Solução_Calculo_Stewart_6e

F.
240 ¤ CHAPTER 5 INTEGRALS
TX.10
25. Note that ∆x = 2 − 1
n
= 1 and xi =1+i ∆x =1+i(1/n) =1+i/n.
n
2
n
n
x 3 dx = lim f(x
n→∞ i) ∆x = lim 1+ i 3
1 1 n n + i
= lim
n→∞ n n n→∞ n n
27.
b
a
1
i=1
1
= lim
n→∞ n 4
= lim
n→∞
= lim
n→∞
i=1
n
n 3 +3n 2 i +3ni 2 + i 3 = lim
i=1
1
n · n 3 +3n 2 n
n 4 i=1
1+ 3 n(n +1) ·
n2 2
= lim 1+ 3
n→∞ 2 · n +1
n + 1 2 · n +1
n
= lim
n→∞
b − a
xdx = lim
n→∞ n
1+ 3 2
n→∞
i +3n n i 2
+ n
i=1
i=1
+ 3 n(n + 1)(2n +1)
·
n3 6
i=1
3
1 n
n 3 + n
3n 2 i + n
3ni 2 + n i 3
n 4 i=1 i=1
i=1
i=1
i 3
+ 1
n · n2 (n +1) 2
4 4
(n +1)2 ·
n 2
2n +1 · + 1 n 4
1+ 1
+ 1
1+ 1
2+ 1
+ 1
1+ 1 2
=1+ 3 n 2 n n 4 n
2 + 1 · 2+ 1 =3.75
2 4
n
a + b − a
n
i
i=1
a(b − a)
= lim n +
n→∞ n
(b − a)2
n 2 ·
a(b − a)
= lim
n→∞ n
n
1+
i=1
(b − a)2
n 2
n
i=1
i
n(n +1)
(b − a) 2
= a (b − a)+ lim
1+ 1
2
n→∞ 2 n
= a(b − a)+ 1 2 (b − a)2 =(b − a) a + 1 2 b − 1 2 a =(b − a) 1 2 (b + a) = 1 2
b 2 − a 2
29. f(x) = x
6 − 2
, a =2, b =6,and∆x = = 4 1+x5 n n . Using Theorem 4, we get x∗ i = x i =2+i ∆x =2+ 4i
n ,
6
x
so
dx = lim
2 1+x R 5
n = lim
n→∞ n→∞
n 2+ 4i
n
i=1
1+ 2+ 4i
n
5 · 4
n .
31. ∆x =(π − 0)/n = π/n and x ∗ i = x i = πi/n.
π
n
π n
sin 5xdx= lim (sin 5x i ) = lim
n
0
n→∞ i=1
n→∞ i=1
sin 5πi
n
π
n
CAS 1 5π
= π lim
n→∞ n cot CAS 2
= π = 2 2n 5π 5
33. (a) Think of 2
f(x) dx as the area of a trapezoid with bases 1 and 3 and height 2. The area of a trapezoid is A = 1
0 2
(b + B)h,
so 2
f(x) dx = 1 (1 + 3)2 = 4.
0 2
(b) 5
f(x) dx = 2
f(x) dx
0 0
trapezoid
+ 3
f(x) dx
2
rectangle
+ 5
f(x) dx
3
triangle
= 1 2 (1 + 3)2 + 3 · 1 + 1 2 · 2 · 3 =4+3+3=10
(c) 7
5 f(x) dx is the negative of the area of the triangle with base 2 and height 3. 7
5 f(x) dx = − 1 2 · 2 · 3=−3.
(d) 9
f(x) dx is the negative of the area of a trapezoid with bases 3 and 2 and height 2, so it equals
7
− 1 (B + b)h = − 1 (3 + 2)2 = −5. Thus,
2 2
9 f(x) dx = 5
f(x) dx + 7
f(x) dx + 9
0 0 5 7
f(x) dx =10+(−3) + (−5) = 2.