Solução_Calculo_Stewart_6e

F.
330 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
TX.10
17. f(y) = 1
1+y , ∆y = 3 − 0 = 1 5 6 2
(a) T 6 = 1
2 · 2 f(0) + 2f 1
2 +2f 2
2 +2f 3
2 +2f 4
2 +2f 5
2 + f(3) ≈ 1.064275
(b) M 6 = 1 2 f 1
4 + f 3
4 + f 5
4 + f 7
4 + f 9
4 + f 11
4 ≈ 1.067416
(c) S 6 = 1
2 · 3
f(0) + 4f
1
2
+2f
2
2
+4f
3
2
+2f
4
2
+4f
5
2
+ f(3)
≈ 1.074915
19. f(x) =cos(x 2 ), ∆x = 1 − 0
8
= 1 8
(a) T 8 = 1
8 · 2 f(0) + 2 f 1
8 + f 2
8 + ···+ f 7
8 + f(1) ≈ 0.902333
M 8 = 1 8 f 1
16 + f 3
16 + f 5
16 + ···+ f 15
16 =0.905620
(b) f(x) =cos(x 2 ), f 0 (x) =−2x sin(x 2 ), f 00 (x) =−2sin(x 2 ) − 4x 2 cos(x 2 ).For0 ≤ x ≤ 1, sin and cos are positive,
so |f 00 (x)| =2sin(x 2 )+4x 2 cos(x 2 ) ≤ 2 · 1+4· 1 · 1=6since sin(x 2 ) ≤ 1 and cos x 2 ≤ 1 for all x,
and x 2 ≤ 1 for 0 ≤ x ≤ 1. Soforn =8,wetakeK =6, a =0,andb =1in Theorem 3, to get
|E T | ≤ 6 · 1 3 /(12 · 8 2 )= 1 =0.0078125 and |E 128 M| ≤ 1 =0.00390625. [A better estimate is obtained by noting
256
from a graph of f 00 that |f 00 (x)| ≤ 4 for 0 ≤ x ≤ 1.]
(c) Take K =6[as in part (b)] in Theorem 3. |E T | ≤
K(b − a)3
12n 2 ≤ 0.0001 ⇔
6(1 − 0)3
12n 2 ≤ 10 −4 ⇔
1
2n ≤ 1 ⇔ 2n 2 ≥ 10 4 ⇔ n 2 ≥ 5000 ⇔ n ≥ 71. Taken =71for T 2 10 4 n .ForE M , again take K =6in
Theorem 3 to get |E M| ≤ 10 −4 ⇔ 4n 2 ≥ 10 4 ⇔ n 2 ≥ 2500 ⇔ n ≥ 50. Taken =50for M n.
21. f(x) =sinx, ∆x = π − 0
10
= π 10
(a) T 10 =
π
10 · 2
f(0) + 2f
π
10
M 10 = π 10
f
π
20
S 10 =
+ f
3π
20
π
10 · 3
f(0) + 4f
π
10
+2f 2π
10
+ f
5π
20
+2f
2π
10
+ ···+2f 9π
10 + f(π) ≈ 1.983524
+ ···+ f
19π
20 ≈ 2.008248
+4f
3π
10
+ ···+4f
9π
10
+ f(π)
≈ 2.000110
Since I = π
sin xdx = − cos x π
=1− (−1) = 2, E 0 0
T = I − T 10 ≈ 0.016476, E M = I − M 10 ≈−0.008248,
and E S = I − S 10 ≈−0.000110.
(b) f(x) =sinx ⇒ f (n) (x) ≤ 1,sotakeK =1forallerrorestimates.
|E T | ≤
|E S | ≤
K(b − a)3 1(π − 0)3
= = π3
12n 2 12(10) 2 1200 ≈ 0.025839. |E M| ≤ |E T |
= π3
2 2400 ≈ 0.012919.
K(b − a)5 1(π − 0)5
=
180n 4 180(10) = π 5
4 1,800,000 ≈ 0.000170.
The actual error is about 64% of the error estimate in all three cases.
(c) |E T | ≤ 0.00001 ⇔ π3
12n ≤ 1 ⇔ n 2 ≥ 105 π 3
2 10 5 12
|E M | ≤ 0.00001 ⇔ π3
24n ≤ 1 ⇔ n 2 ≥ 105 π 3
2 10 5 24
|E S | ≤ 0.00001 ⇔ π5
180n ≤ 1 ⇔ n 4 ≥ 105 π 5
4 10 5 180
Take n =22for S n (since n must be even).
⇒ n ≥ 508.3. Taken = 509 for T n.
⇒ n ≥ 359.4. Taken = 360 for M n.
⇒ n ≥ 20.3.