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1 Chpter Five /SOLUTIONS Since the speed ws between nd mph during this five minute period, the fuel efficienc during this period is between 5 mpg nd 8 mpg. So the fuel used during this period is between gllons nd gllons. Thus, n underestimte of the fuel used is ( Fuel = ) =.7 gllons. 6 An overestimte of the fuel used is ( Fuel = ) =. gllons () The blck curve is for bos, the colored one for girls. The re under ech curve represents the chnge in growth in centimeters. Since men re generll tller thn women, the curve with the lrger re under it is the height velocit of the bos. (b) Ech squre below the height velocit curve hs re cm/r r = cm. Counting squres ling below the blck curve gives bout 4 cm. Thus, on verge, bos grow bout 4 cm between ges nd. (c) Counting squres ling below the blck curve gives bout cm growth for bos during their growth spurt. Counting squres ling below the colored curve gives bout 8 cm for girls during their growth spurt. (d) We cn mesure the difference in growth b counting squres tht lie between the two curves. Between ges nd.5, the verge girl grows fster thn the verge bo. Counting squres ields bout 5 cm between the colored nd blck curves for.5. Counting squres between the curves for.5 8 gives bout 8 squres. Thus, there is net increse of bos over girls b bout 8 5 = cm. 7. On the intervl t b, we hve Averge vlue of v(t) = b Since v(t) = s (t), b the Fundmentl Theorem of Clculus, we get: b v(t) dt. v(t) dt = (s(b) s()) = Averge velocit. b Solutions for Section 5.4 Eercises. () A grph of f () = sin( ) is shown in Figure 5.7. Since the derivtive f () is positive between = nd =, the chnge in f() is positive, so f() is lrger thn f(). Between = nd =.5, we see tht f () is negtive, so the chnge in f() is negtive; thus, f() is greter thn f(.5). Figure 5.7: Grph of f () = sin( )

2 (b) The chnge in f() between = nd = is given b the Fundmentl Theorem of Clculus: Since f() =, we hve Similrl, since we hve Since we hve The results re shown in the tble. f() f() = sin( )d =.. f() = +. =.. f() f() = sin( )d =.85, f() = +.85 =.85. f() f() = sin( )d =.774, f() = =.774. f() SOLUTIONS. We find the chnges in f() between n two vlues of b counting the re between the curve of f () nd the -is. Since f () is liner throughout, this is quite es to do. From = to =, we see tht f () outlines tringle of re / below the -is (the bse is nd the height is ). B the Fundmentl Theorem, so f() + f () d = f() f(), f () d = f() f() = = Similrl, between = nd = we cn see tht f () outlines rectngle below the -is with re, so f() = / = /. Continuing with this procedure (note tht t = 4, f () becomes positive), we get the tble below f() / / / / /. Since F () =, F (b) = f(t) dt. For ech b we determine F (b) grphicll s follows: F () = F () = F () + Are of rectngle = + = F () = F () + Are of tringle ( ) = +.5 =.5 F () = F () + Negtive of re of tringle =.5.5 = F (4) = F () + Negtive of re of rectngle = = F (5) = F (4) + Negtive of re of rectngle = = F (6) = F (5) + Negtive of re of tringle =.5 =.5 The grph of F (t), for t 6, is shown in Figure F (t) Figure 5.8 t

3 4 Chpter Five /SOLUTIONS 4. The grph of = e is bove the line = for. See Figure 5.9. Therefore The integrl ws evluted on clcultor. Are = (e ) d = = e = 5 ln() Figure 5.9 = = 5 Figure The grph of = 5 ln() is bove the line = for 5. See Figure 5.. Therefore The integrl ws evluted on clcultor. 6. Since for, we hve The integrl ws evluted on clcultor. 7. Since / / for, we hve The integrl ws evluted on clcultor. Are = (5 ln() ) d = Are = ( ) d =.8. Are = ( / / ) d = The grph of = sin + is bove the line =.5 for 6. See Figure 5.. Therefore The integrl ws evluted on clcultor. Are = sin +.5 d = = sin + =.5 6 Figure 5. = sin π/4 π = cos Figure 5.

4 5.4 SOLUTIONS 5 9. The grph of = cos t is bove the grph of = sin t for t π/4 nd = cos t is below = sin t for π/4 < t < π. See Figure 5.. Therefore, we find the re in two pieces: Are = π/4 The integrl ws evluted on clcultor. (cos t sin t) dt + π. We hve f(t) = F (t) = t, so b the Fundmentl Theorem of Clculus, (sin t cos t) dt =.88. π/4 t dt = F () F () = 9 = 8.. We hve f(t) = F (t) = /t, so b the Fundmentl Theorem of Clculus, dt = F (5) F () = ln 5 ln = ln 5. t. We hve f(t) = F (t) = cos t, so b the Fundmentl Theorem of Clculus, π/ cos t dt = F (π/) F () = =.. We hve f(t) = F (t) = 4t, so b the Fundmentl Theorem of Clculus, 4t dt = F () F ( ) = =. Notice in this cse the integrl is becuse the function being integrted, f(t) = 4t, is odd: the negtive contribution to the integrl from = to b = ectl cncels the positive. Problems 4. Note tht g() d = g(t) dt. Thus, we hve (f() + g()) d = 5. Note tht f(z) dz = f() d. Thus, we hve cf(z) dz = c f() d + g() d = 8 + =. f(z) dz = 8c. 6. Note tht (g()) d = (g(t)) dt. Thus, we hve ( (f()) (g()) ) d = (f()) d (g()) d = = We hve ( (f()) d f() d) = 8 = 5.

5 6 Chpter Five /SOLUTIONS 8. We write ( cg() + (c f()) ) d = ( cg() + c (f()) ) d = c g() d + c (f()) d = c g() d + c (f()) d = c () + c () = c + c. 9. The grph of = f( 5) is the grph of = f() shifted to the right b 5. Since the limits of integrtion hve lso shifted b 5 (to + 5 nd b + 5), the res corresponding to +5 f( 5) d nd f() d re the sme. Thus, f( 5) d = f() d = 8.. The res we computed re shded in Figure 5.. Since = nd = / re inverse functions, their grphs re reflections bout the line =. Similrl, = nd = / re inverse functions nd their grphs re reflections bout the line =. Therefore, the two shded res in Figure 5. re equl. = / / Figure 5.. We hve 8 = f() d = Since f is odd, f() d =, so f() d = 8.. Since f is even, f() d = (/)6 = nd. We hve 8 = f() d = Thus, since 4 d = 4(5 ) =, we hve f() d + f() d. f() d = (/)4 = 7. Therefore f() d (f() + 4) d = f() d = 7 = 4. f() d + f() d = 8 = 6, so f() d =. 4 d.

6 5.4 SOLUTIONS 7 4. We hve 4 f() d = 8/ = 4 nd f() d = 4 4 f() d = 4 5 f() d + f() d =. Thus 4 f() d = 4 =. 5. () See Figure 5.4. Since the shded region lies within rectngle of re, the re is less thn. (b) Since the re is given b the integrl Are = e / d =.856. = e / Figure (), since the integrnd is n odd function nd the limits re smmetric round. (b), since the integrnd is n odd function nd the limits re smmetric round. 7. () e d >, since e >, nd e d represents the re below the curve = e. (b) Looking t Figure 5.5, we see tht e d represents the re under the curve. This re is clerl greter thn zero, but it is less thn e since it fits inside rectngle of width nd height e (with room to spre). Thus < e d < e <. e = e = f() = + = Figure 5.5 Figure Notice tht f() = + is incresing for, since gets bigger s increses. This mens tht f() f() f(). For this function, f() = nd f() =. Thus, the re under f() lies between the re under the line = nd the re under the line = on the intervl. See Figure 5.6. Tht is, ( ) + d ( ). 9. () The integrnd is positive, so the integrl cnnot be negtive. (b) The integrnd. If the integrl =, then the integrnd must be identicll, which is not true.

7 8 Chpter Five /SOLUTIONS. We know tht we cn divide the integrl up s follows: f() d = f() d + f() d. The grph suggests tht f is n even function for, so f() d = f() d. Substituting this in to the preceding eqution, we hve f() d = f() d + f() d.. () For, we hve Averge vlue = f()d = (6) =. (b) If f() is even, the grph is smmetric bout the -is. For emple, see Figure 5.7. B smmetr, the re between = nd = is twice the re between = nd =, so Thus for, we hve Averge vlue = f()d = (6) =. f()d = () =. () 6 The grph confirms tht the verge vlue between = nd = is the sme s the verge vlue between = nd =, which is. Figure 5.7 Figure 5.8 (c) If f() is odd, then the grph is smmetric bout the origin. For emple, see Figure 5.8. B smmetr, the re bove the -is cncels out the re below the -is, so Thus for, we hve Averge vlue = f()d =. f()d = () =. () The grph confirms tht the verge vlue between = nd = is zero.. () Since the function is odd, the res bove nd below the -is cncel. Thus, so e d = e d = e d + e d, e d =.

8 5.4 SOLUTIONS 9 (b) For with n =, we hve =, =, =, =, nd =. See Figure 5.9. Thus, Left sum = f( ) + f( ) + f( ) = e + e + e =.445. (c) For, with n =, we hve =, =, =, =, nd =. See Figure 5.9. Thus, Left sum = f( ) + f( ) + f( ) = e () e ( ) e ( ) =.449. (d) No. The rectngles between nd re not the sme size s those between nd. See Figure 5.9. There re three rectngles with nonzero height on [, ] nd onl two on [, ]. Figure 5.9. () Yes. (b) No, becuse the sum of the left sums hs subdivisions. The result is the left sum pproimtion with subdivisions to f() d. 4. B the Fundmentl Theorem, f() f() = f () d, Since f () is negtive for, this integrl must be negtive nd so f() < f(). 5. First rewrite ech of the quntities in terms of f, since we hve the grph of f. If A nd A re the positive res shown in Figure 5.4: so Since Are A > Are A, nd therefore f() f() = f(4) f() = f(4) f() = A < f() f() < 4 4 f (t) dt = A f (t) dt = A A + A f A + A (t) dt = < A A + A A < < A f(4) f() < f(4) f(). 4 A A = f () Figure 5.4