Explanation Given information: f ( x ) = x 3 Formula used: Simpson’s rule S N = 1 3 Δ x [ y 0 + 4 y 1 + 2 y 2 + ....... + 2 y N − 2 + 4 y N − 1 + y N ] Where N = 2 is even Calculation: Consider the given function. f ( x ) = x 3 So, the integral can be computed as shown below. I = ∫ a b f ( x ) d x = ∫ a b ( x 3 ) d x = [ x 4 4 ] a b = 1 4 ( b 4 − a 4 ) To approximate I = ∫ a b f ( x ) d x = ∫ a b x 3 d x , using Simpson’s Rule S N . S N = 1 3 Δ x [ y 0 + 4 y 1 + 2 y 2 + ....... + 2 y N − 2 + 4 y N − 1 + y N ] Where N = 2 Δ x = b − a N = b − a 2 y j = f ( a + j Δ x ) = f ( a + j b − a 2 ) = f ( ( 2 − j ) a + j b 2 ) = ( ( 2 − j ) a + j b 2 ) 3 y 0 = a 3 y 1 = ( a + b 2 ) 3 y 2 = b 3 Therefore, the value of S 2 can be calculated as S 2 = 1 3 ( b − a 2 ) [ y 0 + 4 y 1 + y 2 ] = b − a 6 [ a 3 + 4 ( a + b 2 ) 3 + b 3 ] = b − a 12 [ 2 a 3 + ( a + b ) 3 + 2 b 3 ] On further simplification, we get S 2 = b − a 12 [ 2 a 3 + a 3 + b 3 + 3 a 2 b + 3 a b 2 + 2 b 3 ] = b − a 12 [ 3 a 3 + 3 b 3 + 3 a 2 b + 3 a b 2 ] = b − a 4 [ a 3 + b 3 + a 2 b + a b 2 ] = b − a 4 [ ( a 2 + b 2 ) ( a + b ) ] = b 4 − a 4 4 Therefore, S 2 = ∫ a b f ( x ) d x For all f ( x ) = x 3 quadratic polynomials and all endpoints a , b Consider f ( x ) = p x 2 + q x + r be a polynomial We will prove that S 2 = ∫ a b f ( x ) d x Consider the below quadratic polynomial. f ( x ) = p x 2 + q x + r So, the integral can be computed as shown below. I = ∫ a b f ( x ) d x = ∫ a b ( p x 2 + q x + r ) d x = [ p x 3 3 + q x 2 2 + r x ] a b = p 3 ( b 3 − a 3 ) + q 2 ( b 2 − a 2 ) + r ( b − a ) To approximate I = ∫ a b f ( x ) d x = ∫ a b ( p x 2 + q x + r ) d x , using Simpson’s Rule S N . S N = 1 3 Δ x [ y 0 + 4 y 1 + 2 y 2 + ....... + 2 y N − 2 + 4 y N − 1 + y N ] Where N = 2 Δ x = b − a N = b − a 2 y j = f ( a + j Δ x ) = f ( a + j b − a 2 ) = f ( ( 2 − j ) a + j b 2 ) = p ( ( 2 − j ) a + j b 2 ) 2 + q ( ( 2 − j ) a + j b 2 ) + r y 0 = p a 2 + q a + r y 1 = p ( a + b 2 ) 2 + q ( a + b 2 ) + r y 2 = p b 2 + q b + r Therefore, the value of S 2 can be calculated as S 2 = 1 3 ( b − a 2 ) [ y 0 + 4 y 1 + y 2 ] = b − a 6 [ ( p a 2 + q a + r ) + r [ p ( a + b 2 ) 2 + q ( a + b 2 ) + r ] + p b 2 + q b + r ] = p ( b − a ) 6 [ a 2 + ( a + b ) 2 + b 2 ] + q ( b − a ) 6 [ a + 2 ( a + b ) + b ] + r ( b − a ) 6 [ 1 + 4 + 1 ] = p ( b − a ) 6 [ 2 a 2 + 2 a b + 2 b 2 ] + q ( b − a ) 2 ( a + b ) + r ( b − a ) = p ( b 3 − a 3 ) 3 + q ( b 2 − a 2 ) 2 + r ( b − a ) Therefore, S 2 = ∫ a b f ( x ) d x For all f ( x ) = p x 2 + q x + r quadratic polynomials and all endpoints a , b

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Numerical Integration

In a mathematical investigation, numerical integration comprises a wide group of calculations for computing the mathematical estimation of definite integral, and likewise, the term is moreover in some cases used to depict the numerical solution of differ…

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Chapter 8.9, Problem 62E

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Show that S2 also gives the exact value for ∫abx3dx and conclude SN is exact for all cubic polynomials. Show by counterexample that S2 is not exact for integrals of x4.

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Chapter 8.9, Problem 61E

Chapter 8.9, Problem 63E

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Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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Show that S 2 also gives the exact value for ∫ a b x 3 d x and conclude S N is exact for all cubic polynomials. Show by counterexample that S 2 is not exact for integrals of x 4 .

Solution Summary: The author concludes that S_2 is exact for all cubic polynomials. The integral can be computed as shown by counterexample.

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Show that S2 also gives the exact value for ∫abx3dx and conclude SN is exact for all cubic polynomials. Show by counterexample that S2 is not exact for integrals of x4.

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Chapter 8 Solutions