Math 132Midterm Examination 1 Solutions – February 15, 20126 multiple choice, 4 long answer. 100 points.Part I was multiple choice. Only the correct answers are listed here.1. Let xi=3i2n− 1 for i = 0, 1, . . . , n. These xi’s form a partition of the interval:(b) [−1,12]2. Which of the following is equal to19Xi=11i(c) 1 +12+13+ ··· +119.3. Which of the following is an antiderivative of11+4x2?(d)12tan−1(2x)4.ddxZx−πsin t2dx is equal toThis problem had an error, and was intended to beddxZx−πsin t2dt. As this is easy tomiss in reading, we accepted the answer for the intended problem:(a) sin x2as well as:(i) None of the above.5. If f is a continuous function such thatR120f(t) dt = 3,R122f(t) dt = 4, andR42f(t) dt =1, then findR40f(t) dt.(e) 0(SinceR40f(t) dt =R120f(t) dt −R122f(t) dt +R42f(t) dt.)6. Which of the following definite integrals havenXi=1(1 +in)4·2nas an associated Riemann sum?I.Z20(1 +x2)4dxII.R31x4dxIII.R102 · (1 + x)4dx(f) I and III only.Part II was long answer.1. Integration(a) (5 points) EvaluateZ1−1sin πx dx.Solutions 1: Substitute u = πx, so that du = π dx, andZ1−1sin πx dx =Zπ−πsin u1πdu =−cos u ·1ππ−π= 0.Solution 2: Since sin is an odd function, and [−π, π] is symmetric around they-axis, the integral is 0 by symmetry.(b) (5 points) EvaluateZx sin(x2) dx.Substitute u = x2, so that du = 2x dx, andZx sin x2dx =Zsin udu2= −cos u2+ C = −cos x22+ C.(c) (5 points) EvaluateZln 3ln 2e2x√1 + exdx.Substitute u = 1 + ex, so that du = exdx.Zln 3ln 2e2x√1 + exdx =Z43(u − 1) ·√u du =Z43u3/2− u1/2du =25u5/2−23u3/243= (25· 45/2−23· 43/2) − (25· 35/2−23· 33/2)=645−163−18√35+ 2√3.(d) (7 points) Solve the initial value problem: Ifddtf(t) = 8t · (2t2+ 1)4and f(0) = 1,then find f(t).We first integrate 8t · (2t2+ 1)4. Substitute u = 2t2+ 1, so that du = 4t dt, andwe haveZ8t · (2t2+ 1)4dt =Z2u4du = 2u55+ C = 2 ·(2t2+ 1)55+ C.We now solve for C. At t = 0, we havef(0) = 1 = 2 ·155+ C =25+ C,so C =35, and f(t) = 2 ·(2t2+1)55+35.2. Areas and volumes(a) (15 points) Find the volume of the object formed by rotatingy =rx1 + x2about the x-axis for 0 ≤ x ≤ 2.The cross-sectional areas perpendicular to the x-axis have area A(x) = π ·x1+x2,so the volume isV =Z20πx1 + x2dx.We substitute u = 1 + x2(so du = 2x dx) to get=Z51πu12du =π2[ln u]51=π2ln 5(b) (10 points) Find the area between the curves y = sin x and y = cos x for0 ≤ x ≤π2.The curves cross when sin x = cos x, i.e., atπ4. Plotting test points at e.g. 0 andπ2, we determine that cos x > sin x on [0, π/4], and the reverse on [π/4, π/2]. Thearea is thus Zπ/40cos x − sin x dx!+ Zπ/2π/4sin x − cos x dx!= [sin x + cos x]π/40+ [−cos x − sin x]π/2π/4= √22+√22− 0 − 1!+ −1 − 0 +√22+√22!= 2√2 − 2.(The area is symmetric over the line x = π/4, and this could be used to slightlyshorten 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 x2that you are using.By the FTC and continuity of sin x2,ddxZx0sin t2dt = sin x2(see also Problem 4). Hence,Zx0sin t2dtis an antiderivative for sin x2.(b) (8 points) FindZπ/20(ddxex sin x) dx.Since ex sin xis an antiderivative forddxex sin x, the integral is=ex sin x π/20= eπ/2− e0.4. Riemann sums and definite integrals(a) (5 points) The points1 = x0< x1< x2< ··· < xn−1< xn= 3form the uniform partition of [1, 3]. Find xi, and give the length of each part.xi= 1 +3 − 1n· i = 1 +2in; (xi− xi−1) =2n.(b) (6 points) Give any Riemann sum forZ311xdx. Be sure to explain the choice ofpartition and points that you make.There are a number of possible answers to this problem; perhaps the easiest is totake the uniform partition xi= 1 +2infrom part (a) with right endpoints (so thatci= xi), which givesnXi=1f(xi) · (xi− xi−1) =nXi=111 +2in·2n.(c) (4 points) In 1-3 sentences, explain why the functionf(x) =(1 if x is rational0 otherwisefrom Worksheet 1 is not integrable on [0, 1].We showed on the worksheet that different rules for choosing the points ciin theRiemann sums result in different limits. (Specifically, 0 and 1 from the “irrationalrule” and “rational rule”, respectively.) Since the definition of definite integralrequires all choices of Riemann sums to converge to the same limit, the functionis not
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