Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 5.2 THEDEFINITEINTEGRAL ¤ 241
35.
3
0
1
2 x − 1 dx can be interpreted as the area of the triangle above the x-axis
minus the area of the triangle below the x-axis; that is,
1
(1)
1
2 2 −
1
(2)(1) = 1 − 1=− 3 . 0
2 4 4
37.
0
√
−3 1+ 9 − x
2
dx can be interpreted as the area under the graph of
f(x) =1+ √ 9 − x 2 between x = −3 and x =0. This is equal to one-quarter
the area of the circle with radius 3, plus the area of the rectangle, so
0
√
−3 1+ 9 − x
2
dx = 1 π · 4 32 +1· 3=3+ 9 π. 4
39.
2
|x| dx can be interpreted as the sum of the areas of the two shaded
−1
triangles; that is, 1 2 (1)(1) + 1 2 (2)(2) = 1 2 + 4 2 = 5 2 . 0
π
41.
π sin2 x cos 4 xdx=0since the limits of intergration are equal.
43.
45.
47.
1 (5 − 0 6x2 ) dx = 1
5 dx − 6 1
0 0 x2 dx =5(1− 0) − 6
1
3 =5− 2=3
3
1 ex +2 dx = 3
1 ex · e 2 dx = e 2 3
1 ex dx = e 2 (e 3 − e) =e 5 − e 3
2 f(x) dx + 5
f(x) dx − −1
f(x) dx = 5
f(x) dx + −2
f(x) dx [byProperty5andreversinglimits]
−2 2 −2 −2 −1
= 5
f(x) dx [Property 5]
−1
9
49. [2f(x)+3g(x)] dx =2 9
f(x) dx +3 9
g(x) dx = 2(37) + 3(16) = 122
0 0 0
51. Using Integral Comparison Property 8, m ≤ f(x) ≤ M ⇒ m(2 − 0) ≤ 2
f(x) dx ≤ M(2 − 0) ⇒
0
2m ≤ 2
f(x) dx ≤ 2M.
0
53. If −1 ≤ x ≤ 1, then0 ≤ x 2 ≤ 1 and 1 ≤ 1+x 2 ≤ 2, so1 ≤ √ 1+x 2 ≤ √ 2 and
1[1 − (−1)] ≤ 1
√
1+x2 dx ≤ √ 2[1− (−1)] [Property 8]; that is, 2 ≤ 1
√
1+x2 dx ≤ 2 √ 2.
−1
−1
55. If 1 ≤ x ≤ 4,then1 ≤ √ x ≤ 2, so 1(4 − 1) ≤ 4 √ 4 √
1 xdx≤ 2(4 − 1); that is, 3 ≤
1 xdx≤ 6.
57. If π ≤ x ≤ π ,then1 ≤ tan x ≤ √ 3,so1 π
− π π/3
4 3 3 4 ≤ tan xdx≤ √ 3 π
− π
π/4 3 4 or
π
≤ π/3
tan xdx≤ √ π
12 π/4 12 3.
59. The only critical number of f(x) =xe −x on [0, 2] is x =1.Sincef(0) = 0, f(1) = e −1 ≈ 0.368, and
61.
f(2) = 2e −2 ≈ 0.271, we know that the absolute minimum value of f on [0, 2] is 0, and the absolute maximum is e −1 .By
Property 8, 0 ≤ xe −x ≤ e −1 for 0 ≤ x ≤ 2 ⇒ 0(2 − 0) ≤ 2
0 xe−x dx ≤ e −1 (2 − 0) ⇒ 0 ≤ 2
0 xe−x dx ≤ 2/e.
√
x4 +1≥ √ x 4 = x 2 ,so 3
√
x4 +1dx ≥ 3
1
1 x2 dx = 1 3 3 3 − 1 3 = 26 . 3