Solução_Calculo_Stewart_6e

F.
242 ¤ CHAPTER 5 INTEGRALS
TX.10
63. Using right endpoints as in the proof of Property 2, we calculate
b
n
cf(x) dx = lim cf(x
a i ) ∆x = lim c n f(x i ) ∆x = c lim
n→∞
n→∞
i=1
65. Since − |f(x)| ≤ f(x) ≤ |f(x)|, it follows from Property 7 that
i=1
n
n→∞ i=1
f(x i ) ∆x = c b
f(x) dx.
a
− b
|f(x)| dx ≤ b
f(x) dx ≤
b
|f(x)| dx ⇒
b f(x) dx b
a a a a ≤ |f(x)| dx
a
Note that the definite integral is a real number, and so the following property applies: −a ≤ b ≤ a ⇒ |b| ≤ a for all real
numbers b and nonnegative numbers a.
67. To show that f is integrable on [0, 1] , we must show that lim
the interval [0, 1] into n equal subintervals 0, 1
,
n
subinterval, then we obtain the Riemann sum n
i=1
choose x ∗ i to be an irrational number. Then we get
lim
n
n→∞ i=1
n→∞ i=1
1
n , 2 n
, ... ,
n
f(x ∗ i ) ∆x exists. Let n denote a positive integer and divide
n − 1
n
f(x ∗ i ) · 1 =0,so lim
n n→∞
n
i=1
f(x ∗ i ) · 1
n = lim 1=1. Since the value of lim
n→∞ n→∞
limit does not exist, and f is not integrable on [0, 1].
69. lim
n
n→∞ i=1
lim
n→∞ i=1
i 4
n = lim
5 n→∞
n
i=1
i 4
n · 1
4 n = lim
n→∞
n
i=1
i
n
f(x ∗ i ) · 1
n = n
i=1
, 1
. If we choose x ∗ i to be a rational number in the ith
n
i=1
f(x ∗ i ) · 1
n = lim 0=0. Now suppose we
n→∞
1 · 1
n = n · 1 =1for each n, so
n
n
f(x ∗ i ) ∆x depends on the choice of the sample points x ∗ i ,the
i=1
4
1
. At this point, we need to recognize the limit as being of the form
n
n
f(x i ) ∆x,where∆x =(1− 0)/n =1/n, x i =0+i ∆x = i/n,andf(x) =x 4 . Thus, the definite integral
is 1
0 x4 dx.
71. Choose x i =1+ i n and x∗ i = √ x i−1 x i =
2 1
1 x−2 dx = lim
n→∞ n
n
i=1
= lim n n
n→∞ i=1
1+
i − 1
n
1+ i − 1
1+ i
.Then
n n
1
1+
i
n
1
n + i − 1 − 1
n + i
= lim
n→∞ n 1
n + 1
n +1 + ···+ 1
= lim
n→∞ n 1
n − 1
2n
= lim n n
n→∞ i=1
2n − 1
[by the hint]
= lim 1 −
1
n→∞ 2 =
1
2
1
(n + i − 1)(n + i)
n−1
= lim n 1
n→∞ i=0 n + i − n
1
−
n +1 + ···+ 1
2n − 1 + 1
2n
1
i=1 n + i
5.3 The Fundamental Theorem of Calculus
1. Oneprocessundoeswhattheotheronedoes.Thepreciseversion of this statement is given by the Fundamental Theorem of
Calculus. See the statement of this theorem and the paragraph that follows it on page 387.