht for ny continuous function f on [, b], tking the limit of Riemnn sum leds precisely to the definite integrl. Tht is, lim R n = n thus tking the limit s n in Eqution (4.), we hve tht f (x) dx, nd f (x) dx = (b ) f AVG[,b]. (4.2) Solving Eqution (4.2) for f AVG[,b], we hve the following generl principle. Averge vlue of function: If f is continuous function on [, b], then its verge vlue on [, b] is given by the formul f AVG[,b] = b f (x) dx. Observe tht Eqution (4.2) tells us nother wy to interpret the definite integrl: the definite integrl of function f from to b is the length of the intervl (b ) times the verge vlue of the function on the intervl. In ddition, Eqution (4.2) hs nturl visul interprettion when the function f is nonnegtive on [, b]. Consider Figure 4.28, where we see t left the shded region whose re is f (x) dx, t center the shded rectngle whose dimensions re (b ) by f AVG[,b], nd t right these two figures superimposed. Specificlly, note tht in drk green we show the horizontl line y = f AVG[,b]. Thus, the re of the green rectngle is given by (b ) f AVG[,b], which is precisely the vlue of f (x) dx. Sid differently, the re of the blue region in the

4.3. THE DEFINITE INTEGRAL 245 y = f (x) y = f (x) y = f (x) f AVG[,b] A A 2 f (x)dx (b ) f AVG[,b] b b b Figure 4.28: A function y = f (x), the re it bounds, nd its verge vlue on [, b]. left figure is the sme s tht of the green rectngle in the center figure; this cn lso be seen by observing tht the res A nd A 2 in the rightmost figure pper to be equl. Ultimtely, the verge vlue of function enbles us to construct rectngle whose re is the sme s the vlue of the definite integrl of the function on the intervl. The jv pplet 8 t http://gvsu.edu/s/z provides n opportunity to explore how the verge vlue of the function chnges s the intervl chnges, through n imge similr to tht found in Figure 4.28. Activity 4.9. Suppose tht v(t) = 4 (t 2) 2 tells us the instntneous velocity of moving object on the intervl 0 t 4, where t is mesured in minutes nd v is mesured in meters per minute. () Sketch n ccurte grph of y = v(t). Wht kind of curve is y = 4 (t 2) 2? (b) Evlute 4 v(t) dt exctly. 0 (c) In terms of the physicl problem of the moving object with velocity v(t), wht is the mening of 4 v(t) dt? Include units on your nswer. 0 (d) Determine the exct verge vlue of v(t) on [0, 4]. Include units on your nswer. (e) Sketch rectngle whose bse is the line segment from t = 0 to t = 4 on the t-xis such tht the rectngle s re is equl to the vlue of 4 v(t) dt. Wht is 0 the rectngle s exct height? (f) How cn you use the verge vlue you found in (d) to compute the totl distnce trveled by the moving object over [0, 4]? 8 Dvid Austin, http://gvsu.edu/s/5r.

246 4.3. THE DEFINITE INTEGRAL Summry In this section, we encountered the following importnt ides: Any Riemnn sum of continuous function f on n intervl [, b] provides n estimte of the net signed re bounded by the function nd the horizontl xis on the intervl. Incresing the number of subintervls in the Riemnn sum improves the ccurcy of this estimte, nd letting the number of subintervls increse without bound results in the vlues of the corresponding Riemnn sums pproching the exct vlue of the enclosed net signed re. When we tke the just described limit of Riemnn sums, we rrive t wht we cll the definite integrl of f over the intervl [, b]. In prticulr, the symbol f (x) dx denotes the definite integrl of f over [, b], nd this quntity is defined by the eqution f (x) dx = lim n n f (x i ) x, i= where x = b n, x i = + i x (for i = 0,..., n), nd x i stisfies x i x i x i (for i =,..., n). The definite integrl f (x) dx mesures the exct net signed re bounded by f nd the horizontl xis on [, b]; in ddition, the vlue of the definite integrl is relted to wht we cll the verge vlue of the function on [, b]: f AVG[,b] = b f (x) dx. In the setting where we consider the integrl of velocity function v, v(t) dt mesures the exct chnge in position of the moving object on [, b]; when v is nonnegtive, v(t) dt is the object s distnce trveled on [, b]. The definite integrl is sophisticted sum, nd thus hs some of the sme nturl properties tht finite sums hve. Perhps most importnt of these is how the definite integrl respects sums nd constnt multiples of functions, which cn be summrized by the rule [c f (x) ± kg(x)] dx = c f (x) dx ± k g(x) dx where f nd g re continuous functions on [, b] nd c nd k re rbitrry constnts. Exercises. The velocity of n object moving long n xis is given by the piecewise liner function v tht is pictured in Figure 4.29. Assume tht the object is moving to the right when its

4.3. THE DEFINITE INTEGRAL 247 velocity is positive, nd moving to the left when its velocity is negtive. Assume tht the given velocity function is vlid for t = 0 to t = 4. 2 ft/sec 2 3 4 sec - -2 y = v(t) Figure 4.29: The velocity function of moving object. () Write n expression involving definite integrls whose vlue is the totl chnge in position of the object on the intervl [0, 4]. (b) Use the provided grph of v to determine the vlue of the totl chnge in position on [0, 4]. (c) Write n expression involving definite integrls whose vlue is the totl distnce trveled by the object on [0, 4]. Wht is the exct vlue of the totl distnce trveled on [0, 4]? (d) Wht is the object s exct verge velocity on [0, 4]? (e) Find n lgebric formul for the object s position function on [0,.5] tht stisfies s(0) = 0. 2. Suppose tht the velocity of moving object is given by v(t) = t(t )(t 3), mesured in feet per second, nd tht this function is vlid for 0 t 4. () Write n expression involving definite integrls whose vlue is the totl chnge in position of the object on the intervl [0, 4]. (b) Use pproprite technology (such s http://gvsu.edu/s/9 9 ) to compute Riemnn sums to estimte the object s totl chnge in position on [0, 4]. Work to ensure tht your estimte is ccurte to two deciml plces, nd explin how you know this to be the cse. (c) Write n expression involving definite integrls whose vlue is the totl distnce trveled by the object on [0, 4]. 9 Mrc Renult, Shippensburg University.

248 4.3. THE DEFINITE INTEGRAL (d) Use pproprite technology to compute Riemnn sums to estimte the object s totl distnce trvelled on [0, 4]. Work to ensure tht your estimte is ccurte to two deciml plces, nd explin how you know this to be the cse. (e) Wht is the object s verge velocity on [0, 4], ccurte to two deciml plces? 3. Consider the grphs of two functions f nd g tht re provided in Figure 4.30. Ech piece of f nd g is either prt of stright line or prt of circle. 2 y = f (x) 2 y = g(x) 2 3 4 2 3 4 - - -2-2 Figure 4.30: Two functions f nd g. () Determine the exct vlue of [ f (x) + g(x)] dx. 0 (b) Determine the exct vlue of 4 [2 f (x) 3g(x)] dx. (c) Find the exct verge vlue of h(x) = g(x) f (x) on [0, 4]. (d) For wht constnt c does the following eqution hold? 4 4. Let f (x) = 3 x 2 nd g(x) = 2x 2. 0 c dx = 4 0 [ f (x) + g(x)] dx () On the intervl [, ], sketch lbeled grph of y = f (x) nd write definite integrl whose vlue is the exct re bounded by y = f (x) on [, ]. (b) On the intervl [, ], sketch lbeled grph of y = g(x) nd write definite integrl whose vlue is the exct re bounded by y = g(x) on [, ]. (c) Write n expression involving difference of definite integrls whose vlue is the exct re tht lies between y = f (x) nd y = g(x) on [, ]. (d) Explin why your expression in (c) hs the sme vlue s the single integrl [ f (x) g(x)] dx.

4.3. THE DEFINITE INTEGRAL 249 (e) Explin why, in generl, if p(x) q(x) for ll x in [, b], the exct re between y = p(x) nd y = q(x) is given by [p(x) q(x)] dx.

250 4.4. THE FUNDAMENTAL THEOREM OF CALCULUS 4.4 The Fundmentl Theorem of Clculus Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: How cn we find the exct vlue of definite integrl without tking the limit of Riemnn sum? Wht is the sttement of the Fundmentl Theorem of Clculus, nd how do ntiderivtives of functions ply key role in pplying the theorem? Wht is the mening of the definite integrl of rte of chnge in contexts other thn when the rte of chnge represents velocity? Introduction Much of our work in Chpter 4 hs been motivted by the velocity-distnce problem: if we know the instntneous velocity function, v(t), for moving object on given time intervl [, b], cn we determine its exct distnce trveled on [, b]? In the vst mjority of our discussion in Sections 4.-4.3, we hve focused on the fct tht this distnce trveled is connected to the re bounded by y = v(t) nd the t-xis on [, b]. In prticulr, for ny nonnegtive velocity function y = v(t) on [, b], we know tht the exct re bounded by the velocity curve nd the t-xis on the intervl tells us the totl distnce trveled, which is lso the vlue of the definite integrl v(t) dt. In the sitution where velocity is sometimes negtive, the totl re bounded by the velocity function still tells us distnce trveled, while the net signed re tht the function bounds tells us the object s chnge in position. Recll, for instnce, the introduction to Section 4.2, where we observed tht for the velocity function in Figure 4.3, the totl distnce D trveled by the moving object on [, b] is D = A + A 2 + A 3, while the totl chnge in the object s position on [, b] is s(b) s() = A A 2 + A 3. While the res A, A 2, nd A 3, which re ech given by definite integrls, my be computed through limits of Riemnn sums (nd in select specil circumstnces through fmilir geometric formuls), in the present section we turn our ttention to n lternte pproch, similr to the one we encountered in Activity 4.2. To explore these ides further, we consider the following preview ctivity. Preview Activity 4.4. A student with third floor dormitory window 32 feet off the ground tosses wter blloon stright up in the ir with n initil velocity of 6 feet

4.4. THE FUNDAMENTAL THEOREM OF CALCULUS 25 y = v(t) A A 3 A 2 b Figure 4.3: A velocity function tht is sometimes negtive. per second. It turns out tht the instntneous velocity of the wter blloon is given by the velocity function v(t) = 32t + 6, where v is mesured in feet per second nd t is mesured in seconds. () Let s(t) represent the height of the wter blloon bove the ground t time t, nd note tht s is n ntiderivtive of v. Tht is, v is the derivtive of s: s (t) = v(t). Find formul for s(t) tht stisfies the initil condition tht the blloon is tossed from 32 feet bove ground. In other words, mke your formul for s stisfy s(0) = 32. (b) At wht time does the wter blloon rech its mximum height? At wht time does the wter blloon lnd? (c) Compute the three differences s( 2 ) s(0), s(2) s( 2 ), nd s(2) s(0). Wht do these differences represent? (d) Wht is the totl verticl distnce trveled by the wter blloon from the time it is tossed until the time it lnds? (e) Sketch grph of the velocity function y = v(t) on the time intervl [0, 2]. Wht is the totl net signed re bounded by y = v(t) nd the t-xis on [0, 2]? Answer this question in two wys: first by using your work bove, nd then by using fmilir geometric formul to compute res of certin relevnt regions.

252 4.4. THE FUNDAMENTAL THEOREM OF CALCULUS The Fundmentl Theorem of Clculus Consider the setting where we know the position function s(t) of n object moving long n xis, s well s its corresponding velocity function v(t), nd for the moment let us ssume tht v(t) is positive on [, b]. Then, s shown in Figure 4.32, we know two different y = v(t) D = v(t)dt = s(b) s() b Figure 4.32: Finding distnce trveled when we know n object s velocity function v. perspectives on the distnce, D, the object trvels: one is tht D = s(b) s(), which is the object s chnge in position. The other is tht the distnce trveled is the re under the velocity curve, which is given by the definite integrl, so D = v(t) dt. Of course, since both of these expressions tell us the distnce trveled, it follows tht they re equl, so s(b) s() = v(t) dt. (4.3) Furthermore, we know tht Eqution (4.3) holds even when velocity is sometimes negtive, since s(b) s() is the object s chnge in position over [, b], which is simultneously mesured by the totl net signed re on [, b] given by v(t) dt. Perhps the most powerful prt of Eqution (4.3) lies in the fct tht we cn compute the integrl s vlue if we cn find formul for s. Remember, s nd v re relted by the fct tht v is the derivtive of s, or equivlently tht s is n ntiderivtive of v. For exmple, if we hve n object whose velocity is v(t) = 3t 2 + 40 feet per second (which is lwys nonnegtive), nd wish to know the distnce trveled on the intervl [, 5], we hve tht D = 5 v(t) dt = 5 (3t 2 + 40) dt = s(5) s(), where s is n ntiderivtive of v. We know tht the derivtive of t 3 is 3t 2 nd tht the derivtive of 40t is 40, so it follows tht if s(t) = t 3 + 40t, then s is function whose derivtive is v(t) = s (t) = 3t 2 + 40, nd thus we hve found n ntiderivtive of v.

4.4. THE FUNDAMENTAL THEOREM OF CALCULUS 253 Therefore, D = 5 3t 2 + 40 dt = s(5) s() = (5 3 + 40 5) ( 3 + 40 ) = 284 feet. Note the key lesson of this exmple: to find the distnce trveled, we needed to compute the re under curve, which is given by the definite integrl. But to evlute the integrl, we found n ntiderivtive, s, of the velocity function, nd then computed the totl chnge in s on the intervl. In prticulr, observe tht we hve found the exct re of the region shown in Figure 4.33, nd done so without fmilir formul (such s those for the re of tringle or circle) nd without directly computing the limit of Riemnn sum. As we proceed to thinking bout contexts other thn just velocity nd position, it is 40 20 00 80 60 40 20 D = 5 v(t)dt = 284 y = v(t) 3 5 Figure 4.33: The exct re of the region enclosed by v(t) = 3t 2 + 40 on [, 5]. dvntgeous to hve shorthnd symbol for function s ntiderivtive. In the generl setting of continuous function f, we will often denote n ntiderivtive of f by F, so tht the reltionship between F nd f is tht F (x) = f (x) for ll relevnt x. Using the nottion V in plce of s (so tht V is n ntiderivtive of v) in Eqution (4.3), we find it is equivlent to write tht V(b) V() = v(t) dt. (4.4) Now, in the generl setting of wnting to evlute the definite integrl f (x) dx for n rbitrry continuous function f, we could certinly think of f s representing the velocity of some moving object, nd x s the vrible tht represents time. And gin, Equtions (4.3) nd (4.4) hold for ny continuous velocity function, even when v is sometimes negtive. This leds us to see tht Eqution (4.4) tells us something even more importnt thn the chnge in position of moving object: it offers shortcut route to evluting ny definite integrl, provided tht we cn find n ntiderivtive of the integrnd. The Fundmentl

254 4.4. THE FUNDAMENTAL THEOREM OF CALCULUS Theorem of Clculus (FTC) summrizes these observtions. The Fundmentl Theorem of Clculus: If f is continuous function on [, b], nd F is ny ntiderivtive of f, then f (x) dx = F(b) F(). A common lternte nottion for F(b) F() is F(b) F() = F(x) b, where we red the righthnd side s the function F evluted from to b. In this nottion, the FTC sys tht f (x) dx = F(x) b. The FTC opens the door to evluting exctly wide rnge of integrls. In prticulr, if we re interested in definite integrl for which we cn find n ntiderivtive F for the integrnd f, then we cn evlute the integrl exctly. For instnce since d dx [ 3 x3 ] = x 2, the FTC tells us tht 0 x 2 dx = 3 x3 0 = 3 ()3 3 (0)3 = 3. But finding n ntiderivtive cn be fr from simple; in fct, often finding formul for n ntiderivtive is very hrd or even impossible. While we cn differentite just bout ny function, even some reltively simple ones don t hve n elementry ntiderivtive. A significnt portion of integrl clculus (which is the min focus of second semester college clculus) is devoted to understnding the problem of finding ntiderivtives. Activity 4.0. Use the Fundmentl Theorem of Clculus to evlute ech of the following integrls exctly. For ech, sketch grph of the integrnd on the relevnt intervl nd write one sentence tht explins the mening of the vlue of the integrl in terms of the (net signed) re bounded by the curve. () (b) (c) 4 π 2 0 0 (2 2x) dx sin(x) dx e x dx

4.4. THE FUNDAMENTAL THEOREM OF CALCULUS 255 (d) (e) 2 0 x 5 dx (3x 3 2x 2 e x ) dx Bsic ntiderivtives The generl problem of finding n ntiderivtive is difficult. In prt, this is due to the fct tht we re trying to undo the process of differentiting, nd the undoing is much more difficult thn the doing. For exmple, while it is evident tht n ntiderivtive of f (x) = sin(x) is F(x) = cos(x) nd tht n ntiderivtive of g(x) = x 2 is G(x) = 3 x3, combintions of f nd g cn be fr more complicted. Consider such functions s 5 sin(x) 4x 2, x 2 sin(x), sin(x) x 2, nd sin(x 2 ). Wht is involved in trying to find n ntiderivtive for ech? From our experience with derivtive rules, we know tht while derivtives of sums nd constnt multiples of bsic functions re simple to execute, derivtives involving products, quotients, nd composites of fmilir functions re much more complicted. Thus, it stnds to reson tht ntidifferentiting products, quotients, nd composites of bsic functions my be even more chllenging. We defer our study of ll but the most elementry ntiderivtives to lter in the text. We do note tht ech time we hve function for which we know its derivtive, we hve function-derivtive pir, which lso leds us to knowing the ntiderivtive of function. For instnce, since we know tht d [ cos(x)] = sin(x), dx it follows tht F(x) = cos(x) is n ntiderivtive of f (x) = sin(x). It is equivlent to sy tht f (x) = sin(x) is the derivtive of F(x) = cos(x), nd thus F nd f together form the function-derivtive pir. Clerly, every bsic derivtive rule leds us to such pir, nd thus to known ntiderivtive. In Activity 4., we will construct list of most of the bsic ntiderivtives we know t this time. Furthermore, those rules will enble us to ntidifferentite sums nd constnt multiples of bsic functions. For exmple, if f (x) = 5 sin(x) 4x 2, note tht since cos(x) is n ntiderivtive of sin(x) nd 3 x3 is n ntiderivtive of x 2, it follows tht F(x) = 5 cos(x) 4 3 x3 is n ntiderivtive of f, by the sum nd constnt multiple rules for differentition.

256 4.4. THE FUNDAMENTAL THEOREM OF CALCULUS Finlly, before proceeding to build list of common functions whose ntiderivtives we know, we revisit the fct tht ech function hs more thn one ntiderivtive. Becuse the derivtive of ny constnt is zero, ny time we seek n rbitrry ntiderivtive, we my dd constnt of our choice. For instnce, if we wnt to determine n ntiderivtive of g(x) = x 2, we know tht G(x) = 3 x3 is one such function. But we could lterntely hve chosen G(x) = 3 x3 + 7, since in this cse s well, G (x) = x 2. In some contexts lter on in clculus, it is importnt to discuss the most generl ntiderivtive of function. If g(x) = x 2, we sy tht the generl ntiderivtive of g is G(x) = 3 x3 + C, where C represents n rbitrry rel number constnt. Regrdless of the formul for g, including +C in the formul for its ntiderivtive G results in the most generl possible ntiderivtive. Our primry current interest in ntiderivtives is for use in evluting definite integrls by the Fundmentl Theorem of Clculus. In tht sitution, the rbitrry constnt C is irrelevnt, nd thus we usully omit it. To see why, consider the definite integrl 0 x 2 dx. For the integrnd g(x) = x 2, suppose we find nd use the generl ntiderivtive G(x) = 3 x3 + C. Then, by the FTC, 0 x 2 dx = 3 x3 + C = 0 ( ( 3 ()3 + C) 3 (0)3 + C) = 3 + C 0 C = 3. Specificlly, we observe tht the C-vlues pper s opposites in the evlution of the integrl nd thus do not ffect the definite integrl s vlue. In the sme wy, the potentil inclusion of +C with the ntiderivtive hs no bering on ny definite integrl, nd thus we generlly choose to omit this possible constnt whenever we evlute n integrl using the Fundmentl Theorem of Clculus. In the following ctivity, we work to build list of bsic functions whose ntiderivtives we lredy know.

4.4. THE FUNDAMENTAL THEOREM OF CALCULUS 257 given function, f (x) k, (k is constnt) x n, n x, x > 0 sin(x) cos(x) sec(x) tn(x) csc(x) cot(x) sec 2 (x) csc 2 (x) e x x ( > ) +x 2 x 2 ntiderivtive, F(x) Tble 4.: Fmilir bsic functions nd their ntiderivtives. Activity 4.. Use your knowledge of derivtives of bsic functions to complete the bove tble of ntiderivtives. For ech entry, your tsk is to find function F whose derivtive is the given function f. When finished, use the FTC nd the results in the tble to evlute the three given definite integrls. () (b) (c) 0 π/3 0 0 x 3 x e x + 2 dx (2 sin(t) 4 cos(t) + sec 2 (t) π) dt ( x x 2 ) dx The totl chnge theorem As we use the Fundmentl Theorem of Clculus to evlute definite integrls, it is essentil tht we remember nd understnd the mening of the numbers we find. We briefly summrize three key interprettions to dte.

258 4.4. THE FUNDAMENTAL THEOREM OF CALCULUS For moving object with instntneous velocity v(t), the object s chnge in position on the time intervl [, b] is given by v(t) dt, nd whenever v(t) 0 on [, b], v(t) dt tells us the totl distnce trveled by the object on [, b]. For ny continuous function f, its definite integrl f (x) dx represents the totl net signed re bounded by y = f (x) nd the x-xis on [, b], where regions tht lie below the x-xis hve minus sign ssocited with their re. The vlue of definite integrl is linked to the verge vlue of function: for continuous function f on [, b], its verge vlue f AVG[,b] is given by f AVG[,b] = f (x) dx. b The Fundmentl Theorem of Clculus now enbles us to evlute exctly (without tking limit of Riemnn sums) ny definite integrl for which we re ble to find n ntiderivtive of the integrnd. A slight chnge in nottionl perspective llows us to gin even more insight into the mening of the definite integrl. To begin, recll Eqution (4.4), where we wrote the Fundmentl Theorem of Clculus for velocity function v with ntiderivtive V s V(b) V() = v(t) dt. If we insted replce V with s (which represents position) nd replce v with s (since velocity is the derivtive of position), Eqution (4.4) equivlently reds s(b) s() = s (t) dt. (4.5) In words, this version of the FTC tells us tht the totl chnge in the object s position function on prticulr intervl is given by the definite integrl of the position function s derivtive over tht intervl. Of course, this result is not limited to only the setting of position nd velocity. Writing the result in terms of more generl function f, we hve the Totl Chnge Theorem. The Totl Chnge Theorem: If f is continuously differentible function on [, b] with derivtive f, then f (b) f () = f (x) dx. Tht is, the definite integrl of the derivtive of function on [, b] is the totl chnge of the function itself on [, b]. The Totl Chnge Theorem tells us more bout the reltionship between the grph of function nd tht of its derivtive. Recll Figure.8, which provided one of the first times we sw tht heights on the grph of the derivtive function come from slopes on the grph of the function itself. Tht observtion occurred in the context where we knew f

4.4. THE FUNDAMENTAL THEOREM OF CALCULUS 259 nd were seeking f ; if now insted we think bout knowing f nd seeking informtion bout f, we cn insted sy the following: differences in heights on f correspond to net signed res bounded by f. 4 3 2 3 3 4 4 3 2 (2,4) (3,3) (,3) (0,0) (4,0) - -2-3 2 y = f (x) 3 - -2-3 2 3 4 y = f (x) -4-4 Figure 4.34: The grphs of f (x) = 4 2x (t left) nd n ntiderivtive f (x) = 4x x 2 t right. Differences in heights on f correspond to net signed res bounded by f. To see why this is so, sy we consider the difference f () f (0). Note tht this vlue is 3, in prt becuse f () = 3 nd f (0) = 0, but lso becuse the net signed re bounded by y = f (x) on [0, ] is 3. Tht is, f () f (0) = 0 f (x) dx. A similr pttern holds throughout, including the fct tht since the totl net signed re bounded by f on [0, 4] is 0, 4 0 f (x) dx = 0, so it must be tht f (4) f (0) = 0, so f (4) = f (0). Beyond this generl observtion bout re, the Totl Chnge Theorem enbles us to consider interesting nd importnt problems where we know the rte of chnge, nd nswer key questions bout the function whose rte of chnge we know. Exmple 4.. Suppose tht pollutnts re leking out of n underground storge tnk t rte of r(t) gllons/dy, where t is mesured in dys. It is conjectured tht r(t) is given by the formul r(t) = 0.0069t 3 0.25t 2 +.079 over certin 2-dy period. The grph of y = r(t) is given in Figure 4.35. Wht is the mening of 0 r(t) dt nd wht is 4 its vlue? Wht is the verge rte t which pollutnts re leving the tnk on the time intervl 4 t 0?

260 4.4. THE FUNDAMENTAL THEOREM OF CALCULUS 2 gl/dy 0 8 6 4 2 y = r(t) dys 2 4 6 8 0 2 Figure 4.35: The rte r(t) of pollution leking from tnk, mesured in gllons per dy. We know tht since r(t) 0, the vlue of 0 r(t) dt is the re under the curve on 4 the intervl [4, 0]. If we think bout this re from the perspective of Riemnn sum, the rectngles will hve heights mesured in gllons per dy nd widths mesured in dys, thus the re of ech rectngle will hve units of gllons dy dys = gllons. Thus, the definite integrl tells us the totl number of gllons of pollutnt tht lek from the tnk from dy 4 to dy 0. The Totl Chnge Theorem tells us the sme thing: if we let R(t) denote the function tht mesures the totl number of gllons of pollutnt tht hve leked from the tnk up to dy t, then R (t) = r(t), nd 0 4 r(t) dt = R(0) R(4), which is the totl chnge in the function tht mesures totl gllons leked over time, thus the number of gllons tht hve leked from dy 4 to dy 0. To compute the exct vlue, we use the Fundmentl Theorem of Clculus. Antidifferentiting r(t) = 0.0069t 3 0.25t 2 +.079, we find tht 0 4 (0.0069t 3 0.25t 2 +.079) dt = ( 0.0069 44.282. 4 t4 0.25 ) 0 3 t3 +.079t Thus, pproximtely 44.282 gllons of pollutnt leked over the six dy time period. To find the verge rte t which pollutnt leked from the tnk over 4 t 0, we 4

4.4. THE FUNDAMENTAL THEOREM OF CALCULUS 26 wnt to compute the verge vlue of r on [4, 0]. Thus, r AVG[4,0] = 0 4 0 4 r(t) dt 44.282 6 = 7.380, which hs its units mesured in gllons per dy. Activity 4.2. During 40-minute workout, person riding n exercise mchine burns clories t rte of c clories per minute, where the function y = c(t) is given in Figure 4.36. On the intervl 0 t 0, the formul for c is c(t) = 0.05t 2 + t + 0, while on 30 t 40, its formul is c(t) = 0.05t 2 + 3t 30. 5 cl/min y = c(t) 0 5 0 20 30 40 min Figure 4.36: The rte c(t) t which person exercising burns clories, mesured in clories per minute. () Wht is the exct totl number of clories the person burns during the first 0 minutes of her workout? (b) Let C(t) be n ntiderivtive of c(t). Wht is the mening of C(40) C(0) in the context of the person exercising? Include units on your nswer. (c) Determine the exct verge rte t which the person burned clories during the 40-minute workout. (d) At wht time(s), if ny, is the instntneous rte t which the person is burning clories equl to the verge rte t which she burns clories, on the time intervl 0 t 40?

262 4.4. THE FUNDAMENTAL THEOREM OF CALCULUS Summry In this section, we encountered the following importnt ides: We cn find the exct vlue of definite integrl without tking the limit of Riemnn sum or using fmilir re formul by finding the ntiderivtive of the integrnd, nd hence pplying the Fundmentl Theorem of Clculus. The Fundmentl Theorem of Clculus sys tht if f is continuous function on [, b] nd F is n ntiderivtive of f, then f (x) dx = F(b) F(). Hence, if we cn find n ntiderivtive for the integrnd f, evluting the definite integrl comes from simply computing the chnge in F on [, b]. A slightly different perspective on the FTC llows us to restte it s the Totl Chnge Theorem, which sys tht f (x) dx = f (b) f (), for ny continuously differentible function f. This mens tht the definite integrl of the instntneous rte of chnge of function f on n intervl [, b] is equl to the totl chnge in the function f on [, b]. Exercises. The instntneous velocity (in meters per minute) of moving object is given by the function v s pictured in Figure 4.37. Assume tht on the intervl 0 t 4, v(t) is given by v(t) = 4 t3 + 3 2 t2 +, nd tht on every other intervl v is piecewise liner, s shown. () Determine the exct distnce trveled by the object on the time intervl 0 t 4. (b) Wht is the object s verge velocity on [2, 24]? (c) At wht time is the object s ccelertion gretest? (d) Suppose tht the velocity of the object is incresed by constnt vlue c for ll vlues of t. Wht vlue of c will mke the object s totl distnce trveled on [2, 24] be 20 meters?

4.4. THE FUNDAMENTAL THEOREM OF CALCULUS 263 5 2 9 6 3 m/min y = v(t) min 4 8 2 6 20 24 Figure 4.37: The velocity function of moving body. 2. A function f is given piecewise by the formul f (x) = x 2 + 2x +, if 0 x < 2 x + 3, if 2 x < 3 x 2 8x + 5, if 3 x 5 () Determine the exct vlue of the net signed re enclosed by f nd the x-xis on the intervl [2, 5]. (b) Compute the exct verge vlue of f on [0, 5]. (c) Find formul for function g on 5 x 7 so tht if we extend the bove definition of f so tht f (x) = g(x) if 5 x 7, it follows tht 7 f (x) dx = 0. 0 3. When n ircrft ttempts to climb s rpidly s possible, its climb rte (in feet per minute) decreses s ltitude increses, becuse the ir is less dense t higher ltitudes. Given below is tble showing performnce dt for certin single engine ircrft, giving its climb rte t vrious ltitudes, where c(h) denotes the climb rte of the irplne t n ltitude h. h (feet) 0 000 2000 3000 4000 5000 6000 7000 8000 9000 0,000 c (ft/min) 925 875 830 780 730 685 635 585 535 490 440 Let new function clled m(h) mesure the number of minutes required for plne t ltitude h to climb the next foot of ltitude. () Determine similr tble of vlues for m(h) nd explin how it is relted to the tble bove. Be sure to explin the units. (b) Give creful interprettion of function whose derivtive is m(h). Describe wht the input is nd wht the output is. Also, explin in plin English wht the function tells us.

264 4.4. THE FUNDAMENTAL THEOREM OF CALCULUS (c) Determine definite integrl whose vlue tells us exctly the number of minutes required for the irplne to scend to 0,000 feet of ltitude. Clerly explin why the vlue of this integrl hs the required mening. (d) Use the Riemnn sum M 5 to estimte the vlue of the integrl you found in (c). Include units on your result. 4. In Chpter, we showed tht for n object moving long stright line with position function s(t), the object s verge velocity on the intervl [, b] is given by AV [,b] = s(b) s(). b More recently in Chpter 4, we found tht for n object moving long stright line with velocity function v(t), the object s verge vlue of its velocity function on [, b] is v AVG[,b] = v(t) dt. b Are the verge velocity on the intervl [, b] nd the verge vlue of the velocity function on [, b] the sme thing? Why or why not? Explin.