Page 192 - Calculus Workbook For Dummies

176 Part IV: Integration and Infinite Series

2. Plug these values into the formula.
π - 0 π 2 π 3 π 7 π
T 8 = c sin0 + 2 sin + 2 sin + 2 sin + ... + 2 sin + sinπm
2 8 $ 8 8 8 8
π
. ^ 0 + .765 + . 1 414 + . 1 848 ... + h . 1 974
0 .
16
The percent error is 1.3%.
p Estimate the same area with 16 and 24 trapezoids and compute the percent error.
π - 0 π 2 π 3 π 15 π
T 16 = c sin0 + 2 sin + 2 sin + 2 sin + ... + 2 sin + sinπm
$
2 16 16 16 16 16
π
0 .
. ^ 0 + .390 + .765 + ... + h . 1 994
32
The percent error for 16 trapezoids is 0.321%.
π - 0 π 2 π 3 π 23 π
T 24 = c sin0 + 2 sin + 2 sin + 2 sin + ... + 2 sin + sinπm
$
2 24 24 24 24 24
π
0 .
. ^ 0 + .261 + .518 + ... + h . 1 997
48
The percent error for 24 trapezoids is 0.143%.
q Approximate the same area with eight Simpson’s rule “trapezoids” and compute the percent
error. The area for 8 “trapezoids” is 2.00001659 and the error is 0.000830%.
For 8 Simpson’s “trapezoids”:
1. List the values for a, b, and n, and determine the x-values x 0 through x 16 , the 9 edges and the
8 base midpoints of the 8 curvy-topped “trapezoids.”
a = 0
b π
=
n = 16
π 2 π 3 π
x 0 = 0 , x 1 = , x 2 = , x 3 = , ... x 16 = π
16 16 16
2. Plug these values into the formula.
π - 0 π 2 π 3 π 4 π 15 π
S 16 = c sin0 + 4 sin + 2 sin + 4 sin + 2 sin + ... + 4 sin + sinπm
$
3 16 16 16 16 16 16
π
0 .
. ^ 0 + .7804 + .7654 + . 2 2223 + . 1 4142 + ... + . 0 7804 + h . 2 00001659
48
The percent error for eight Simpson “trapezoids” is 0.000830%.
r Use the following shortcut to figure S 20 for the area under lnx from 1 to 6.
Using the formula in the problem, you get:
M n + M n + T n
S n2 =
3
M 10 + M 10 + T 10
S 20 =
3
. 5 759 + . 5 759 + . 5 733
.
3
. . 5 750
This agrees (except for a small round-off error) with the result obtained the hard way in the
Simpson’s rule example problem.