page 1 of 8 Math 1B First Exam Tuesday 7 July 2009 Instructor Theo Johnson Freyd http math berkeley edu theojf 09Summer1B Name Problem Number Score Maximum ANSWERS 1a 1b 1c 1d 2a 2b 3 4a 4b 4c Total 10 10 10 15 5 15 100 10 10 10 5 Please do not begin this test until 2 10 p m You may work on the exam until 4 p m Please do not leave during the last 15 minutes of the exam time You must always justify your answers show your work show it neatly and when in doubt use words and pictures to explain your reasoning Please box your final answers Calculators are not allowed Please sign the following honor code I the student whose name and signature appear on this midterm have completed the exam by myself without any help during the exam from other people or from sources other than my allowed one page hand written cheat sheet Moreover I have not provided any aid to other students in the class during the exam I understand that cheating prevents me from learning and hurts other students by creating an atmosphere of distrust I consider myself to be an honorable person and I have not cheated on this exam in any way I promise to take an active part in seeing to it that others also do not cheat Signature Name ANSWERS Math 1B First Exam 7 July 2009 1 40 pts 4 questions 10 pts each Evaluate the following integrals Z dx a x x2 4 We substitute x 2 tan Then Z Z 2 sec2 d dx x x2 4 2 tan 4 tan2 4 Z 2 sec2 d 2 tan 2 sec Z 1 sec d 2 tan Z 1 cos 1 d 2 sin cos Z 1 csc d 2 1 ln csc cot C 2 1 x2 4 2 ln C 2 x x x2 4 2 1 C ln 2 x Z b x 1 dx x2 6x 5 We factor the denominator and expand into partial fractions Z Z x 1 x 1 dx dx 2 x 6x 5 x 1 x 5 Z 3 2 1 2 dx x 1 x 5 1 3 ln x 1 ln x 5 C 2 2 page 2 of 8 Name ANSWERS Z c Math 1B First Exam 7 July 2009 page 3 of 8 3 x sec x tan x dx 0 We integrate by parts with u x and dv sec x tan x dx Z 3 x sec x tan x dx x sec x 0 3 0 Z 3 tan x dx 0 3 3 x sec x 0 ln sec x tan x 0 sec 0 ln sec tan ln 1 0 3 3 3 3 2 ln 2 3 3 2 ln 2 3 3 Z d cos x sin x 2 dx We expand and use trig identities Z Z cos x sin x 2 dx cos2 x 2 cos x sin x sin2 x dx Z 1 2 cos x sin x dx Z 1 sin 2x dx x 1 cos 2x C 2 Name ANSWERS Math 1B First Exam 7 July 2009 page 4 of 8 2 20 pts 2 questions 10 pts each Determine whether each of the following improper integrals is convergent or divergent You do not need to prove your answer and you do not need to evaluate theR integrals although you are welcome to do either or both For example if I asked you about 1 x2 x 1 dx you could evaluate the integral partial fractions or you could 2 2 use the comparison test to prove that it converges but R it s 2enough to write x x x as x so we expect the integral to converge since 1 x dx converges by the p test Z 1 cos x dx Hint do not try to evaluate a x 0 The only potential divergence is at x 0 When x 0 we have cos x 1 and so Z 0 small cos x dx x Z 0 small 1 dx x which converges by the P test with p 1 2 Thus we expect the integral to converge Z b dx Hint u substitution x ln x 2 We substitute u ln x Then du dx x and ln so the integral is Z Z du dx x ln x 2 ln 2 u which diverges by the P test Thus the original integral diverges Name ANSWERS Math 1B First Exam 7 July 2009 page 5 of 8 R3 3 15 pts Let s say you want to know the numerical value of ln 3 1 x 1 dx and you decide to approximate the integral using the trapezoid rule How many subintervals do you need to take in order to assure that your answer is accurate to within an error of 0 00001 10 5 We recall the error formula for the Trapezoid Rule E K b a 3 12 n2 where K is any number such that K f 00 x for all x a b where a b 1 3 and f x x 1 Then f 00 x 2x 3 which is a decreasing function so it is maximized when x 1 whence K 2 works Thus we have E 2 23 4 n 2 12 n2 3 We want this number to be less than 10 5 Thus we solve 4 2 n 10 5 3 3 2 n 105 4 4 n2 105 3 r 4 5 n 10 3 p 40 3 102 We round 40 3 13 13 up to 16 a perfect square and conclude that n 4 102 400 works Name ANSWERS Math 1B First Exam 7 July 2009 page 6 of 8 4 Each of the following questions required the answers to the previous ones If you answer one incorrectly I will grade the later problems as if your answer was correct so that you will not be penalized twice If you cannot answer one question you may ask me for the answer and I will mark your test as such a 5 pts Let a be an arbitrary positive real number Sketch the curve y x 0 x a Write an expression for the arclength as an integral in terms of x We sketch the curve y x 4 3 2 1 1 0 1 2 3 4 5 6 7 8 9 1 Recalling the arclength formula the length of the curve for 0 x a is Z r 1 0 since if f x a 1 dx 4x x then f 0 x 1 2 x b 5 pts Explain why your expression in part a is an improper integral 1 x 0 4x We remark that the integral nevertheless converges which we prove byqevaluating it R 1 3 1 1 in the next question Indeed 1 4x x1 for x 1 3 and so 0 1 4x dx R 1 3 1 2 x dx which converges by the P test 0 The integral is improper at x 0 The integrand …
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