Math 132 Midterm Examination 1 Solutions February 15 2012 6 multiple choice 4 long answer 100 points Part I was multiple choice Only the correct answers are listed here 1 Let xi 3i 1 for i 0 1 n These xi s form a partition of the interval 2n 1 b 1 2 2 Which of the following is equal to 19 X 1 i 1 c 1 12 13 i 1 19 3 Which of the following is an antiderivative of 1 tan 1 2x 2 d d 4 dx 1 1 4x2 Z x sin t2 dx is equal to Z x d This problem had an error and was intended to be sin t2 dt As this is easy to dx miss in reading we accepted the answer for the intended problem a sin x2 as well as i None of the above 5 If f is a continuous function such that R4 1 then find 0 f t dt R 12 0 f t dt 3 R 12 2 e 0 Since R4 0 f t dt R 12 0 f t dt R 12 2 f t dt R4 2 f t dt f t dt 4 and R4 2 f t dt 6 Which of the following definite integrals have n X i 1 1 i 4 2 n n as an associated Riemann sum Z 2 x 1 4 dx I 2 0 R3 4 II 1 x dx R1 III 0 2 1 x 4 dx f I and III only Part II was long answer 1 Integration Z 1 sin x dx a 5 points Evaluate 1 Solutions 1 Substitute u x so that du dx and Z 1 Z 1 1 sin x dx sin u du cos u 0 1 Solution 2 Since sin is an odd function and is symmetric around the y axis the integral is 0 by symmetry Z b 5 points Evaluate x sin x2 dx Substitute u x2 so that du 2x dx and Z Z du cos u cos x2 2 x sin x dx sin u C C 2 2 2 Z ln 3 c 5 points Evaluate e2x 1 ex dx ln 2 Substitute u 1 ex so that du ex dx Z ln 3 Z 4 Z 2x x e 1 e dx u 1 u du ln 2 3 3 4 u 3 2 u 1 2 2 2 du u5 2 u3 2 5 3 2 2 2 2 45 2 43 2 35 2 33 2 5 3 5 3 64 16 18 3 2 3 5 3 5 4 3 d 7 points Solve the initial value problem If then find f t d f t 8t 2t2 1 4 and f 0 1 dt We first integrate 8t 2t2 1 4 Substitute u 2t2 1 so that du 4t dt and we have Z Z u5 2t2 1 5 2 4 8t 2t 1 dt 2u4 du 2 C 2 C 5 5 We now solve for C At t 0 we have f 0 1 2 so C 53 and f t 2 2t2 1 5 5 2 15 C C 5 5 35 2 Areas and volumes a 15 points Find the volume of the object formed by rotating r x y 1 x2 about the x axis for 0 x 2 The cross sectional areas perpendicular to the x axis have area A x so the volume is Z 2 x V dx 2 0 1 x We substitute u 1 x2 so du 2x dx to get Z 5 1 du ln u 51 ln 5 2 2 1 u 2 x 1 x2 b 10 points Find the area between the curves y sin x and y cos x for 0 x 2 The curves cross when sin x cos x i e at 4 Plotting test points at e g 0 and we determine that cos x sin x on 0 4 and the reverse on 4 2 The 2 area is thus Z 4 Z 2 cos x sin x dx sin x cos x dx 0 4 2 sin x cos x 4 cos x sin x 4 0 2 2 2 2 0 1 1 0 2 2 2 2 2 2 2 The area is symmetric over the line x 4 and this could be used to slightly shorten the calculations 3 The Fundamental Theorem of Calculus a 6 points Using definite integrals give an antiderivative of sin x2 For full credit explain clearly what theorems and or properties of sin x2 that you are using By the FTC and continuity of sin x2 Z x d sin t2 dt sin x2 dx 0 see also Problem 4 Hence Z x sin t2 dt 0 2 is an antiderivative for sin x Z 2 d ex sin x dx b 8 points Find dx 0 Since ex sin x is an antiderivative for d x sin x e dx the integral is 2 ex sin x 0 e 2 e0 4 Riemann sums and definite integrals a 5 points The points 1 x0 x1 x2 xn 1 xn 3 form the uniform partition of 1 3 Find xi and give the length of each part 3 1 2i 2 i 1 xi xi 1 n n n Z 3 1 b 6 points Give any Riemann sum for dx Be sure to explain the choice of 1 x partition and points that you make xi 1 There are a number of possible answers to this problem perhaps the easiest is to take the uniform partition xi 1 2i from part a with right endpoints so that n ci xi which gives n X i 1 f xi xi xi 1 n X i 1 1 1 2i n 2 n c 4 points In 1 3 sentences explain why the function 1 if x is rational f x 0 otherwise from Worksheet 1 is not integrable on 0 1 We showed on the worksheet that different rules for choosing the points ci in the Riemann sums result in different limits Specifically 0 and 1 from the irrational rule and rational rule respectively Since the definition of definite integral requires all choices of Riemann sums to converge to the same limit the function is not integrable
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