Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 7.8 IMPROPER INTEGRALS ¤ 335
45. Since the Trapezoidal and Midpoint approximations on the interval [a, b] are the sums of the Trapezoidal and Midpoint
approximations on the subintervals [x i−1 ,x i ], i =1, 2,...,n, we can focus our attention on one such interval. The condition
f 00 (x) < 0 for a ≤ x ≤ b means that the graph of f isconcavedownasinFigure5.Inthatfigure, T n is the area of the
trapezoid AQRD, b
a f(x) dx istheareaoftheregionAQP RD, andM n is the area of the trapezoid ABCD, so
T n < b
a f(x) dx < Mn. In general, the condition f 00 < 0 implies that the graph of f on [a, b] lies above the chord joining the
points (a, f(a)) and (b, f(b)). Thus, b
a f(x) dx > Tn. SinceMn is the area under a tangent to the graph, and since f 00 < 0
implies that the tangent lies above the graph, we also have M n > b
f(x) dx. Thus, a Tn < b
f(x) dx < Mn.
a
47. T n = 1 2 ∆x [f(x 0)+2f(x 1 )+···+2f(x n−1 )+f(x n )] and
M n = ∆x [f(x 1)+f(x 2)+···+ f(x n−1)+f(x n)],wherex i = 1 (xi−1 + xi). Now
2
T 2n = 1 2
1
2 ∆x [f(x 0 )+2f(x 1 )+2f(x 1 )+2f(x 2 )+2f(x 2 )+···+2f(x n−1 )+2f(x n−1 )+2f(x n )+f(x n )]
so
1
2 (T n + M n )= 1 2 T n + 1 2 M n
= 1 4 ∆x[f(x0)+2f(x1)+···+2f(xn−1)+f(xn)] + 1 4 ∆x[2f(x1)+2f(x2)+···+2f(xn−1)+2f(xn)]
= T 2n
7.8 Improper Integrals
1. (a) Since ∞
1
x 4 e −x4 dx has an infinite interval of integration, it is an improper integral of Type I.
(b) Since y =secx has an infinite discontinuity at x = π 2 , π/2
0
sec xdxis a Type II improper integral.
(c) Since y =
x
2
(x − 2)(x − 3) has an infinite discontinuity at x =2, x
dx is a Type II improper integral.
x 2 − 5x +6
0
1
(d) Since
dx has an infinite interval of integration, it is an improper integral of Type I.
−∞ x 2 +5
0
3. Theareaunderthegraphofy =1/x 3 = x −3 between x =1and x = t is
A(t) = t
1 x−3 dx = − 1 x−2 t
= − 1
2 1 2 t−2 −
− 1 2 =
1
− 1 2t 2 . So the area for 1 ≤ x ≤ 10 is
2
A(10) = 0.5 − 0.005 = 0.495, theareafor1 ≤ x ≤ 100 is A(100) = 0.5 − 0.00005 = 0.49995, and the area for
1 ≤ x ≤ 1000 is A(1000) = 0.5 − 0.0000005 = 0.4999995. The total area under the curve for x ≥ 1 is
lim A(t) = lim 1 −
t→∞ t→∞ 2 1/(2t2 ) = 1 . 2