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Exam 2 – Answers on Second Page Math 2423 Instructions Give thorough explanations of your work to guarantee getting full credit. Use of calculators is not permitted. Please write on the back of the pages if you need more space. 1. (5 points) Sketch the graphs of y = ex and y = ln(x) in the same picture. 2. (15 points) Determine the intervals of monotonicity for the function f (x) = ln(x2 − 2x + 2) and identify any local maximums or minimums for f . (Hint: The domain of f consisits of all real numbers.) 3. (5 points) One of the two integrals Z ex + sin(x) dx ex + cos(x) Z and ex + cos(x) dx ex + sin(x) is easy to compute and the other is very difficult. Identify the easier one and work it out. (a) sin(tan−1 ( 32 )) 4. (10 points) Use a right triangle analysis to simplify each of: 5. (15 points) If f (x) = 3−x then find a formula for the inverse function f −1 (x). 5 + 2x 6. (15 points) Differentiate (and simplify): (a) g(x) = x arctan(x4 ), 7. (20 points) Compute each of the following integrals: Z (b) Z √ (b) cos(tan−1 ( 23 )) (a) R (b) h(x) = ln(sec(x) + tan(x)) e3x+1 dx 1 dx 1 − x2 e2 1 dx 1/e x Z p ln(5x) (d) dx x (c) 8. (10 points) Let R be the triangular region underneath the line y = 3 − 2x in the first quadrant. If R is rotated about the x-axis express the volume of the resulting solid as an integral using: (a) the disk/washer method (b) the shell method. 9. (5 points) Let R be the triangular region described in the previous problem. If R is rotated about the horizontal line y = 5 then express the volume of the resulting solid as an integral. ANSWERS: 1. Since the two functions are inverses of each other, their graphs will be mirror images across the line y = x. Check the textbook for the graphs of these very important functions. 2. To find the critical points of f we set f 0 (x) = 2(x − 1)/(x2 − 2x + 2) equal to 0, and solve for x. This gives x = 1 as the only critical point. (NOTE: the only way a quotient can equal zero is if its numerator equals zero.) The intervals of monotonicity are: f is increasing on (1, ∞) and f is decreasing on (−∞, 1). (These are determined by observing that f 0 (x) > 0 when x > 1, and f 0 (x) < 0 when x < 1 because the numerator x2 − 2x + 2 = (x − 1)2 + 1 is always positive.) There is one local extreme, it is a local minimum at x = 1. 3. The second integral can be worked by making the substitution u = ex + sin(x) (and du = (ex + cos(x))dx), which results in Z ex + cos(x) dx = ln(| ex + sin(x) |) + C ex + sin(x) 4. The right traingle would legs with lengths 3 and 2, and the hypotenuse would have length √ √ 2/ 13 and cos(tan−1 ( 23 )) = 3/ 13 5. f −1 (x) = √ 13, so sin(tan−1 ( 32 )) = 3 − 5x 2x + 1 6. (a) g 0 (x) = arctan(x4 ) + (b) h0 (x) = sec(x) 4x4 1 + x8 7. Z (a) Z e3x+1 dx = 1 3x+1 e +C 3 1 dx = sin−1 (x) 1 − x2 Z e2 1 dx = 3 (c) x 1/e Z p ln(5x) 2 (d) dx = (ln(5x))3/2 + C x 3 (b) √ 8. The volume works out to be 9π/2 using either method below (however the problem doesn’t require that the integral be computed). (a) Using the disk method, the volume is (b) Using the shell method, the volume is R 3/2 R03 0 π(3 − 2x)2 dx. πy(3 − y) dy. R 3/2 9. Using the washer method, the volume equals 0 π(25 − (2 + 2x)2 )) dx. Using the shell method, the volume R3 equals 0 π(5 − y)(3 − y) dy. (Either integral would be considered to be a correct answer.) With either method the volume works out to be 18π (however the problem doesn’t require that the integral be computed).