Chapter 6 Notes, Larson/Hostetler 3e

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1 Contents 6. Antiderivtives nd the Rules of Integrtion Are nd the Definite Integrl Are Reimnn Sums nd Definite Integrls Reimnn Sums Properties of the definite integrls The Fundmentl Theorem of Clculus Integrtion by Substitution

2 6. Antiderivtives nd the Rules of Integrtion Consider the following derivtives: F () = F () = + 7 F () = F () = F () = F () = If we wnt to work bckwrds nd sk the question, Wht function hs derivtive F () = we cn see tht there re mny possible nswers. When we sk this question we re solving for n ntiderivtive. This is more often clled n integrl. So if f() =, then F () = + C is the ntiderivtive (or integrl) of f(). NOTE: Since the ntiderivtive is generl epression you MUST include the +C in every ntiderivtive. Nottion: We would like to be ble to epress these integrls (ntiderivtives) mthemticlly without hving to write out the words so we hve the integrl sign:. In the previous emple we would write d = + C A word bout nottion nd the use of d: You will remember tht we cn write the derivtive of y = f() mny equivlent wys: y = f () = dy d With this we cn get n ide of why the integrl hs the d: dy d = f () dy = f ()d If we consider dy to be the derivtive of y then the integrl (ntiderivtive) of dy is y. Similrly f () is the derivtive of f() then the integrl (ntiderivtive) of f () is f(). Now integrte: dy = f ()d y = f() + C

3 Formuls. Integrl of constt:. The Power Rule. Constnt Multiple Rule k d = k + C n d = n+ n + + C; k f()d = k f()d n R, n. The Sum Rule f() ± g()d = f()d ± g() d. Eponentil Function e d = e + C 6. Integrl of f() = d = ln + C Emple 6... Integrte the following:. d = d = d = d = / d = ( + + )d = ( + + e ) d = Initil Vlue Problems If you know something obut the eqution you cn solve for C. This piece of informtion is clled n initil vlue. Emple 6... f () = +, f() = 9

4 6. Are nd the Definite Integrl 6.. Are Consider the problem of finding the re under the curve on the function y = + over the domin [, ]. We cn pproimte this re by using fmilir shpe, the rectngle. If we divide the domin intervl into severl pieces, then drw rectngles hving the width of the pieces, nd the height of the curve, we cn get rough ide of the totl re. For emple suppose we divide the intervl [, ] into equl subintervls of length = b n, i.e., ech of width. The intervls re [,.], [.,.8], [.8,.], [.,.6], [.6,.]. The tble below shows the vlues obtined when y () is evluted t the corresponding points. y = Plotting these points yields the following grph. y y = +

5 If we find the minimum vlue in the subintervl, nd use this s our height for tht rectngle, we hve wht is known s n inscribed rectngle. See the grph below. y y = + Now ech of the bove rectngles hs the ect sme width, nmely /. For this function the height of ech rectngle is given by clculting the vlue of the function t the right hnd endpoint of ech subintervl. The re under the curve, cn then be pproimted by dding the res of ll the rectngles together. Notice tht when using the minimum vlues, i.e. using inscribed rectngles, we rrive t n estimte tht is lower thn the ctul re under the curve. Hence, this method results in wht is known s the lower sum or n underestimte. Let s clculte this estimte using the right endpoints R = F ( i ) i i= ( ) ( ) ( ) ( ) 6 8 = f + f + f + f + f () = [ ( ) ( ) ( ) ( ) ] 6 8 f + f + f + f + f () = [ ] = [6.] = 6.8

6 You cn lso clculte n estimte using the mimum vlue in the subintervl nd using it s the height of the rectngles. These rectngles re known s circumscribed rectngles. The resulting re pproimtion will be greter thn the re under the curve. Consequently, we cll this type of sum n upper sum or n oversetimte. y y = + Clculuting the sum with the left endpoints we get: L = F ( i ) i i= = f () ( ) ( ) ( ) ( ) f + f + f + f = [ ( ) ( ) ( ) ( )] 6 8 f () + f + f + f + f = [ ] = [.] = 8.8 From the two clcultions bove we cn conclude tht the re of the curve lies some where between the two pproimtions, i.e. 6.8 < re of region < 8.8 6

7 Another method tht cn yield better pproimtion is known s the midpoint rule. In the midpoint rule, you choose the vlue ectly in the middle of the subintervl to use in clculting the height of the rectngle; resulting in some rectngles being both inscribed nd circumscribed. y y = + Let s clculte the bove estimte: The Midpoint Sum. M = F ( i ) i= ( ) ( ) ( ) ( ) ( ) 7 9 = f + f + f + f + f = [ ( ) ( ) ( ) ( ) ( )] 7 9 f + f + f + f + f = [ ] = [8.] = 7.6 NOTE: The smller the subintervls, the better the pproimtion will be. This is becuse, the function s vlues re chnging less in the subintervl, i.e. the vlue of the function is firly constnt in ech subintervl. Consequently, we re not pproimting by such rough mount ech time. For emple, here is the sme region divided into rectngle insted of. Note tht the error is minute compred with the previous work. y y = + Ech of the bove processes (lower sum, upper sum, midpoint sum) re just pproimtions. They re not ect. 7

8 Emple 6... Let f() = nd compute the upper nd lower sums of f over the intervl [, ], using. two subintervls of equl length (n = ). five subintervls of equl length (n = ) 8

9 6. Reimnn Sums nd Definite Integrls 6.. Reimnn Sums Definition 6.. Let f() be continuous function defined on [, b] where [, b] is prtitioned into n subintervls ech of width = b, then the definite integrl of f from to b is denoted by n b f()d = lim n [f( ) + f( ) + + f( n ) ], where = b n, i = + i. Emple 6... Set up definite integrl tht yields the re of the region shown for f() = y y = 6 Are = 6 d Emple 6... Set up definite integrl tht yields the re of the region f(y) = y y Are = y dy 9

10 6.. Properties of the definite integrls. c. e. f. b b f()d = b. cd = c(b ); c = constnt function. d. c f() ± g()d = f()d + b b f()d = f()d ± b c b f()d g() d b f()d = b b c f()d = c b f()d = f()d Emple 6... Given tht evlute the following: f()d = 7, f()d =, g()d =, nd g()d = 8, ) f()d = ) f()d = ) 6g()d = ) [f() + g()]d = ) d =

11 6. The Fundmentl Theorem of Clculus Fundmentl Theorem : If f() is continuous on [, b], then b f()d = F (b) F () where F () is ny ntiderivtive of f() on [, b], tht is, function such tht F () = f(). Emple 6... Evlute the following integrls ) ( + )d = ) (y + e y )dy = ) (8 )d = ) t dt =

12 Emple 6... Find the re of the region under the grph of the function f on the intervl [, b].. f() = + ; [, ]. f() = ; [, ]

13 6. Integrtion by Substitution Recll: The Chin Rule Suppose we wnt to find the derivtive of y = f() = ( + 7). To do this we need to use the chin rule becuse we hve n inside function u = + 7 nd n outside function f(u) = u. dy d = ( + 7) 9 () = ( + 7) 9 () When we integrte we cn use the chin rule in reverse. ( + 7) 9 ()d = ( + 7) + C The trick is to recognize the integrl. To do this we use trick known s u - substitution. We will solve the integrl gin but this time without knowing the nswer. ( + 7) 9 ()d Here we hve function inside the prentheses nd we will cll tht function u: u = + 7 Emple 6... ( + ) d

14 Emple 6... ( ) 9 d Emple 6... e d

15 Emple 6... (e + e ) d Emple d

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