Solução_Calculo_Stewart_6e

F.
258 ¤ CHAPTER 5 INTEGRALS
TX.10
In Exercises 39 and 40, let f(x) denote the integrand and F (x) its antiderivative (with C =0).
39. Let u =1+sinx. Thendu =cosxdx,so
cos xdx
√ = u −1/2 du =2u 1/2 + C =2 √ 1+sinx + C.
1+sinx
41. From the graph, it appears that the area under the curve y = x √ x between x =0
and x =4is somewhat less than half the area of an 8 × 4 rectangle, so perhaps
about 13 or 14. Tofind the exact value, we evaluate
4
0 x √ xdx= 4
0 x3/2 dx =
4
2
5 x5/2 = 2 5 (4)5/2 = 64 =12.8.
5
0
x
43. F (x) =
0
t 2
1+t dt ⇒ F 0 (x) = d x
3 dx 0
t 2 x2
dt =
1+t3 1+x 3
45. Let u = x 4 .Then du
dx =4x3 .Also, dg
dx = dg du
du dx ,so
g 0 (x) = d x
4
cos(t 2 ) dt = d u
cos(t 2 ) dt · du
dx 0
du 0
dx =cos(u2 ) du
dx =4x3 cos(x 8 ).
47. y =
x
e t 1
√ x t dt = e t x
√ x t dt +
1
e t
t dt = − √ x
1
e t
t dt + x
dy
dx = − d
√
x
e t
dx 1 t dt + d x
e t
dx 1 t dt .Letu = √ x.Then
1
e t
t dt
⇒
so dy
dx = −e√ x
2x + ex x .
d
√ x
e t
dx 1 t dt = d u
e t
dx 1 t dt = d u
e t du
du 1 t dt dx = eu u ·
1
2 √ x = e√ x
√ ·
x
1
2 √ x = e√ x
2x ,
49. If 1 ≤ x ≤ 3, then √ 1 2 +3≤ √ x 2 +3≤ √ 3 2 +3 ⇒ 2 ≤ √ x 2 +3≤ 2 √ 3,so
2(3 − 1) ≤ 3
√
x2 +3dx ≤ 2 √ 3(3 − 1); thatis,4 ≤ 3
√
x2 +3dx ≤ 4 √ 3.
1
1
51. 0 ≤ x ≤ 1 ⇒ 0 ≤ cos x ≤ 1 ⇒ x 2 cos x ≤ x 2 ⇒ 1
0 x2 cos xdx≤ 1
0 x2 dx = 1 3 x
3 1
= 1 [Property 7].
0 3
53. cos x ≤ 1 ⇒ e x cos x ≤ e x ⇒ 1
0 ex cos xdx≤ 1
0 ex dx =[e x ] 1 0 = e − 1
55. ∆x =(3− 0)/6 = 1 2 , so the endpoints are 0, 1 2 , 1, 3 2 , 2, 5 2 ,and3, and the midpoints are 1 4 , 3 4 , 5 4 , 7 4 , 9 4 ,and 11 4 .
The Midpoint Rule gives
3
0 sin(x3 ) dx ≈
6
i=1
f(x i) ∆x = 1 2
sin 1
4
3
+sin 3
4
3
+sin 5
4
3
+sin 7
4
3
+sin 9
4
3
+sin
11 3
≈ 0.280981.
4