Solução_Calculo_Stewart_6e

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