Math 132 Fall 2007 Final Exam 2 3 1 Calculate cos x sin x dx 0 a 1 f 2 3 1 2 3 g 4 1 3 3 h 2 b c 1 4 4 i 3 d 1 5 1 j 6 e Solution d J Int cos x sin x 3 x 0 Pi 2 2 J cos x sin x 3 dx 0 K student changevar u sin x J u 1 K u3 du 0 value K 1 4 2 5 t4 2 Let F x dt 1 t3 Calculate the derivative D F 2 of F at 2 x a 4 f 4 b 5 g 5 Solution i c 6 h 6 d 7 i 7 e 8 j 8 F x Int 5 t 4 sqrt 1 t 3 t x 2 2 5 t4 F x dt 1 t3 x D F x 5 x4 1 x3 D F 2 7 9 3 simplify D F 2 7 1 x 3 Calculate dx x 1 x 2 0 9 a ln 8 7 b ln 6 5 c ln 4 4 d ln 3 9 f ln 5 8 g ln 3 9 h ln 4 16 i ln 3 3 e ln 2 16 j ln 9 Solution a J Int x x 1 x 2 x 0 1 1 x J dx x 1 x 2 0 R student integrand J R x x 1 x 2 PFE convert R parfrac x PFE 1 x 1 antiderivative int PFE x 2 x 2 antiderivative ln x 1 2 ln x 2 definiteIntegral subs x 1 antiderivative subs x 0 antiderivative definiteIntegral 3 ln 2 2 ln 3 ln 1 Answer combine definiteIntegral ln 8 Answer ln 9 1 2 8x 2x 6 4 Calculate dx 2 1 x 1 x 0 a 1 ln 2 4 f 4 ln 2 b 1 ln 2 2 g 5 ln 2 c ln 2 d 2 ln 2 e 3 ln 2 h 6 ln 2 i 7 ln 2 j 8 ln 2 Solution i J Int 8 x 2 2 x 6 1 x 1 x 2 x 0 1 1 8 x2 2 x 6 J dx 2 x 1 1 x 0 R student integrand J R 8 x2 2 x 6 x 1 1 x2 PFE convert R parfrac x PFE 2x 1 x antiderivative int PFE x 2 6 x 1 antiderivative ln 1 x2 6 ln x 1 definiteIntegral subs x 1 antiderivative subs x 0 antiderivative definiteIntegral 7 ln 2 7 ln 1 Answer combine definiteIntegral ln Answer 7 ln 2 e 2 5 Calculate x ln x dx 1 a f 1 3 e3 1 e3 2 3 b g 1 2 e3 1 3 2 e3 1 3 c h 1 e3 2 3 d 1 2 e3 1 9 i 2 e3 1 3 e 1 e3 2 9 j 1 2 e3 1 3 2 e3 1 9 Solution h J Int x 2 ln x x 1 exp 1 e J x2 ln x dx 1 K student intparts J ln x Integration by Parts with u ln x e 2 x 1 3 K e dx 3 3 1 value K 2 9 6 What is the derivative of a ln 2 f 1 ln 2 2 b 1 ln 2 2 g 1 ln 2 x 1 x c e 3 1 9 with respect to x at x 1 ln 2 h 1 1 ln 2 2 d 1 i 1 2 1 ln 2 2 1 ln 2 4 e ln 2 j 1 4 Solution g restart eqn1 f x x 1 x 1 x eqn1 f x x eqn2 map z simplify ln z symbolic eqn1 eqn2 ln f x ln x x eqn3 map z diff z x eqn2 d eqn3 dx f x ln x 1 f x x x2 eqn4 D f x solve eqn3 diff f x x eqn4 D f x 2 f x ln x 1 x2 eqn5 subs x 1 2 eqn4 1 1 1 eqn5 D f 4 f ln 1 2 2 2 eqn6 subs x 1 2 eqn1 1 1 eqn6 f 2 4 subs eqn6 eqn5 1 D f ln 2 1 2 7 If y 0 0 a sin cos x and dy cos x dx 2 1 y then what is y x cos x b sin sin x c cos 2 f arcsin arcsin x g sin tan x h tan sin x arcsin arctan x Solution b By the Method of Separation of Variables d cos sin x 1 i arcsin tan x 2 e arcsin x j eqn1 Int 1 sqrt 1 t 2 t 0 y x Int cos t t 0 x eqn1 y x x dt cos t dt 1 t2 0 1 0 eqn2 map value eqn1 eqn2 arcsin y x sin x Answer y x solve eqn2 y x Answer y x sin sin x For those who are interested here is how to get MAPLE to solve this differential equation without the user supplying any guidance ode diff y x x cos x sqrt 1 y x 2 ode d y x cos x dx initialCondition y 0 0 1 y x 2 initialCondition y 0 0 IVP ode initialCondition d IVP y 0 0 y x cos x 1 y x 2 dx dsolve IVP y x Using Maple s differential equation solver y x sin sin x 8 Consider the following three statements about a series I The series converges because a with positive terms n 1 n lim a 0 n n a II The series converges because n 1 1 1 b n lim n a III The series converges because lim n n 1 1 a n and b converges n 1 n For each statement determine whether the reasoning is correct or incorrect a I correct II correct b I correct II correct c I correct II incorrect d I correct II incorrect e I incorrect II correct f I incorrect II correct g I incorrect II incorrect h I incorrect II incorrect i Wrong answer j Bonus wrong answer III correct III incorrect III correct III incorrect III correct III incorrect III correct III incorrect Solution f I Incorrect When an 1 n the terms of the series satisfy lim a 0 but the series diverges n n II Correct The assertion follows from the Limit Comparison Test III Incorrect The limit condition is the inconclusive case of the Ratio Test When an lim n an 1 an 1 n we have 1 but the series diverges 9 Consider the following three statements about a series a with positive terms n 1 n 1 I The series converges because a n 10 n II The series diverges because 1 a n 2 n a III The series converges because lim n n 1 0 a n For each statement determine whether the reasoning is correct C or incorrect F a I C II …
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