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M161, Test 2, Spring 2007 Name: Section: Instructor: Time: 75 minutes. You may not use calculators on this exam d sin(x) = cos(x), dx d csc(x) = − csc(x) cot(x), dx d sinh(x) = cosh(x), dx d 1 asin(x) = √ , dx 1 − x2 d 1 acsc(x) = − √ 2 , dx x x −1 d 1 asinh(x) = √ , dx 1 + x2 sin(2x) = 2 sin(x) cos(x) tan2 (x) + 1 = sec2 (x) d cos(x) = − sin(x), dx d sec(x) = sec(x) tan(x), dx d cosh(x) = sinh(x), dx 1 d acos(x) = − √ , dx 1 − x2 d 1 asec(x) = √ 2 , dx x x −1 d 1 , acosh(x) = √ 2 x −1 Zdx ln xdx = x ln x − x + C 1 + cos(2x) cos2 (x) = 2 Problem Points Score 1 20 2ab 16 2cd 16 3 16 4 16 5 16+5 X 100+5 d tan(x) = sec2 (x), dx d cot(x) = − csc2 (x), dx d tanh(x) = sech2 (x), dx 1 d atan(x) = , dx 1 + x2 d 1 acot(x) = − dx 1 + x2 d 1 atanh(x) = , 1 − x2 Zdx sec(x)dx = ln(sec(x) + tan(x)) + C 1 − cos(2x) sin2 (x) = 2 Theorem (The Derivative Rule for Inverses) If f has an interval I as domain and f 0 (x) exists and is never zero on I, then f −1 is differentiable at every point in its domain. The 1 value of (f −1 )0 at a point b = f (a) in the domain of f −1 is given by (f −1 )0 (b) = 0 . f (a) 1) The following terms are prototypes of the terms resulting from a partial fraction decomposition. In each case determine an antiderivative: Z 1 dx a) (x − 1) b) c) d) Z 1 dx (x − 1)2 Z x+1 dx (x2 + 1) Z (x2 x+1 dx + 2x + 2) 2) Z Evaluate the following integrals. Show your work. sin5 x cos2 xdx a) b) Z x5 ln xdx c) Z 0 d) Z ∞ (x2 √ x dx + 4)3/2 x2 dx 4 − x2 3) For each of the following improper integrals explain whether the integral converges or diverges. (You do not need to calculate the values of convergent integrals.) Z 1 1 dx a) 3 0 x +x b) ∞ 1 dx x3 + x ∞ 1 dx x2 ∞ x dx ex Z 1 c) Z 0 d) Z 1 4) We want to perform a partial fraction decomposition of the quotient x11 − 4x10 + 8x9 − 17x8 + 4x7 + 56x6 − 112x5 + 256x4 − 320x3 + 128x2 − 255x + 259 (x + 2) (x − 2)3 (x2 + 4)2 (x2 + x + 1) Circle below which of the listed terms will (for suitable values of the coefficients) occur in this decomposition (there are 8 terms occuring in total. You do not need to solve for the values of the variables, perform an actual partial fraction decomposition or find an antiderivative). You get 2 points for every correct term, -1 points for every wrong term. You cannot get less than 0 points in this problem. A) Ax + B B) A Bx C) A x−1 D) E) A x−2 F) Ax + B x−2 G) A (x − 2)2 H) Ax + B (x − 2)2 I) A (x − 2)3 J) Ax + B (x − 2)3 K) A x+2 L) Ax + B x+2 x5 A +x+7 M) A (x + 2)2 N) Ax + B (x + 2)2 O) A (x + 2)3 P) Ax + B (x + 2)3 Q) A (x2 + 4) R) Ax + B (x2 + 4) S) A (x2 + 4)2 T) Ax + B (x2 + 4)2 A + 4)3 V) Ax + B (x2 + 4)3 W) A (x2 + x + 1)2 Z) Ax + B (x2 + x + 1)2 U) Y) (x2 (x2 A + x + 1) X) Ax + B + x + 1) (x2 5) In this problem we want to see that asinh(x) really is not a “new” function deserving its own name. Z 1 a) We know that asinh(x) + C = √ 2 dx. Integrate this using a suitable trigonometric x +1 substitution to end up with a solution not involving hyperbolic functions. b) Let F (x) be the antiderivative you obtained in Part a). We know that F (x)+C = asinh(x) for some constant C. Determine the value of C. c (bonus) ) Verify your calculation by showing that F (sinh(x)) + C = x.