The Fundamental Theorem of Calculus

Transcription

1 Chpter 3 The Fundmentl Theorem of Clculus In this chpter we will formulte one of the most importnt results of clculus, the Fundmentl Theorem. This result will link together the notions of n integrl nd derivtive. Using this result will llow us to replce the technicl clcultions of Chpter 2 b much simpler procedures involving ntiderivtives of function. 3. The definite integrl In Chpter 2, we defined the definite integrl, I, of function f() > on n intervl [, b] s the re under the grph of the function over the given intervl b. We used the nottion I = f()d to represent tht quntit. We lso set up technique for computing res: the procedure for clculting the vlue of I is to write down sum of res of rectngulr strips nd to compute limit s the number of strips increses: I = f()d = lim N k= N f( k ), (3.) where N is the number of strips used to pproimte the region, k is n inde ssocited with the k th strip, nd = k+ k is the width of the rectngle. As the number of strips increses (N ), nd their width decreses ( ), the sum becomes better nd better pproimtion of the true re, nd hence, of the definite integrl, I. Emple of such clcultions (tedious s the were) formed the min theme of Chpter 2. We cn generlize the definite integrl to include functions tht re not strictl positive, s shown in Figure 3.. To do so, note wht hppens s we incorporte strips corresponding to regions of the grph below the is: These re ssocited with negtive vlues of the function, so tht the quntit f( k ) in the bove sum would be negtive for ech rectngle in the negtive portions of the function. This mens tht regions of the grph below the is will contribute negtivel to the net vlue of I. 43

2 44 Chpter 3. The Fundmentl Theorem of Clculus If we refer to A s the re corresponding to regions of the grph of f() bove the is, nd A 2 s the totl re of regions of the grph under the is, then we will find tht the vlue of the definite integrl I shown bove will be I = A A 2. Thus the notion of re under the grph of function must be interpreted little crefull when the function dips below the is. =f() =f() () (b) =f() =f() (c) (d) b c Figure 3.. () If f() is negtive in some regions, there re terms in the sum (3.) tht crr negtive signs: this hppens for ll rectngles in prts of the grph tht dip below the is. (b) This mens tht the definite integrl I = f()d will correspond to the difference of two res, A A 2 where A is the totl re (drk) of positive regions minus the totl re (light) of negtive portions of the grph. Properties of the definite integrl: (c) illustrtes Propert. (d) illustrtes Propert Properties of the definite integrl The following properties of definite integrl stem from its definition, nd the procedure for clculting it discussed so fr. For emple, the fct tht summtion stisfies the distributive

3 3.3. The re s function 45 propert mens tht n integrl will stisf the sme the sme propert. We illustrte some of these in Fig c f()d =, f()d = Cf()d = C f()d + (f()+g())d = f()d = b c b f()d, f()d. f()+ f()d, g()d, Propert sttes tht the re of region with no width is zero. Propert 2 shows how region cn be broken up into two pieces whose totl re is just the sum of the individul res. Properties 3 nd 4 reflect the fct tht the integrl is ctull just sum, nd so stisfies properties of simple ddition. Propert 5 is obtined b noting tht if we perform the summtion in the opposite direction, then we must replce the previous rectngle width given b = k+ k b the new width which is of opposite sign: k k+. This ccounts for the sign chnge shown in Propert The re s function In Chpter 2, we investigted how the re under the grph of function chnges s one of the endpoints of the intervl moves. We defined function tht represents the re under the grph of function f, from some fied strting point, to n endpoint. A() = f(t) dt. This endpoint is considered s vrible 2, i.e. we will be interested in the w tht this re chnges s the endpoint vries (Figure 3.2()). We will now investigte the interesting connection between A() nd the originl function, f(). We would like to stud how A() chnges s is incresed ever so slightl. Let = h represent some (ver smll) increment in. (Cution: do not confuse h with height here. It is ctull step size long the is.) Then, ccording to our definition, A( + h) = +h f(t) dt. 2 Recll tht the dumm vrible t inside the integrl is just plce holder, nd is used to void confusion with the endpoint of the integrl ( in this cse). Also note tht the vlue of A() does not depend in n w on t, so n letter or smbol in its plce would do just s well.

4 46 Chpter 3. The Fundmentl Theorem of Clculus =f() =f() A() A(+h) () (b) +h =f() =f() A(+h) A() f() (c) +h (d) h Figure 3.2. When the right endpoint of the intervl moves b distnce h, the re of the region increses from A() to A( + h). This leds to the importnt Fundmentl Theorem of Clculus, given in Eqn. (3.2). In Figure 3.2()(b), we illustrte the res represented b A() nd b A( + h), respectivel. The difference between the two res is thin sliver (shown in Figure 3.2(c)) tht looks much like rectngulr strip (Figure 3.2(d)). (Indeed, if h is smll, then the pproimtion of this sliver b rectngle will be good.) The height of this sliver is specified b the function f evluted t the point, i.e. b f(), so tht the re of the sliver is pproimtel f() h. Thus, A( + h) A() f()h or A( + h) A() f(). h As h gets smll, i.e. h, we get better nd better pproimtion, so tht, in the limit, A( + h) A() lim = f(). h h The rtio bove should be recognizble. It is simpl the derivtive of the re function, i.e. f() = da d = lim h A( + h) A(). (3.2) h

5 3.4. The Fundmentl Theorem of Clculus 47 We hve just given simple rgument in support of n importnt result, clled the Fundmentl Theorem of Clculus, which is restted below The Fundmentl Theorem of Clculus 3.4. Fundmentl theorem of clculus: Prt I Let f() be bounded nd continuous function on n intervl [, b]. Let Then for < < b, A() = f(t) dt. da d = f(). In other words, this result ss tht A() is n ntiderivtive of the originl function, f() 3. Proof See bove rgument. nd Figure Emple: n ntiderivtive Recll the connection between functions nd their derivtives. Consider the following two functions: g () = 2 2, g 2 = Clerl, both functions hve the sme derivtive: g () =g 2 () =. We would s tht 2 /2 is n ntiderivtive of nd tht ( 2 /2) + is lso n ntiderivtive of. In fct, n function of the form g() = C where C is n constnt is lso n ntiderivtive of. This emple illustrtes tht dding constnt to given function will not ffect the vlue of its derivtive, or, stted nother w, ntiderivtives of given function re defined onl up to some constnt. We will use this fct shortl: if A() nd F () re both ntiderivtives of some function f(), then A() =F ()+C. 3 We often write ntiderivtive, with no hphen.

6 48 Chpter 3. The Fundmentl Theorem of Clculus Fundmentl theorem of clculus: Prt II Let f() be continuous function on [, b]. Suppose F () is n ntiderivtive of f(). Then for b, A() = f(t) dt = F () F (). Proof From comments bove, we know tht function f() could hve mn different ntiderivtives tht differ from one nother b some dditive constnt. We re told tht F () is n ntiderivtive of f(). But from Prt I of the Fundmentl Theorem, we know tht A() is lso n ntiderivtive of f(). It follows tht A() = However, b propert of definite integrls, f(t) dt = F ()+ C, where C is some constnt. (3.3) A() = f(t) =F ()+C =. Thus, C = F (). Replcing C b F () in eqution 3.3 leds to the desired result. Thus A() = f(t) dt = F () F (). Remrk : Implictions This theorem hs tremendous implictions, becuse it llows us to use powerful new tool in determining res under curves. Insted of the drudger of summtions in order to compute res, we will be ble to use shortcut: find n ntiderivtive, evlute it t the two endpoints, b of the intervl of interest, nd subtrct the results to get the re. In the cse of elementr functions, this will be ver es nd convenient. Remrk 2: Nottion We will often use the nottion F (t) = F () F () to denote the difference in the vlues of function t two endpoints.

7 3.5. Review of derivtives (nd ntiderivtives) Review of derivtives (nd ntiderivtives) B remrks bove, we see tht integrtion is relted to nti-differentition. This motivtes review of derivtives of common functions. Tble 3. lists functions f() nd their derivtives f () (in the first two columns) nd functions f() nd their ntiderivtives F () in the subsequent two columns. These will prove ver helpful in our clcultions of bsic integrls. function derivtive function ntiderivtive f() f () f() F () C C C C n n n m m+ m+ sin() cos() cos(b) (/b) sin(b) cos() sin() sin(b) (/b) cos(b) tn() sec 2 () sec 2 (b) (/b) tn(b) e k ke k e k e k /k ln() rctn() rcsin() ln() rctn() 2 2 rcsin() Tble 3.. Common functions nd their derivtives (on the left two columns) lso result in corresponding reltionships between functions nd their ntiderivtives (right two columns). In this tble, we ssume tht m,b,k. Also, when using ln() s ntiderivtive for /, we ssume tht >. As n emple, consider the polnomil p() = This polnomil could hve mn other terms (or even n infinite number of such terms, s we discuss much lter, in Chpter ). Its ntiderivtive cn be found esil using the power rule together with the properties of ddition of terms. Indeed, the ntiderivtive is F () =C

8 5 Chpter 3. The Fundmentl Theorem of Clculus This cn be checked esil b differentition Emples: Computing res with the Fundmentl Theorem of Clculus 3.6. Emple : The re under polnomil Consider the polnomil p() = (Here we hve tken the first few terms from the emple of the lst section with coefficients ll set to.) Then, computing I = p() d leds to I = ( ) d =( ) Emple 2: Simple res = Determine the vlues of the following definite integrls b finding ntiderivtives nd using the Fundmentl Theorem of Clculus:. I = 2. I = 3. I = 4. I = Solutions π 2 d, ( 2 ) d, e 2 d, ( ) sin d, 2. An ntiderivtive of f() = 2 is F () = ( 3 /3), thus I = 2 d = F () 4 In fct, it is ver good prctice to perform such checks. = (/3)( 3 ) = 3 (3 ) = 3.

9 3.6. Emples: Computing res with the Fundmentl Theorem of Clculus 5 2. An ntiderivtive of f() = ( 2 ) is F () = ( 3 /3), thus I = ( 2 ) d = F () = ( ( 3 /3) ) = ( ( 3 /3) ) ( ( ) (( ) 3 /3) ) =4/3 See comment below for simpler w to compute this integrl. 3. An ntiderivtive of e 2 is F () = ( /2)e 2. Thus, I = e 2 d = F () =( /2)(e 2 ) =( /2)(e 2 e 2 ). 4. An ntiderivtive of sin(/2) is F () = cos(/2)/(/2) = 2 cos(/2). Thus π ( ) π I = sin d = 2 cos(/2) 2 2(cos(π/2) cos()) = 2( ) = 2. Comment: The evlution of Integrl 2. in the emples bove is trick onl in tht signs cn esil get grbled when we plug in the endpoint t -. However, we cn simplif our work b noting the smmetr of the function f() = 2 on the given intervl. As shown in Fig 3.3, the res to the right nd to the left of =rethe sme for the intervl. This stems directl from the fct tht the function considered is even 5. Thus, we cn immeditel write I = ( 2 ) d =2 ( 2 ) d =2 ( ( 3 /3) ) =2 ( ( 3 /3) ) =4/3 Note tht this clcultion is simpler since the endpoint t =is trivil to plug in. = 2 Figure 3.3. We cn eploit the smmetr of the function f() = 2 in the second integrl of Emples We cn integrte over nd double the result. We stte the generl result we hve obtined, which holds true for n function with even smmetr integrted on smmetric intervl bout =: If f() is n even function, then f() d =2 f() d (3.4) 5 Recll tht function f() is even if f() =f( ) for ll. A function is odd if f() = f( ).

10 52 Chpter 3. The Fundmentl Theorem of Clculus Emple 3: The re between two curves The definite integrl is n re of somewht specil tpe of region, i.e., n is, two verticl lines ( = nd = b) nd the grph of function. However, using dditive (or subtrctive) properties of res, we cn generlize to computing res of other regions, including those bounded b the grphs of two functions. () Find the re enclosed between the grphs of the functions = 3 nd = /3 in the first qudrnt. (b) Find the re enclosed between the grphs of the functions = 3 nd = in the first qudrnt. (c) Wht is the reltionship of these two res? Wht is the reltionship of the functions = 3 nd = /3 tht leds to this reltionship between the two res? = = /3 = A A Figure 3.4. In Emple 3, we compute the res A nd A 2 shown bove. Solution () The two curves, = 3 nd = /3, intersect t =ndt =in the first qudrnt. Thus the intervl tht we will be concerned with is < <. On this intervl, /3 > 3, so tht the re we wnt to find cn be epressed s: Thus, A = A = 4/3 4/3 ( /3 3) d. 4 4 = = 2. (b) The two curves = 3 nd = lso intersect t =ndt =in the first qudrnt, nd on the intervl < < we hve > 3. The re cn be represented s ( A 2 = 3 ) d.

11 3.7. Qulittive ides 53 A 2 = = 2 4 = 4. (c) The re clculted in () is twice the re clculted in (b). The reson for this is tht /3 is the inverse of the function 3, which mens geometricll tht the grph of /3 is the mirror imge of the grph of 3 reflected bout the line =. Therefore, the re A between = /3 nd = 3 is twice s lrge s the re A 2 between = nd = 3 clculted in prt (b): A =2A 2 (see Figure 3.4) Emple 4: Are of lnd Find the ect re of the piece of lnd which is bounded b the is on the west, the is in the south, the lke described b the function = f() = + (/) 2 in the north nd the line = in the est. Solution The re is A = ( ( ) ) 2 + d. = ( ( ) ) + 2 d. Note tht the multiplictive constnt (/) is not ffected b integrtion. The result is A = + 3 ( ) 3 = Qulittive ides In some cses, we re given sketch of the grph of function, f(), from which we would like to construct sketch of the ssocited function A(). This sketching skill is illustrted in the figures shown in this section. Suppose we re given function s shown in the top left hnd pnel of Figure 3.5. We would like to ssemble sketch of A() = f(t)dt which corresponds to the re ssocited with the grph of the function f. As moves from left to right, we show how the re ccumulted long the grph grdull chnges. (See A() in bottom pnels of Figure 3.5): We strt with no re, t the point = (since, b definition A() = ) nd grdull build up to some net positive mount, but then we encounter portion of the grph of f below the is, nd this subtrcts from the mount ccrued. (Hence the grph of A() hs little pek tht corresponds to the point t which f =.) Ever time the function f() crosses the is, we see tht A() hs either mimum or minimum vlue. This fits well with our ide of A() s the ntiderivtive of f(): Plces where A() hs criticl point coincide with plces where da/d = f() =.

12 54 Chpter 3. The Fundmentl Theorem of Clculus f() f() A() A() () f() (b) f() A() A() (c) (d) Figure 3.5. Given function f(), we here show how to sketch the corresponding re function A(). (The reltionship is tht f() is the derivtive of A() Sketching the function A() is thus nlogous to sketching function g() when we re given sketch of its derivtive g (). Recll tht this ws one of the skills we built up in lerning the connection between functions nd their derivtives in first semester clculus course. Remrks The following remrks m be helpful in gining confidence with sketching the re function A() = f(t) dt, from the originl function f():. The endpoint of the intervl, on the is indictes the plce t which A() =. This follows from Propert of the definite integrl, i.e. from the fct tht A() = f(t) dt =. 2. Whenever f() is positive, A() is n incresing function - this follows from the fct tht the re continues to ccumulte s we sweep cross positive regions of f().

13 3.7. Qulittive ides 55 f() g() Figure 3.6. Given function f() (top, solid line), we ssemble plot of the corresponding function g() = f(t)dt (bottom, solid line). g() is n ntiderivtive of f(). Whether f() is positive (+) or negtive (-) in portions of its grph, determines whether g() is incresing or decresing over the given intervls. Plces where f() chnges sign correspond to mim nd minim of the function g() (Two such plces re indicted b dotted verticl lines). The bo in the middle of the sketch shows configurtions of tngent lines to g() bsed on the sign of f(). Where f() =, those tngent lines re horizontl. The function g() is drwn s smooth curve whose direction is prllel to the tngent lines shown in the bo. While the function f() hs mn ntiderivtives (e.g., dshed curve prllel to g()), onl one of these stisfies g() = s required b Propert of the definite integrl. (See dshed verticl line t = ). This determines the height of the desired function g(). 3. Wherever f(), chnges sign, the function A() hs locl minimum or mimum. This mens tht either the re stops incresing (if the trnsition is from positive to negtive vlues of f), or else the re strts to increse (if f crosses from negtive to positive vlues). 4. Since da/d = f() b the Fundmentl Theorem of Clculus, it follows tht (tk-

14 56 Chpter 3. The Fundmentl Theorem of Clculus ing derivtive of both sides) d 2 A/d 2 = f (). Thus, when f() hs locl mimum or minimum, (i.e. f () = ), it follows tht A () =. This mens tht t such points, the function A() would hve n inflection point. Given function f(), Figure 3.6 shows in detil how to sketch the corresponding function g() = 3.7. Emple: sketching A() f(t)dt. Consider the f() whose grph is shown in the top prt of Figure 3.7. Sketch the corresponding function g() = f()d. f() + + g() Figure 3.7. The originl functions, f() is shown bove. The corresponding functions g() is drwn below. Solution See Figure Some fine print The Fundmentl Theorem hs number of restrictions tht must be stisfied before its results cn be pplied. In this section we look t some emples in which cre must be used.

15 3.8. Some fine print Function unbounded I Consider the definite integrl 2 d. The function f() = is undefined t =, nd unbounded on n intervl tht contins the point =. Hence, we cnnot evlute this integrl using the Fundmentl theorem, nd indeed, we s tht this integrl does not eist Function unbounded II Consider the definite integrl 2 d. This function is lso undefined (nd hence not continuous) t =. The Fundmentl Theorem of Clculus cnnot be pplied. Technicll, lthough one cn go through the motions of computing n ntiderivtive, evluting it t both endpoints, nd getting numericl nswer, the result so obtined would be simpl wrong. We s tht his integrl does not eist Emple: Function discontinuous or with distinct prts Suppose we re given the integrl I = 2 d. This function is ctull mde up of two distinct prts, nmel { if > f() = if <. The integrl I must therefore be split up into two prts, nmel We find tht I = I = d = ( ) d + 2 d. [ = ] [ ] = Function undefined Now let us emine the integrl /2 d.

16 58 Chpter 3. The Fundmentl Theorem of Clculus = 2 Figure 3.8. In this emple, to compute the integrl over the intervl 2, we must split up the region into two distinct prts. We see tht there is problem here. Recll tht /2 =. Hence, the function is not defined for < nd the intervl of integrtion is inpproprite. Hence, this integrl does not mke sense Infinite domin ( improper integrl ) Consider the integrl I = e r d, where r>, nd b> re constnts. Simple integrtion using the ntiderivtive in Tble 3. (for k = r) leds to the result I = e r r b = ( e rb e ) = ( e rb ). r r This is the re under the eponentil curve between =nd = b. Now consider wht hppens when b, the upper endpoint of the integrl increses, so tht b. Then the vlue of the integrl becomes I = lim b e r ( d = lim e rb ) = b r r ( ) = r. (We used the fct tht e rb s b.) We hve, in essence, found tht I = e r d = r. (3.5) An integrl of the form (3.5) is clled n improper integrl. Even though the domin of integrtion of this integrl is infinite, (, ), observe tht the vlue we computed is finite, so long s r. Not ll such integrls hve bounded finite vlue. Lerning to distinguish between those tht do nd those tht do not will form n importnt theme in Chpter.

17 3.9. Summr 59 Regions tht need specil tretment So fr, we hve lerned how to compute res of regions in the plne tht re bounded b one or more curves. In ll our emples so fr, the bsis for these clcultions rests on imgining rectngles whose heights re specified b one or nother function. Up to now, ll the rectngulr strips we considered hd bses (of width ) on the is. In Figure 3.9 we observe n emple in which it would not be possible to use this technique. We re Δ =g() Figure 3.9. The re in the region shown here is best computed b integrting in the direction. If we do so, we cn use the curved boundr s single function tht defines the region. (Note tht the curve cnnot be epressed in the form of function in the usul sense, = f(), but it cn be epressed in the form of function = f().) sked to find the re between the curve 2 + =nd the is. However, one nd the sme curve, 2 + =forms the boundr from both the top nd the bottom of the region. We re unble to set up series of rectngles with bses long the is whose heights re described b this curve. This mens tht our definite integrl (which is rell just convenient w of crring out the process of re computtion) hs to be hndled with cre. Let us consider this problem from new ngle, i.e. with rectngles bsed on the is, we cn chieve the desired result. To do so, let us epress our curve in the form = g() = 2. Then, plcing our rectngles long the intervl < < on the is (ech hving bse of width ) leds to the integrl ( ) I = g() d = ( 2 2 )d = 2 3 = = Summr In this chpter we first recpped the definition of the definite integrl in Section 3., reclled its connection to n re in the plne under the grph of some function f(), nd emined its bsic properties. If one of the endpoints, of the integrl is llowed to vr, the re it represents, A(), becomes function of. Our construction in Figure 3.2 showed tht there is connection between the derivtive A () of the re nd the function f(). Indeed, we showed tht A () =f() nd rgued tht this mkes A() n ntiderivtive of the function f().

18 6 Chpter 3. The Fundmentl Theorem of Clculus This importnt connection between integrls nd ntiderivtives is the cru of Integrl Clculus, forming the Fundmentl Theorem of Clculus. Its significnce is tht finding res need not be s tedious nd lbored s the clcultion of Riemnn sums tht formed the bulk of Chpter 2. Rther, we cn tke shortcut using ntidifferentition. Motivted b this ver importnt result, we reviewed some common functions nd derivtives, nd used this to relte functions nd their ntiderivtives in Tble 3.. We used these ntiderivtives to clculte res in severl emples. Finll, we etended the tretment to include qulittive sketches of functions nd their ntiderivtives. As we will see in upcoming chpters, the ides presented here hve much wider rnge of pplicbilit thn simple re clcultions. Indeed, we will shortl show tht the sme concepts cn be used to clculte net chnges in continull vring processes, to compute volumes of vrious shpes, to determine displcement from velocit, mss from densities, s well s host of other quntities tht involve process of ccumultion. These ides will be investigted in Chpters 4, nd 5.