prob_06_03_06

# prob_06_03_06 - Problem 6.3-6 Find: Solutions to Integrals...

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Problem 6.3-6 Find: Solutions to Integrals using 1, 2 and 3 point quadrature. Solution: a) I exact 1 1 ξ ξ 2 ξ 3 + d 0.66667 = := f ξ () ξ 2 ξ 3 + := One Point Rule: ξ 1 0. := W 1 2. := I 1 W 1 f ξ 1 0 = := I 1 I exact I exact 100 100 = % error Two Point Rule: ξ 1 1 3 := W 1 1 := ξ 2 1 3 := W 2 1 := I 2 W 1 f ξ 1 W 2 f ξ 2 + 0.66667 = := I 2 I exact I exact 100 0 = % error Three Point Rule: ξ 1 0.6 := W 1 5 9 := ξ 2 0 := W 2 8 9 := ξ 3 0.6 := W 3 5 9 := I 3 W 1 f ξ 1 W 2 f ξ 2 + W 3 f ξ 3 + 0.66667 = := I 3 I exact I exact 100 0 = % error b) I exact 1 1 ξ cos 1.5 ξ d 1.32999 = := f ξ ( ) cos 1.5 ξ := One Point Rule: ξ 1 0. := W 1 2. := I 1 W 1 f ξ 1 2 = := I 1 I exact I exact 100 50.377 = % error Two Point Rule: ξ 1 1 3 := W 1 1 := ξ 2 1 3 := W 2 1 := I 2 W 1 f ξ 1 W 2 f ξ 2 + 1.296 = := I 2 I exact I exact 100 2.577 = % error Three Point Rule: ξ 1 0.6 := W 1 5 9 := ξ 2 0 := W 2 8 9 := ξ 3 0.6 := W 3 5 9 := I 3 W 1 f ξ 1 W 2 f ξ 2 + W 3 f ξ 3 + 1.331 = := I 3 I exact I exact 100 0.051 = % error

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c) I exact 1 1 ξ 1 ξ 2 ξ + d 1.296 = := f ξ () 1 ξ 2 ξ + := One Point Rule: ξ 1 0. := W 1 2. := I 1 W 1 f ξ 1 1 = := I 1 I exact I exact 100 22.83 = % error Two Point Rule: ξ 1 1 3 := W 1 1 := ξ 2 1 3 := W 2 1 := I 2 W 1 f ξ 1 W 2 f ξ 2 + 1.273 = := I 2 I exact I exact 100 1.783 = % error Three Point Rule: ξ 1 0.6 := W 1 5 9 := ξ 2 0 := W 2 8 9 := ξ 3 0.6 := W 3 5 9 := I 3 W 1 f ξ 1 W 2 f ξ 2 + W 3 f ξ 3 + 1.294 = := I 3 I exact I exact 100 0.133 = % error d) I exact 1 7 x 1 x d 1.946 = := x 1 ξ
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## This note was uploaded on 01/11/2011 for the course MAE 5020 taught by Professor Folkman during the Fall '10 term at Utah State University.

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prob_06_03_06 - Problem 6.3-6 Find: Solutions to Integrals...

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