Math 10A midterm exam October 26 2017 You put away all books calculators cell phones and other devices You consulted a single two sided sheet of notes You wrote care fully and clearly USING WORDS not just symbols The paper you handed in was your only representative when your work is graded The corrected point counts Problem 1 2 3 4 5 6 7 Total Points 5 5 6 6 6 6 6 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 attempt to explain the main ideas 1 The perimeter of a regular n gon inscribed in the circle of radius 1 is 2n sin n Find the limit as n of this expression Explain in words what you are doing this requirement applies to each of the questions on this midterm This was my attempt to placate Archimedes We calculate lim n 2n sin 2 lim n n sin n n 2 lim h 0 sin h h This last limit is known to be 1 from the beginning of the course so the answer is 2 2a Show that 1 1 1 for m 0 m2 3m 2 m 1 m 2 1 1 cid 88 n 0 If we write the di erence as a single fraction the common denominator will be m 1 m 2 m2 3m 2 and the numerator will be m 2 m 1 1 m 2 m 1 b Find the sum of the in nite series by considering the partial 1 n2 3n 2 sums of the series You acted with honesty integrity and respect for others Math 10A midterm exam October 26 2017 1 2 1 6 1 12 The rst terms are and so the rst three partial sums are 3 4 and This suggests that the partial sums are of the form the sum of the series will therefore be lim k sum as k 1 k 2 1 Write the nth partial 1 2 2 3 and that k 1 k 2 1 1 2 1 2 1 3 1 n 1 n 1 1 1 n 1 n 2 There is massive cancellation between adjacent terms leaving us with the formula that the nth partial sum is 1 1 n 1 n 2 n 2 3 Referring to the diagram below explain carefully why 1 22 1 32 1 n2 1 1 n for n 2 cid 90 2 The graph passes through 1 1 2 25 3 11 This suggests that we re looking at the graph of y 1 x2 and in fact we are I know this because I draw the graph using technology The picture shows that the rectangle extending from 1 to 2 has height 1 4 and area 1 4 and that it s less than than cid 90 3 2 1 x2 dx Similarly the next rectangle has area 1 9 and is less 1 1 x2 dx And as they say et cetera The sum of the areas of the You acted with honesty integrity and respect for others 12345600 20 40 60 81 Math 10A midterm exam October 26 2017 rectangles is the left hand sum in the question and the sum of the integrals is cid 21 n 1 1 x 1 x2 dx 1 1 1 n which is the right hand di erence in the question We get the desired in equality because each rectangle is less than a corresponding integral and therefore the sum of the areas of the rectangles is less than the sum of the integrals which is 1 1 n It follows by the way that 1 i2 lim n 1 1 n 1 cid 90 n 1 cid 88 i 2 You may know that the in nite sum is less than 1 1 0 64 this number is indeed 2 6 4 Determine the volume of the solid obtained by revolving the area under y sin x from x 0 to x about the x axis it may be helpful to know that cos 2x 1 2 sin2 x Hint cid 90 b a We just have to apply the formula Volume y2 dx with a 0 b y sin x The volume is then cid 90 0 sin2 x dx 1 cos 2x dx 2 2 2 Here I ve used the fact that the antiderivative of cos 2x is sin 2x up to some factor the function sin 2x vanishes at 0 and at cid 90 0 2 dy dx 5 Find y as a function of x given y 2x 1 and y 0 2 You acted with honesty integrity and respect for others 0 511 522 530 20 40 60 81 Math 10A midterm exam October 26 2017 This problem is similar to ones you ve seen before Formally we write cid 90 cid 90 1 y dy 2x 1 dx and get ln y x2 x C Exponentiating gives y Kex2 x 2 Ke0 where K is some constant When x 0 y 2 Plugging in this info we get so K 2 Note that the general solution includes the solution y 0 which we get from K 0 This solution disappears when we do the separation of variables manipulation we divided by y assuming implicitly that y cid 54 0 As I have explained in class and on piazza the case where K is negative corresponds to the case where y y and the case where K is positive corresponds to the situation where y y 6 Use integration by parts twice to nd an antiderivative of ex sin x This problem was done out in class on October 12 For details see the notes for that class session They re available on bCourses and also from https math berkeley edu ribet 10A 7 Use the chain rule and the fundamental theorem of calculus to nd cid 32 cid 90 x3 x2 d dx cid 33 sin t2 dt By the fundamental theorem of calculus the integral may be written as F x3 F x2 where F is an antiderivative of the integrand i e where F cid 48 x sin x2 By the chain rule cid 0 F x3 F x2 cid 1 F cid 48 x3 3x2 F cid 48 x2 2x 3x2 sin x6 2x sin x4 d dx You acted with honesty integrity and respect for others