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Calculus I Exam 3, Summer 2002, Answers 1. Integrate: 5 x 1 dx Answer. Let u x 3x 5 du 3 x 1 dx. Then 1 x 3x 5 x 1 dx 3 b) sin x 1 cos xdx Answer. Let u sin x du cos xdx. Then sin x 1 cos xdx u 1 du a) x 3 3x 3 2 3 2 3 3 2 u3 du 1 3 x 12 3x 5 C 4 2 2 2 1 3 u 3 u C 1 3 sin x 3 C sinx y 1 2 dy 1 x dx. Integrating: 1 x x C y 2 Putting in the initial values we get 1 2 1 1 2 C, so C 2. Putting in this value of C, and solving for y: x y 2 x 2. Solve the differential equation: dy 1 x y2 dx Answer. As an equation of differentials, this becomes y 2 2 2 1 2 3. Calculate the definite integrals: 4 a) x 3 0 3x 1 dx Answer. π 2 b) 0 4 x 1 dx x 4 4 3 3x 0 3x2 2 x 4 0 64 24 4 92 sin x cos x dx Answer. π 2 sin x cos x dx 0 sin2 x 2 π 2 0 1 2 4. Find the area of the region in the third quadrant bounded by the curves y x 3 and y 2x Answer. Draw the graph. The region in question is that in the third quadrant running from x Since the upper curve is y x 3 , the area is 0 x 2 3 2x x dx 2 x4 4 x 2 x3 3 164 4 83 0 2 8 3 x2 . 2 to x 0. PSfrag replacements 30 20 10 3 1 2 0 0 1 2 3 10 20 30 5. The region in the first quadrant bounded by the curves y is the volume of the solid so produced? x and y x is rotated about the y-axis. What Answer. Draw the graph. Sweeping along the x axis and using the shell method we obtain 1 PSfrag replacements V 2π x x x dx 2π 1 x x dx 3 2 2 2π 2 1 5 3 0 30 Sweeping along the y-axis2and using the washer method we first rewrite the curves as x 1 1 0 1 1 1 2π π y2 π y4 dy π y2 y4 dy π V 3 5 15 0 0 2 3 30 12 20 10 1 10 200 8 30 06 04 02 10 0 0 02 04 06 08 1 12 2π 15 y and x y 2 . Then