Solução_Calculo_Stewart_6e

F.
256 ¤ CHAPTER 5 INTEGRALS
TX.10
11. False. The function f(x) =1/x 4 is not bounded on the interval [−2, 1]. It has an infinite discontinuity at x =0,soitis
not integrable on the interval. (If the integral were to exist, a positive value would be expected, by Comparison
Property 6 of Integrals.)
13. False. For example, the function y = |x| is continuous on R, but has no derivative at x =0.
15. False.
b d
b
f(x) dx is a constant, so
a
dx
f(x) dx =0, not f(x) [unless f(x) =0]. Compare the given statement
a
carefully with FTC1, in which the upper limit in the integral is x.
1. (a) L 6 = 6
i=1
f(x i−1 ) ∆x [∆x = 6 − 0
6
=1]
= f(x 0 ) · 1+f(x 1 ) · 1+f(x 2 ) · 1+f(x 3 ) · 1+f(x 4 ) · 1+f(x 5 ) · 1
≈ 2+3.5+4+2+(−1) + (−2.5) = 8
(b)
The Riemann sum represents the sum of the areas of the four rectangles
above the x-axis minus the sum of the areas of the two rectangles below the
x-axis.
M 6 = 6
i=1
f(x i ) ∆x [∆x = 6 − 0
6
=1]
= f(x 1) · 1+f(x 2) · 1+f(x 3) · 1+f(x 4) · 1+f(x 5) · 1+f(x 6) · 1
= f(0.5) + f(1.5) + f(2.5) + f(3.5) + f(4.5) + f(5.5)
≈ 3+3.9+3.4+0.3+(−2) + (−2.9) = 5.7
3.
1
√
0 x + 1 − x
2
dx = 1
xdx+ 1
√
0 0 1 − x2 dx = I 1 + I 2 .
I 1 can be interpreted as the area of the triangle shown in the figure
and I 2 can be interpreted as the area of the quarter-circle.
Area = 1 2 (1)(1) + 1 4 (π)(1)2 = 1 2 + π 4 .
5.
6 f(x) dx = 4
f(x) dx + 6
f(x) dx ⇒ 10 = 7 + 6
f(x) dx ⇒ 6
0 0 4 4 4
f(x) dx =10− 7=3
7. First note that either a or b must be the graph of x
f(t) dt,since 0
f(t) dt =0,andc(0) 6= 0. Now notice that b>0 when c
0 0
is increasing, and that c>0 when a is increasing. It follows that c is the graph of f(x), b is the graph of f 0 (x),anda is the
graph of x
f(t) dt.
0
2
9.
1 8x 3 +3x 2 dx = 8 · 1
4 x4 +3· 1 x3 2
=
3
2x 4 + x 3 2
= 2 · 2 4 +2 3 − (2 + 1) = 40 − 3=37
1 1
1
11.
0 1 − x
9
dx = x − 1 x10 1
=
1 − 1 10 0 10 − 0=
9
10