Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 5.2 THEDEFINITEINTEGRAL ¤ 239
13. In Maple, we use the command with(student); to load the sum and box commands, then
m:=middlesum(sin(xˆ2),x=0..1,5); which gives us the sum in summation notation, then M:=evalf(m); which
gives M 5 ≈ 0.30843908,confirming the result of Exercise 11. The command middlebox(sin(xˆ2),x=0..1,5)
generates the graph. Repeating for n =10and n =20gives M 10 ≈ 0.30981629 and M 20 ≈ 0.31015563.
15. We’ll create the table of values to approximate π
sin xdxby using the
0
program in the solution to Exercise 5.1.7 with Y 1 =sinx, Xmin = 0,
Xmax = π,andn =5, 10, 50,and100.
The values of R n appear to be approaching 2.
17. On [2, 6], lim
n
n→∞ i=1
x i ln(1 + x 2 i ) ∆x = 6
2 x ln(1 + x2 ) dx.
n
R n
5 1.933766
10 1.983524
50 1.999342
100 1.999836
19. On [1, 8], lim
n
n→∞ i=1
21. Note that ∆x = 5 − (−1)
n
5
−1
23. Note that ∆x = 2 − 0
n
2
0
2x
∗
i +(x ∗ i )2 ∆x = 8
1
√
2x + x2 dx.
(1 + 3x) dx = lim
2 − x
2 dx = lim
= 6 n and x i = −1+i ∆x = −1+ 6i
n .
n→∞ i=1
6 n
= lim
n→∞ n
n
f(x i) ∆x = lim
i=1
(−2) + n
6
= lim −2n + 18
n→∞ n n
= lim
n→∞
n
n→∞ i=1
i=1
−12 + 54 n +1
n
= 2
n and x i =0+i ∆x = 2i
n .
n→∞ i=1
= lim
n→∞
n
f(x i ) ∆x = lim
2
n
2n − 4 n 2
n
i=1
= lim 4 − 4
n→∞ 3 · n +1
n
18i
n
1+3
= lim
n→∞
n(n +1) · = lim
2
n→∞
n→∞ i=1
−1+ 6i
n
6
−2n + 18 n n
6
n = lim 6
n→∞ n
n
i
i=1
−12 + 108 n(n +1) ·
n2 2
n
−2+ 18i
n
i=1
= lim −12 + 54 1+ 1
= −12 + 54 · 1=42
n→∞
n
n
2 − 4i2 2
n 2 n
i 2
= lim
n→∞
· 2n +1
n
= lim
n→∞
4 − 8 n(n + 1)(2n +1)
·
n3 6
2 n
2 − 4
n
i 2
n i=1 n 2 i=1
= lim 4 − 4
1+ 1
2+ 1
=4− 4
n→∞
3
3 n n
· 1 · 2= 4 3