Solução_Calculo_Stewart_6e

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F. TX.10 SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ¤ 245 49. It appears that the area under the graph is about 2 3 oftheareaoftheviewing rectangle, or about 2 π ≈ 2.1. The actual area is 3 π sin xdx=[− cos 0 x]π 0 =(− cos π) − (− cos 0) = − (−1) + 1 = 2. 51. 2 −1 x3 dx = 1 4 x4 2 −1 =4− 1 4 = 15 4 =3.75 53. g(x) = 3x 2x u 2 − 1 0 u 2 +1 du = g 0 (x) =− (2x)2 − 1 (2x) 2 +1 · 2x u 2 − 1 3x u 2 +1 du + d − 1 dx (2x)+(3x)2 (3x) 2 +1 · 0 u 2 − 1 2x u 2 +1 du = − 0 u 2 − 1 3x u 2 +1 du + d dx (3x) =−2 · 4x2 − 1 4x 2 +1 +3· 9x2 − 1 9x 2 +1 0 u 2 − 1 u 2 +1 du ⇒ 55. y = x 3 √ x √ t sin tdt= 1√x √ t sin tdt+ x 3 y 0 = − 4√ x (sin √ x ) · 57. F (x) = =3x 7/2 sin(x 3 ) − sin √ x 2 4√ x x 1 F 00 (x) =f 0 (x) = 1 √ √ x √ x 3 √ t sin tdt= − 1 t sin tdt+ 1 t sin tdt d dx (√ x )+x 3/2 sin(x 3 d ) · x 3 = − 4√ x sin √ x dx 2 √ + x 3/2 sin(x 3 )(3x 2 ) x f(t) dt ⇒ F 0 (x) =f(x) = 1+(x2 ) 4 x 2 · x 2 d √ x 2 1+x 8 = dx 1 √ 1+u 4 du since f(t) = u · 2x = 2 √ 1+x 8 x 2 x 59. By FTC2, 4 1 f 0 (x) dx = f(4) − f(1),so17 = f(4) − 12 ⇒ f(4) = 17 + 12 = 29. t 2 1 ⇒ √ 1+u 4 du u ⇒ .SoF 00 (2) = √ 1+2 8 = √ 257. 61. (a) The Fresnel function S(x) = x 0 sin π 2 t2 dt has local maximum values where 0=S 0 (x) =sin π 2 t2 and S 0 changes from positive to negative. For x>0, this happens when π 2 x2 =(2n − 1)π [odd multiples of π] ⇔ x 2 =2(2n − 1) ⇔ x = √ 4n − 2, n any positive integer. For x 0. Differentiating our expression for S 0 (x), weget S 00 (x) =cos π x2 2 π x = πx cos π x2 .Forx>0, S 00 (x) > 0 where cos( π 2 2 2 2 x2 ) > 0 ⇔ 0 < π 2 x2 < π or 2 2n − 1 2 π< π 2 x2 < √ √ 2n + 1 2 π, n any integer ⇔ 0
F. 246 ¤ CHAPTER 5 INTEGRALS TX.10 − √ 4n − 3 >x>− √ 4n − 1, so the intervals of upward concavity for x9. (b) We can see from the graph that 1 fdt 3 0 < fdt 5 1 < fdt 7 3 < fdt 9 . 5 < fdt 1 , 7 Sog(1) = fdt 0 g(5) = 5 0 fdt= g(1) − 3 1 fdt + 5 3 fdt , andg(9) = 9 0 fdt= g(5) − 7 5 fdt + 9 7 fdt . Thus, g(1)
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