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Calculus I Exam 3, Fall 2002, Answers 1. Integrate: 5 x 1 dx Answer. Let u x 4x 5 !du 4 x 1 dx. Then 1 u x 4x 5 x 4x 5 x 1 dx u du C 4 16 16 b) tan x 1 sec xdx Answer. Let u tan x du sec xdx. Then u tan x tan x 1 sec xdx u 1 du u C tanx 3 3 a) x 4 3 4x 3 4 3 4 2 3 3 4 3 4 4 C 2 2 2 2 dy dx 2. Solve the differential equation: 3 2 xy2 y2 3 C 0 Answer. We can separate the variables, getting y which integrates to 2 dy xdx x2 C 1 y x 2 C 1 0 2 C y so that 2 1 2 Substituting the initial conditions: which has no solution. Thus there is no function of this type satisfying this initial condition. Recall, however, that when we separated variables, we divided by y 2 ; this can only be done if y 0. Thus the alternative y 0 remains, and provides the solution: the function which is identically zero. 3. Calculate the definite integrals: 1 dx x 3 4 3 4 Answer. x x 4 4 0 3 2 3 2 3 b) cos x sin x dx Answer. Let u sin x du cos xdx. Then when x 0 u cos x sin x dx 4 a) x2 3x 0 3 2 4 0 3 2 π 2 0 0 and when x π 2 0 1 udu 0 1 2 π 2 u 1, and we have 4. Find the area of the region in the first quadrant bounded by the curves y x1 x and y 4 4x 2 . Answer. The second curve is always above the first, and both leave the first quadrant when x 1. Thus the area is 1 1 x2 5 1 Area 4 4x2 x x2 dx 4 x 3x2 dx 4x x3 0 2 2 0 0 5. The region in the first quadrant bounded by the curves y is the volume of the solid so produced? x 2 and x 1 is rotated about the y-axis. What Answer. We sweep out the volume along the x-axis, as x ranges from 0 to 1. The volume of the shell of width dx at the point x is dV 2π xydx 2π x 3 dx. Thus the volume is Volume 1 0 2π x3 dx π 2