Math 132 Exam 1 Fall 2016• 16 multiple choice questions worth 4.5 points each.• 2 hand graded questions worth 14 points each.• Exam covers sections 4.9 through 6.2• No calculators!• For the multiple choice questions, mark your answer on the answer card.• Show all your work for the written problems. Your ability to make your solution clearwill be part of the grade.Useful FormulasnXi=1i =n(n + 1)2nXi=1i2=n(n + 1)(2n + 1)6nXi=1i3=n(n + 1)22(Area Circle Radius r) = πr2(Area Ellipse With Semi-Major Axis a, Semi-Minor Axis b) = πabMath 132 Exam 1 Page 2 of 191. Let f(x) be the function satisfying f00(x) = sin x + ex, f0(0) = 2, and f(0) = 0. Findf(π/2).A. eπ/2+ π/2 − 1B. eπ/2+ πC. 2eπ/2+ π/2D. eπ+ π/2 − 2E. eπ/2− 1F. eπ/2+ π − 2G. eπ/2− π + 1Solution: Integrate twice. f0(x) = −cos x + ex+ C1. Use f0(0) = 2 to get C1= 2and f0(x) = −cos x + ex+ 2.f(x) = −sin x + ex+ 2x + C2. Use f(0) = 0 to get C2= −1 and f(x) = −sin x +ex+ 2x − 1.Plugging in gives f(π/2) = eπ/2+ π − 2.2. Suppose you know the following about a function f(x):•Z41f(x) dx = 4•Z82f(x) dx = −3•Z81f(x) dx = 4FindZ212f(x) − 6dx.A. −5B. −3C. 0D. 2E. 4Math 132 Exam 1 Page 3 of 19F. 6G. 8H. None of the aboveSolution:Z212f(x) − 6dx =2Z21f(x) dx − 6Z21dx=2Z21f(x) dx − 6(1) = −6 + 2Z21f(x) dx= − 6 + 2Z81f(x) dx −Z82f(x) dx= − 6 + 2(4 − (−3)) = 8Math 132 Exam 1 Page 4 of 193. Suppose f(x), f0(x), and f00(x) are all continuous functions and you know the followingadditional information about f(x):f(2) = 4 f(6) = 12f0(2) = 5 f0(6) = 11f00(2) = 6 f00(6) = 10If possible, findZ62f00(x) dx.A. 0B. 2C. 4D. 6E. 8F. 10G. 12H. We do not have enough information to determineZ62f00(x) dx.Solution: Use the FTC, Part 2:Z62f00(x) dx = f0(6) − f0(2) = 11 − 5 = 64. ComputeZ621 + |x − 4|dx.A. 0B. 2C. 4D. 6Math 132 Exam 1 Page 5 of 19E. 8F. 10G. 12H. None of the aboveSolution: If you draw the graph you can just compute the area. Looking at thegraph you should be able to see that the area is 8.(You could also split the integral up, that might be slightly easier.)2 3 4 5 6 712345y = 1 + |x − 4|xyMath 132 Exam 1 Page 6 of 195. FindZπ/20sin5x cos x dx.A. 0B.16C. −16D.13E. −13F. 1G. −1H. None of the above.Solution:Zπ/20sin5x cos x dx =Zx=π/2x=0u5du Let u = sin x, du = cos x dx=u66x=π/2x=0=sin6x6x=π/2x=0=sin6(π/2)6−sin6(0)6=16− 06. Let u = ln x and rewrite the following integral in the variable u:Zx2ln x dx.A.Zu e3uduB.Zu duC.Zu e2uduD.Zu2ln u duE.Zu2euduMath 132 Exam 1 Page 7 of 19F.Zeuln u duG.Zu3ln u duH.Zu2e2uduSolution: If u = ln x then du =1xdx, dx = x du and x = eu:Zx2ln x dx =Zx2u · x du=Zux3du=Zue3uduMath 132 Exam 1 Page 8 of 197. ComputeZ30√36 − 4x2dx.A. 4.5B. 4.5πC. 9D. 9πE. 18F. 18πG. 36H. None of the aboveSolution:Z30√36 − 4x2dx = 2Z30√9 − x2dxGraphing y =√9 − x2is a quarter of a circle of radius 3 and thus has area14π32=94π.Multiply this by two to get the answer.8. A function f(x) and a number b satisfy the question3 +Zxbf(t)t4dt = 24x−3.What is b?A. 0B. 1C. 2D. 3E. 4F. ln 3G. ln 8H. It is not possible to determine b.Math 132 Exam 1 Page 9 of 19Solution: Plugging in x = b makes the integral equal to 0 and gives the equation3 + 0 = 24b−3. Solving gives b = 2Math 132 Exam 1 Page 10 of 199. Let f(x) = x2+ 1. Compute L4over the interval [−1, 3].(L4is the Riemann sum using left endpoints as sample points, with 4 subdivisions.)A. 0B. 2.75C. 3D. 5E. 9F. 10G. 12.5H. 18Solution: With 4 subdivisions x0= −1, x1= 0, x2= 1, x3= 2 and x4= 3.∆x =b−a4=3−(−1)4= 1.L4=4Xi=1f(xi−1)∆x=f(x0)(1) + f(x1)(1) + f(x2)(1) + f(x3)(1)=f(−1)(1) + f(0)(1) + f(1)(1) + f(2)(1)=(1 + 1) + (0 + 1) + (1 + 1) + (4 + 1) = 1010. Identify the definite integral that is equal to the limit of Riemann Sums:limn→∞nXi=11 +4in84nA.Z50(x + 1)7dxB.Z84(x + 4)7dxC.Z62x8dxMath 132 Exam 1 Page 11 of 19D.Z40x8dxE.Z52(x + 1)8dxF.Z51x8dxG. None of the aboveSolution: You should be able to see that ∆x =4n, which means b−a = 4. Then, youcan see f(xi) =1 +4in8. While there are several possibilities, one is that f(x) = x8and xi= 1 + 4i/n. Since xi= a + i∆x this means that a = 1. Putting all thesetogether givesZ51x8dx.Math 132 Exam 1 Page 12 of 1911.Z30(3x2− 2) dx = limn→∞Rn, where Rnis the right hand Riemann sum. Find Rn.A.27(n + 1)(2n + 1)2n2−6nB.9(n + 1)(2n + 1)2n2− 6C.18(n + 1)2n2−2nD.81(n + 1)(2n + 1)2n2− 2E.27(n + 1)(2n + 1)2n2− 6F.18(n + 1)3n+ 15G.9(n + 1)22n2− 3H. None of the aboveSolution:Rn=nXi=1f(xi)∆x =nXi=1f(0 + i∆x)∆x=nXi=1f(3i/n)3n=nXi=13(3i/n)2− 23n=nXi=181i2− 6n2n3=81n3nXi=1i2−6nnXi=11=81n3n(n + 1)(2n + 1)6−6n· n=27(n + 1)(2n + 1)2n2− 6Math 132 Exam 1 Page 13 of 1912. Let F (x) =Zx3sin(x2)ln(3t + 5)dt. Find F0(x).A. ln(3x3+ 5)B. ln(3 sin(x2) + 5)C. ln(3x3+ 5) − ln(3 sin(x2) + 5)D. 3x2ln(3x3+ 5)E. 2x cos(x2) ln(3 sin(x2) + 5)F. 3x2ln(3x3+ 5) − 2x cos(x2) ln(3 sin(x2) + 5)G. 3x2ln(3x3+ 5) + 2x cos x2ln(3 sin(x2) + 5)H. 0Solution:ddxZx3sin(x2)ln(3t + 5)dt =ddx Z0sin(x2)ln(3t + 5)dt +Zx30ln(3t + 5)dt!=ddx −Zsin(x2)0ln(3t + 5)dt +Zx30ln(3t + 5)dt!=ddx −Zsin(x2)0ln(3t + 5)dt!+ddx Zx30ln(3t + 5)dt!= − 2x cos(x2) ln(3 sin(x2) + 5) + 3x2ln(3x3+ 5)Math 132 Exam 1 Page 14 of 1913. If g(x) =Zx1312t2+ 2tdt, what is g0(1)?A. 0B. 1C. 3D. 4E. 6F. 10G. 12H. None of the aboveSolution: g0(x) =12x2+2xand g0(1) =121+2= 4.14. Find the area of the region enclosed by the graphs of f(x) = 1 − x2and g(x) = x2− 1.A. 2B.12C.32D.35E.43F. 4G.83H. None of the aboveSolution: First solve f(x) = g(x) and get x = −1, 1. Note that when x is in [−1, 1]we have f(x) > g(x). SoArea =Z1−1(f(x) − g(x)) dx =Z1−1(2 − 2x2) dx =83Math 132 Exam 1 Page 15 of 1915. A solid is formed with a base that is a triangle with vertices at (0, 0), (5, 0) and (0, 1).Cross sections of this solid, perpendicular to the x axis are squares.Find the volume of the solid.A. 0B.56C.53D.52E. 3F. 5G. 6H. ∞Solution: Note the equation of the line forming the hypotenuse of the triangle isy = 1 −15x. Thus:V =Z50(Area of Cross Sectional Area) dx=Z50y2dx =Z50(1 − x/5)2dx =5316. Let R be the region in the plane enclosed by y = x3, y = 0, and x = 1.Find the volume of the solid formed by rotating R about the axis x = 2.A.3π5B.8π3C.2π3D.3π4E.7π4F.33π5G.5π3H. None of the aboveMath 132 Exam 1 Page 16 of 19Solution: Graph the region.1 2 3 41RxyIf you
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