1b-2008-exam_1_solutions

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

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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 = π 2 . Answer. On the given interval, sin x ≥ sin 3 x , so the area between the graphs is A = Z π/ 2 (sin x- sin 3 x ) dx = Z π/ 2 sin x (1- sin 2 x ) dx = Z π/ 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 θ seems appropriate here, al- though a u-substitution would also work. Z 2 √ 3 x 3 √ 16- x 2 dx = Z π/ 3 64sin 3 θ 4cos θ 4cos θ dθ = 64 Z π/ 3 sin 3 θ dθ = 64 Z π/ 3 sin θ (1- cos 2 θ ) dθ =- 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...

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