Math 10A midterm exam October 27 2016 Please put away all books calculators cell phones and other de vices You may consult a single two sided sheet of notes Please write carefully and clearly USING WORDS not just symbols Remember that the paper you hand in will be your only represen tative when your work is graded Please write your name clearly on each page of your exam Your paper will be scanned and will be processed using Gradescope It is essential that you hand in all pages that you have received including this cover sheet and that the order of the pages be preserved Point counts Problem 1 2 3 4 5 Total Points 9 8 8 7 8 40 These quick and dirty solutions were written by me Ribet just before the exam These are not necessarily perfect model solutions but rather my at tempt to explain what s going on You probably picked up on the fact that a number of these problems come from old exams or from the textbook s exercises That s laziness on my part To put a positive spin on my choice we can say that I selected problems that you were likely to have seen before thereby increasing the chances that you could do them without much strain 1 Discuss the convergence of each of the following in nite series cid 88 n 1 a ln n 2n cid 18 1 cid 88 n 1 b 1 n n 1 cid 19 If an ln n 2n then an 1 an ln n 1 ln n 1 ln n 1 as n so 1 2 can see that series converges by the ratio test ln n By l H opital s rule or whatever you ln n 1 ln n 1 2 1 and thus the You acted with honesty integrity and respect for others Math 10A midterm exam October 27 2016 The sum of the rst three terms is 1 1 2 1 2 1 3 1 3 1 4 1 1 4 In the same vein the sum of the rst N terms is 1 1 quantity approaches 1 Thus the series converges its sum is 1 N 1 As N this cid 88 n 1 nn n c nn n cid 18 n cid 88 i 1 cid 19 cid 18 2 cid 19 n 1 2i n For each n the fraction is bigger than 1 Hence the nth term of this series does not approach 0 It follows that the series diverges The sum of the rst N terms is bigger than N 2a Express lim n as an integral of the form f x dx This problem is a lot like problem 6 of 5 3 which was part of HW 7 In fact it s problem 4 of 5 3 The idea is this if you use right endpoints then you see that cid 90 1 0 Matching things up we see that we want cid 90 1 0 f x dx lim n n cid 88 i 1 1 n cid 18 cid 19 f cid 18 i cid 19 n 2 1 2i n cid 18 i cid 19 n f To achieve this we take f x 2 1 2x 2 4x Then the limit is 2 4x dx b Express cos x dx as a limit of Riemann sums cid 90 1 0 cid 90 1 1 2 Math 10A midterm exam October 27 2016 There s no single correct answer to this problem The problem by the way is 12 of 5 3 with the absolute value signs removed James told me that the absolute value signs would be distracting We can for example divide the interval 1 1 into n equal segments and use left endpoints this time around The intervals have length so the Riemann sum becomes cid 18 cid 19 2 n cos 1 2i n 2 n n 1 cid 88 cid 90 i 0 cid 90 cid 90 cid 90 0 The integral is the limit of this sum as n approaches 3a Find the inde nite integral sin3 x dx Use that sin2 cos2 1 The hint nudges us to write the integrand as sin x 1 cos2 x Set u cos x du sin x dx The integral is then u3 1 u2 du u 3 C cos x cos3 x C 3 b Evaluate ex 1 ex 2 dx Let u ex du ex dx In u land the integral becomes 0 1 2 1 u 2 du 1 1 u 1 1 cid 18 cid 19 1 2 cid 21 1 4 Calculate the volume of the football shaped solid obtained by rotating the interior of the ellipse 1 about the x axis x2 9 y2 4 cid 90 b y2 dx for the volume In this case y2 4 We use the formula The possible values of x range from 3 to 3 because the quantity 1 x2 9 to be non negative The volume is a cid 90 3 cid 18 cid 19 4 3 1 x2 9 dx 4 cid 18 cid 19 cid 21 3 3 x x3 27 3 8 3 1 16 cid 18 cid 19 1 x2 9 needs Math 10A midterm exam October 27 2016 5a Evaluate this function of t se s ds cid 90 t 0 The integral is ripe for integration by parts In the inde nite integral we set u s dv e s ds v e s Then the formula cid 90 cid 90 u dv uv v du cid 90 se s ds becomes cid 90 The answer is then se s ds se s e s ds se s e s se s e s 1 te t e t Alternatively we can carry along the limits of integration as we work se s ds se s e s ds te t e s ds etc cid 90 t 0 cid 90 t 0 cid 90 t2 0 d dt e x2 cid 21 t cid 90 t 0 0 cid 90 u 0 form G t2 where G u e x2 dx b Find dx Hint write the function to be di erentiated in the Following the hint we write the function to be di erentiated as G t2 Use the chain rule if u t2 the derivative is G cid 48 u of calculus G cid 48 u e u2 be 2te t4 2t Thus the derivative appears to By the fundamental theorem e t4 du dx du dx also cid 90 cid 21 t 0 4