hw9sol - 2. Evaluate the improper integral Z 1 t + 1 t 2 +...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Fall 2010, MAT21B, Paper Homework 9 Solutions 1. Estimate the minimum number of subintervals needed to approximate the integral Z 2 0 sin ( x + 1) dx with an error of magnitude less than 10 - 4 by (a) the Trapezoidal Rule and (b) Simpson’s Rule. Solution: Let f ( x ) = sin ( x + 1), then f 0 ( x ) = cos ( x + 1) f 00 ( x ) = - sin ( x + 1) f 000 ( x ) = - cos ( x + 1) f (4) ( x ) = sin ( x + 1) . (a) With a = 0, b = 2, M = max | f 00 ( x ) | = max |- sin ( x + 1) | ≤ 1 , the error estimate of the trapezoidal rule gives | E T | ≤ ( b - a ) 3 12 n 2 M (2 - 0) 3 12 n 2 = 2 3 n 2 . The error’s absolute value will therefore be less than 10 - 4 if 2 3 n 2 < 10 - 4 , 2 3 × 10 4 < n 2 , 100 r 2 3 < n, or 81 . 650 < n The first integer beyond 81 . 650 is n = 82. So, the minimum number of subintervals needed is 82. (b) With a = 0, b = 2, M = max | f (4) ( x ) | = max | sin ( x + 1) | ≤ 1 , the error estimate of Simpson’s rule gives | E S | ≤ ( b - a ) 5 180 n 4 M (2 - 0) 5 180 n 4 = 32 180 n 4 = 8 45 n 4 . The error’s absolute value will therefore be less than 10 - 4 if 8 45 n 4 < 10 - 4 , 8 45 × 10 4 < n 4 , 10 ± 8 45 ² 1 / 4 < n, or 6 . 4936 < n The first even number beyond 6 . 4936 is n = 8. So, using Simpson’s rule, the minimum number of subintervals needed is 8.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2. Evaluate the improper integral Z 1 t + 1 t 2 + 2 t dt Solution: Let u = t 2 + 2 t , then du = 2 ( t + 1) dt . When t = 0, u = 0; when t = 1, u = 3. So, Z 1 t + 1 t 2 + 2 t dt = Z 3 du 2 u = u | 3 = 3-0 = 3. 1 3. Determine whether the integral Z 1 dx 2 x 3 + 10 x-5 is divergent or convergent. Solution. Let f ( x ) = 1 2 x 3 +10 x-5 and g ( x ) = x-3 2 . Then, lim x f ( x ) g ( x ) = lim x x 3 / 2 2 x 3 + 10 x-5 = lim x r x 3 2 x 3 + 10 x-5 = lim x r 1 2 + 10 x-2-5 x-3 = r 1 2 (0 , ) . By the limit comparison test, R 1 f ( x ) dx and R 1 g ( x ) dx either both converge or both diverge. Since 3 2 &gt; 1, we know R 1 g ( x ) dx is convergent. Therefore, R 1 f ( x ) dx is also convergent. 2...
View Full Document

Page1 / 2

hw9sol - 2. Evaluate the improper integral Z 1 t + 1 t 2 +...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online