1b-2008-exam_1_solutions - Math 1B Exam 1 Solutions Fall...

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Unformatted text preview: Math 1B Exam 1 Solutions Fall Program for Freshmen 2008 Fred Bourgoin 1. (8 points) Find the area of the region bounded by y = sin x, y = sin 3 x, x = 0 , x = 2 . Answer. On the given interval, sin x sin 3 x , so the area between the graphs is A = Z / 2 (sin x- sin 3 x ) dx = Z / 2 sin x (1- sin 2 x ) dx = Z / 2 sin x cos 2 x dx =- Z 1 u 2 du = Z 1 u 2 du = 1 3 u 3 1 = 1 3 . 2. (10 points) Evaluate Z 2 3 x 3 16- x 2 dx . Answer. The trigonometric substitution x = 4sin seems appropriate here, al- though a u-substitution would also work. Z 2 3 x 3 16- x 2 dx = Z / 3 64sin 3 4cos 4cos d = 64 Z / 3 sin 3 d = 64 Z / 3 sin (1- cos 2 ) d =- 64 Z 1 / 2 1 (1- u 2 ) du = 64 Z 1 1 / 2 (1- u 2 ) du = 64 u- 1 3 u 3 1 1 / 2 = 40 3 . 3. (12 points) Evaluate Z 3 dx x 2- x- 2 . Answer. Be careful! this is an improper integral of type II....
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1b-2008-exam_1_solutions - Math 1B Exam 1 Solutions Fall...

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