sample_final_solutions

# sample_final_solutions - Problem 1 Evaluate the following...

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Problem 1. Evaluate the following integrals: (a) (4pts) Z x ln x dx = solution: Apply integration by parts with u = ln x , du = 1 x dx , v = 1 2 x 2 , dv = x dx . Then Z x ln x dx = 1 2 x 2 ln x - 1 2 Z x dx = 1 2 x 2 ln x - 1 4 x 2 + C. (b) (4pts) Z sin 10 θ cos θ dθ = solution: Let u = sin θ . Then we have R u 10 du = 1 11 u 11 + C = 1 11 sin 11 θ + C . (c) (4pts) Z 2 ( x + 4)( x + 6) dx = solution: Apply partial fractions to write 2 ( x + 4)( x + 6) = 1 x + 4 - 1 x + 6 . Then Z 2 ( x + 4)( x + 6) dx = Z ( 1 x + 4 - 1 x + 6 ) dx = ln | x + 4 | - ln | x + 6 | + C = ln | x + 4 x + 6 | + C. 2

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Problem 2. (15pts) Find the area of the region enclosed by the graphs of y = 4 x and y = 4 x 4 . solution: First find the intersection points by solving 4 x = 4 x 4 to get x = 0 , 1. Picturing the graphs tells us to find the area for 0 x 1. Since 4 x 4 x 4 on that interval, we integrate Z 1 0 (4 x - 4 x 4 ) dx = [2 x 2 - 4 5 x 5 ] 1 0 = 6 5 . 3
Problem 3. Consider the function f ( x ) = x 1+ x 3 and the region R bounded by the graph of y = f ( x ) and the x -axis for 0 x < . (a) (6pts) Write an integral that represents the area of region R . Is this area finite? Justify your answer. solution: The area is given by R 0 x 1+ x 3 dx . This integral is finite. It is equal to R 1 0 x 1+ x 3 dx + R 1 x 1+ x 3 dx . The first integral is proper, and the second integral converges by direct (or limit) comparison with 1 x 2 . (b) (6pts) Write an integral that represents the volume of the solid obtained by revolving region R about the x-axis . Is this volume finite? Justify your answer. solution: By the disk method, the volume is given by π R 0 ( x 1+ x 3 ) 2 dx . This integral is finite. Again it can be split into two integrals with the first proper and the second convergent by comparison with 1 x 4 .

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