Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 5.5 THE SUBSTITUTION RULE ¤ 253
53. Let u =1+2x 3 ,sodu =6x 2 dx. Whenx =0, u =1;whenx =1, u =3.Thus,
1 x2 1+2x 35 dx = 3
u5 1
du
= 1 1 u6 3
= 1
0 1 6 6 6 1 36 (36 − 1 6 )= 1 728
(729 − 1) = = 182 .
36 36 9
55. Let u = t/4, sodu = 1 dt. Whent =0, u =0;whent = π, u = π/4. Thus,
4
π
0 sec2 (t/4) dt = π/4
sec 2 u (4 du) =4 tan u π/4
=4 tan π − tan 0 =4(1− 0) = 4.
0 0
4
π/6
57.
−π/6 tan3 θdθ=0by Theorem 7(b), since f(θ) =tan 3 θ is an odd function.
59. Let u =1/x, sodu = −1/x 2 dx. Whenx =1, u =1;whenx =2, u = 1 2 . Thus,
2
e 1/x
1
x 2 dx = 1/2
1
e u (−du) =− e u 1/2
1
= −(e 1/2 − e) =e − √ e.
61. Let u =1+2x,sodu =2dx. Whenx =0, u =1;whenx =13, u =27.Thus,
13
dx
27
= u (1 −2/3
1
du 1
= · 3u1/3 27
= 3 (3 − 1) = 3.
+ 2x) 2 2 2 2
1
0
3
1
63. Let u = x 2 + a 2 ,sodu =2xdxand xdx= 1 2 du. Whenx =0, u = a2 ;whenx = a, u =2a 2 . Thus,
a
0
1
x 2a
2
x 2 + a 2 dx = u 1/2 1
du
a 2 2
= 1 2
0
0
2
3 u3/2 2a 2
a 2 =
1
u3/2 2a 2
3
= 1 (2a 2
a 2 3
) 3/2 − (a 2 ) 3/2
= 1 2 √
3
2 − 1 a 3
65. Let u = x − 1,sou +1=x and du = dx. Whenx =1, u =0;whenx =2, u =1. Thus,
2
x √ 1
x − 1 dx = (u +1) √ 1
udu= (u 3/2 + u 1/2 2
) du =
5 u5/2 + 2 u3/2 1
= 2 + 2 = 16 .
3 5 3 15
67. Let u =lnx, sodu = dx x .Whenx = e, u =1;whenx = e4 ; u =4.Thus,
0
e
4
e
0
dx
4
x √ ln x = u −1/2 du =2
u 1/2 4
=2(2− 1) = 2.
1
1
69. Let u = e z + z,sodu =(e z +1)dz. Whenz =0, u =1;whenz =1, u = e +1. Thus,
1
e z
+1
e+1
e z + z dz = 1
e+1
u du = ln |u| =ln|e +1| − ln |1| =ln(e +1).
1
1
71. From the graph, it appears that the area under the curve is about
1+ a little more than 1 2 · 1 · 0.7 , or about 1.4. The exact area is given by
A = 1 √
0 2x +1dx. Letu =2x +1,sodu =2dx. The limits change to
2 · 0+1=1and 2 · 1+1=3,and
A = √
3
3
1 u 1 du √ √
2
= 1 2
2 3 u3/2 = 1 3 3 3 − 1 = 3 −
1
≈ 1.399.
3
1
73. First write the integral as a sum of two integrals:
I = 2
−2 (x +3)√ 4 − x 2 dx = I 1 + I 2 = 2
−2 x √ 4 − x 2 dx + 2
−2 3 √ 4 − x 2 dx. I 1 =0by Theorem 7(b), since
f(x) =x √ 4 − x 2 is an odd function and we are integrating from x = −2 to x =2. We interpret I 2 as three times the area of
a semicircle with radius 2,soI =0+3· 1
2
π · 2
2 =6π.