2008_Solutions_Practice_Test 1

2008_Solutions_Practice_Test 1 - Solutions for Practice...

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Solutions for Practice Midterm 1 1. Evaluate the integral () 2 23 4 sin cos x xd x π . Solution . Substitution sin ux = 1 1 35 2 2 2 2 2 4 2 2 1 1 22 42 sin cos 1 3 5 3 8 5 3 2 222 1 6 1 0 2 3 2 1 6 7 2 15 12 40 120 120 uu x x dx u u du ⎛⎞ ⋅= = = ⎜⎟ ⎜⎟ ⎜⎟ ⋅⋅ ⎝⎠ −+ =− + = = ∫∫ = 2. Evaluate the integral cos5 x x . Solution . Integration by parts: 1 cos 5 sin 5 5 u x du dx dv x dx v x == 11 1 1 sin 5 sin 5 sin 5 cos 5 55 5 2 5 x x x x x x x x C = + + 3. Find the area of the region that is bounded by the curve 3s i n r θ = + . 22 2 2 00 2 2 2 2 0 0 0 2 0 i n 1c o s2 9 6sin sin 18 6 cos 2 1 1 1 18 1 cos 2 18 2 sin 2 18 2 2 Area r d d dd d d ππ θθ θπ 9 2 πθθ πθ + = =+ + = + + = = 4. Evaluate the integral 32 2 41 61 3 2 27 3 xxx dx xx + . Solution . Long division: 2 2 2 21 2 7 3 4 16 13 2 4 6 5 2 3 x x x x x + + ++ +−
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32 2 22 2 41 61 3 2 5 5 1 21 73 27 3 3 2 xxx dx x dx x x dx xx −+ + =− + = + ∫∫ = Try to factor the denominator: the equation
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This note was uploaded on 04/24/2008 for the course CALCULUS 142 taught by Professor Brodsky during the Spring '08 term at University of Tennessee.

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2008_Solutions_Practice_Test 1 - Solutions for Practice...

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