Solução_Calculo_Stewart_6e

F.
328 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
TX.10
47. Since f(x) =sin 4 x cos 6 x is everywhere positive, we know that its antiderivative F is increasing. Maple gives
f(x) dx = −
1
10 sin3 x cos 7 x − 3 sin x 80 cos7 x + 1
160 cos5 x sin x + 1
128 cos3 x sin x + 3
3
cos x sin x + x
256 256
and this expression is 0 at x =0.
F has a minimum at x =0and a maximum at x = π.
F has inflection points where f 0 changes sign, that is, at x ≈ 0.7, x = π/2,
and x ≈ 2.5.
7.7 Approximate Integration
1. (a) ∆x =(b − a)/n =(4− 0)/2 =2
L 2 = 2 f(x i−1) ∆x = f(x 0) · 2+f(x 1) · 2=2[f(0) + f(2)] = 2(0.5+2.5) = 6
i=1
R 2 = 2 f(x i ) ∆x = f(x 1 ) · 2+f(x 2 ) · 2=2[f(2) + f(4)] = 2(2.5+3.5) = 12
i=1
M 2 = 2 f(x i)∆x = f(x 1) · 2+f(x 2) · 2=2[f(1) + f(3)] ≈ 2(1.6+3.2) = 9.6
i=1
(b)
L 2 is an underestimate, since the area under the small rectangles is less than
the area under the curve, and R 2 is an overestimate, since the area under the
large rectangles is greater than the area under the curve. It appears that M 2
is an overestimate, though it is fairly close to I. See the solution to
Exercise 45 for a proof of the fact that if f is concave down on [a, b],then
the Midpoint Rule is an overestimate of b
f(x) dx.
a
(c) T 2 = 1
∆x [f(x
2 0 )+2f(x 1 )+f(x 2 )] = 2 [f(0) + 2f(2) + f(4)] = 0.5+2(2.5) + 3.5 =9.
2
This approximation is an underestimate, since the graph is concave down. Thus, T 2 =9