prelim1_fa04_solutions

prelim1_fa04_solutions - Math 191 – Fall, 2004 – Prelim...

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Unformatted text preview: Math 191 – Fall, 2004 – Prelim I “Solutions” 9/21/04 1a. By the Fundamental Theorem of Calculus we have d dx Z x cos t dt = cos x. 1b. Let u = sin x ⇒ du dx = cos x . Then by the Fundamental Theorem and the Chain Rule we have d dx Z sin x t 2 dt = d du Z u t 2 dt ¶ du dx = u 2 du dx = sin 2 x cos x. 2a. Let u = 1 + x 3 ⇒ du = 3 x 2 dx , so Z 2 3 x 2 p 1 + x 3 dx = Z 9 1 √ u du = • 2 3 u 3 / 2 ‚ 9 1 = 2 3 (9 3 / 2- 1 3 / 2 ) = 2 3 (27- 1) = 52 3 . 2b. Let u = sin x ⇒ du = cos x dx , so Z π/ 2 cos x cos(sin x ) dx = Z 1 cos u du = [sin u ] 1 = sin1 . 3. The error formula for the trapezoidal rule is | E T | ≤ b- a 12 h 2 M where M is any upper bound on | f 00 | . In this problem we have b = 2, a = 0, h = b- a n = 2 n , and f 00 ( x ) = 6 x + 4. Therefore, since | f 00 | ≤ 6 · 2 + 4 = 16 on [0 , 2] we have M = 16. In order to make | E T | < 1 100 , we need b- a 12 h 2 M = 2 12 2 n ¶ 2 · 16 < 1 100 , 2 5 3 n 2 < 1 2 2 · 5 2 , n 2 > 2 7 · 5 2 3 , n > r 2 7 · 5...
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This note was uploaded on 06/01/2008 for the course MATH 1910 taught by Professor Berman during the Fall '07 term at Cornell.

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prelim1_fa04_solutions - Math 191 – Fall, 2004 – Prelim...

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