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

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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.
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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 > 1, we know R ∞ 1 g ( x ) dx is convergent. Therefore, R ∞ 1 f ( x ) dx is also convergent. 2...
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This note was uploaded on 10/08/2010 for the course MATH 21B taught by Professor Vershynin during the Spring '08 term at UC Davis.

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

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