HW8solution - 5 ± d = |-2082 1765 2-1765 2 | × 100 = 18 0...

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

View Full Document Right Arrow Icon
Homework 8 Solution 1. Evaluate the following integral Z 6 0 ( - 2 + x - 3 x 2 + 4 x 4 - x 5 ) dx (a) [1pt] Analytically. (b) [1pt] Simpson’s 1/3 rule where h = 3( n = 2); (c) [1pt] Simpson’s 1/3 rule where h = 1( n = 6); (d) [1pt] Simpson’s 3/8 rule where h = 2( n = 3); (e) [1pt] Simpson’s 3/8 rule where h = 1( n = 6), where h is the step size and n is the number of segments. (f) [2pt] Compute the true percent relative error for each (b) - (e) based on the analytical result from (a): ± t = ± ± ± ± I t - I a I t ± ± ± ± × 100% where I t is the true value and I a is an estimated integral. Q1: (a) Analytically: Z 6 0 ( - 2 + x - 3 x 2 + 4 x 4 - x 5 ) dx = ( - 2 x + 1 2 x 2 - x 3 + 4 5 x 5 - 1 6 x 6 ) | 6 0 = - 1765 . 2 (b) Simpson’s 1/3 rule with h=3,n=2 I = 3 3 ( f (0) + 4 f (3) + f (6)) = - 2478 (c) Simpson’s 1/3 rule with h=1,n=6 I = 1 3 [ f (0) + 4( f (1) + f (3) + f (5)) + 2( f (2) + f (4)) + f (6)] = - 1774 (d) Simpson’s 3/8 rule with h=2,n=3 I = 3 × 2 8 ( f (0) + 3 f (2) + 3 f (4) + f (6)) = - 2082 (e) Simpson’s 3/8 rule with h=1,n=6 I = 3 8 ( f (0) + 3( f (1) + f (4)) + 3 f ( f (2) + f (5)) + 2 f (3) + f (6)) = - 1785 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
(f) percent relative error: ± b = | - 2478 + 1765 . 2 - 1765 . 2 | × 100% = 40 . 4% ± c = | - 1774 + 1765 . 2 - 1765 . 2 | × 100% = 0
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . 5% ± d = |-2082 + 1765 . 2-1765 . 2 | × 100% = 18 . 0% ± e = |-1785 + 1765 . 2-1765 . 2 | × 100% = 1 . 1% 2. [3pt] Use Romberg integration to evaluate Z 2 e 2 x sin( x ) 1 + x 2 dx to an accuracy less than 0.5%. Use an estimate of the percent relative error defined by ± a = ± ± ± ± I 1 ,k-I 2 ,k-1 I 1 ,k ± ± ± ± × 100% . Q2: I 1 , 1 ( h = 2) = 2 2 ( f (0) + f (2)) = 9 . 9292 I 2 , 1 ( h = 1) = 1 2 ( f (0) + 2 f (1) + f (2)) = 8 . 0734 I 3 , 1 ( h = 0 . 5) = . 5 2 ( f (0) + 2( f (0 . 5) + f (1) + f (1 . 5)) + f (2)) = 7 . 6403 I 1 , 2 = 4 I 2 , 1-I 1 , 1 4-1 = 7 . 4548 I 2 , 2 = 4 I 3 , 1-I 2 , 1 4-1 = 7 . 4959 I 1 , 3 = 4 2 I 2 , 2-I 1 , 2 4 2-1 = 7 . 4986 ± = | I 1 , 3-I 2 , 2 I 1 , 3 | = | 7 . 4986-7 . 4959 7 . 4986 | = 0 . 036% < . 5% so we have Z 2 e 2 x sin( x ) 1 + x 2 dx ≈ 7 . 4986 2...
View Full Document

This note was uploaded on 11/14/2011 for the course EAME 250 taught by Professor Lee during the Spring '11 term at Case Western.

Page1 / 2

HW8solution - 5 ± d = |-2082 1765 2-1765 2 | × 100 = 18 0...

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