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Calculus I Final Exam, Spring 2003, Answers 1. Find the integrals: a) esinx 2 cos xdx Answer. Let u cos xdx. Then e e cos xdx sin x du sinx 2 Alternatively, let v b) x x u 2 esinx dv du e2u du e2u 2 e2 sinx 2 C esinx cos xdx, so that esin x 2 cos xdx esinx esinx cos xdx vdv v2 2 C C esin x 2 2 C 1dx x 1 so that x u 1 and du dx. Then 2 2 x x 1dx u 1 u du u u du 5 x 1 3 x 1 C Answer. Let u 1 2 2. Integrate t2 3 2 1 2 5 2 3 2 2 dt t2 1 t 1 t 2 B t 1 t 2 C t 1 t 1 t 2 Answer. . We expand the function in partial fractions. The roots are -1,1,2, so we write 2 t 1 t 1 t 2 Equate the numerators at the roots. 1 A 2 3 so A t 1 : 12 t 1 : 1 B 2 1 so B 4 t 2 : 2 C 4 1 so C 3 t2 t2 1 t A B C At 2 2 2 1 6 2 2 This gives us 3. Integrate t2 dt 2 16 ln t 1 t2 1 t t 1 1 ln t 1 2 1 6 dt 1 2 dt 4 3 dt 1 t 4 ln t 2 C 3 t 2 x ln xdx Answer. We integrate by parts so as to get rid of the ln term: u ln x dv xdx, so that du and x2 x2 dx x2 x2 x ln xdx ln x ln x C 2 2 x 2 4 dx x v x 2 2, 4. A certain compound transforms from state A to state B at a (per minute) rate proportional to the concentration of B in the mixture: dcA 02cB dt where cA and cB are the concentrations of A and B respectively (and, assuming no other material is present, cA cB 1). If at time t 0 the mixture is 90% in state A how long will it take to be 10% A? Answer. Substitute cB 1 cA in the differential equation, and separate variables, obtaining dcA 1 cA 02dt ln 1 c 02t C exponentiating to 1 c Ke Solve for K using c 9 when t 0, getting K 1. Now solve for c : c 1 1e (Of course this makes sense so long as c 0; once A is gone, the process stops.) Now set c 1 and solve for t: 1e 9, so ln 9 t 02 109 86 minutes Integrate both sides and exponentiate: 02t A A A A 02t A A 02t A 5. Find the limit. Show your work. Answer. At x ln x 1 sin π x a) lim x 1, both numerator and denominator are zero, so l’Hò‚pital’s rule applies: x b) lim lim 1 x 1 π cos π x π ln x 1 sin π x lim lH x 1 1 xex x 0 e2x Answer. Again both numerator and denominator are zero at x lim x 3x6 7x4 ∞ 2 x3 12 c) lim 0, so lim e xe 1 2e xex x 0 e2x x lH x 2x x 0 1 2 3 2 since the factors have the same degree. 6. Do the integrals converge? If so, evaluate: ∞ a) Answer. 0 xe x dx lim A ∞ 0 A xe x dx lim xe e lim e A 1 1 1 x A ∞ x A 0 A A ∞ ∞ b) 2 dx x ln x 25 Answer. Let u ln x du dx x. Then dt x ln x which converges to 1 24 ln2 as A A 24 du 25 ln 2 u 25 2 u24 A 1 24 A ln 2 1 ln 2 ln A1 24 24 ∞. 24 7. Do the series converge or diverge? Give a valid reason for your answer. n 1 n 1 Answer. The series diverges by comparison with the p-test with p ∞ a) ∑ 2 n 1 3 1: the denominator is only of degree 1 more than the numerator. ∞ b) ln n n 1 2 ∑ n Answer. The series converges by comparison with a geometric series: ln n 2n 2n 2 2n 12 n 2 1. n! 1 c) ∑ n 1 ! series converges by comparison with the p-test with p 2. Divide both numerator and denomAnswer. The inator by n! : n! 1 1 2 ! " n 1! n 1 and 1 ∞ 2 2 n 1 2 1 2 n! n 1! 2 n! 2 2 2 since the numerator is bounded by 2. 8. Find the vertices of the conic given by the equation 4x 2 1 y Answer. Complete the square: 4 y 2 8x 4y 12 4x 1 y 2 4 4 0 or This gives the standard form x 31 y 122 1 This is a hyperbola with center at (-1,-2) and axis the line x 1. Setting x the vertices: y 2 12 1 or y 2 # 12 Thus the vertices are at 1 2 2 3 and 1 2 2 3 . 4x 2 2x 2 4y 12 2 0 2 2 12 2 1 gives the y coordinates of 2 9. Find the area of the region enclosed by the curve given in polar coordinates by r the segment of the x axis between x 2 and x 2e 2π . θ 2e θ 0 2π and PSfrag replacements 200 50 $ 200$ 0 50 $ 100 $ 150 $ 200 $ 250 $ 300 $ 350 0 200 400 600 800 1000 1200 Answer. From the diagram we see that the area is Area 1 2 2π 0 r2 d θ 2π e2θ d θ 2 0 e2θ 2π 0 e4π 10. a) Find the general solution of the homogeneous differential equation y Answer. The roots of the equation r 2 3r 2 yh % %& 3y% 2y 0. 0 are 1,2. Thus the general solution is Aex Be2x b) Find a particular solution of the homogeneous differential equation y 1 % %' 3y% 2y sin x. Answer. We use the method of undetermined coefficients. Try a solution of the form y A cos x Bsin x: 3 A sin x Bcos x 2 A cos x Bsin x sin x leading to the equations A 3B 2A 0 ( B 3A 2B 1. The solutions are B 1 10 A 3 10, so a particular solution is y 0 3 cos x 0 1 sin x A cos x Bsin x p