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- Texas A&M University
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- Mathematics 152
- Albrecht/austin
- Exam 1 Solutions (Form A) Spring 2007- Detailed.pdf

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Spring 2007 Math 152 Exam 1A: Problems and Solutions Mon, 19/Feb c©2007, Art Belmonte 1. (e) Compute integraldisplay 2 0 5x + 7 x2 + 4x + 3 dx. • Split the rational integrand into a sum of partial fractions. 5x + 7 (x + 1)(x + 3) = A x + 1 + B x + 3 5x + 7 = A (x + 3)+ B (x + 1) 5x + 7 = (A + B) x + (3A + B) • Equate coefficients of like terms. A + B = 5 3A + B = 7 Subtracting the second equation from the first gives −2A = −2 or A = 1. Thus B = 5 − A = 4. • Integrate term-by-term. integraldisplay 2 0 1 x + 1 + 4 x + 3 dx = (ln|x + 1|+ 4 ln|x + 3|)vextendsinglevextendsingle20 = (ln 3 + 4 ln 5)− (ln 1 + 4 ln 3) = 4 ln 5 − 3 ln 3 Here we have recalled that ln 1 = 0. 2. (a) Find the volume of the solid generated when the region in the first quadrant bounded by x = 0, y = 0, and y = 4 − x2 is revolved about the y-axis. 0 1 2 0 1 2 3 4 x y X1A/P2 −2 −1 0 1 2 −2−1 01 20 1 2 3 4 x X1A/P2: 3−D z y • Via disks, the volume is V = integraldisplay pir2 dy = integraldisplay pix2 dy = pi integraldisplay 4 0 4 − y dy = parenleftBig −pi2 (4 − y)2 parenrightBigvextendsinglevextendsingle vextendsingle40 = 0 − (−8pi) = 8pi. 3. (b) Find the value of integraldisplay 2 0 x3ex2 dx. • Use integration by parts. Compute an antiderivative, then apply the FTC. Let u = x2 and dv = ex2 x dx. Then du = 2x dx and v = 12 ex2. Henceintegraltext x3ex2 dx = 12 x2ex2 −integraltext ex2 x dx = 12 parenleftbigx2 − 1parenrightbigex2. • Thus integraltext 20 x3ex2 dx = 12 parenleftbigx2 − 1parenrightbigex2 vextendsinglevextendsingle vextendsingle20 = 32 e4 − parenleftBig −12 parenrightBig or 3e 4 + 1 2 . 4. (c) Find the average value of f (x) = xradicalbig x2 + 16 on the interval [0, 3]. • The average value is given by fave = 1b − a integraldisplay b a f (x) dx = 13 − 0 integraldisplay 3 0 parenleftBig x2 + 16 parenrightBig−1/2 x dx = parenleftBig1 3 parenrightBigparenleftBig1 2 parenrightBig (2)parenleftbigx2 + 16parenrightbig1/2 vextendsinglevextendsingle vextendsingle30 = 53 − 43 = 13. 5. (d) The region in the first quadrant bounded by y = sin parenleftbigx2parenrightbig and y = cosparenleftbigx2parenrightbig, 0 ≤ x ≤ 12√pi, is revolved about the y-axis. Find the volume of the resulting solid. 0 0.5 1 0 0.5 1 x y X1A/P5 −0.5 0 0.5 −0.500.5 0 0.5 1 x X1A/P5: 3−D z y • Via cylindrical shells, the volume swept out by the revolving the stated region about the y-axis is given by V = integraldisplay 2pirh dx = pi integraldisplay √pi/2 0 2x parenleftBig cos parenleftBig x2 parenrightBig − sin parenleftBig x2 parenrightBigparenrightBig dx = pi parenleftBig sin parenleftBig x2 parenrightBig + cos parenleftBig x2 parenrightBigparenrightBigvextendsinglevextendsingle vextendsingle √pi/2 0 = √ 2 pi − pi 6. (e) The base of a solid is bounded by y = √cos x, −12pi ≤ x ≤ 12pi, and the x-axis. Each cross-section perpendicular to the x-axis is a square region whose bottom is sitting on this base. Find the volume of the solid. 1 −2 −1 0 1 2 0 1 x y X1A/P6 −1 0 10 0.5 1 0 0.5 1 x X1A/P6: 3−D boundaries and cross−sections y z −1 0 10 0.50 0.5 1 x X1A/P6: 3−D picture of solid; surface patches y z • The volume by slicing is V = integraldisplay y2 dx = integraldisplay pi/2 −pi/2 cos x dx = sin x vextendsinglevextendsingle vextendsinglepi/2−pi/2 = 1 − (−1) = 2. 7. (c) A 10-lb object hangs over a ledge at the end of a 20-ft chain that weighs 12 lb per foot. Find the total work (in ft-lb) done hauling the object up to the ledge. • The work done lifting the object itself is Wobject = (10)(20) = 200 ft-lb. • The work done lifting the rope is Wrope = integraldisplay 20 0 1 2 x dx = 1 4 x 2 vextendsinglevextendsingle vextendsingle200 = 100 ft-lb. • The total amount of work is 200 + 100 = 300 ft-lb. 8. (d) Compute integraldisplay e 1 ln x x2 dx. • Use integration by parts. Compute an antiderivative, then apply the FTC. Let u = ln x and dv = x−2 dx. Then du = 1x dx and v = −x−1. Accordingly,integraldisplay ln x x2 dx = − ln x x + integraldisplay x−2 dx = −1 + ln xx . • Thus integraldisplay e 1 ln x x2 dx = parenleftbigg −1 + ln xx parenrightbiggvextendsinglevextendsingle vextendsinglee1 = −2e − (−1) or 1 − 2e . 9. (b) For a certain type of linear spring, the force required to keep it stretched 2 feet beyond its natural length is 5 lb. How much work (in ft-lb) is done stretching this spring 4 feet beyond its natural length? • Via Hooke’s Law, we have F (x) = kx or 5 = 2k, whence k = 52 . • The work done is W = integraltext ba F (x) dx or integraldisplay 4 0 5 2 x dx = 5 4 x 2 vextendsinglevextendsingle vextendsingle40 = 20 − 0 = 20 ft-lb. 10. (d) Find the area of the region bounded by the line y = x + 4 and the parabola y = x2 − 2. −2 0 2 −2 0 2 4 6 8 x y X1A/P10 • When the curves intersect, their y-coordinates are equal. This yields x + 4 = x2 − 2 0 = x2 − x − 6 0 = (x + 2)(x − 3) x = −2, 3. • At x = 0, an interior point of [−2, 3], the line’s y-coordinate is 4, whereas that of the parabola is −2. Thus the line lies above the parabola. • Hence the area is given by A = integraldisplay 3 −2 (x + 4)− parenleftBig x2 − 2 parenrightBig dx = integraldisplay 3 −2 6 + x − x2 dx = parenleftBig 6x + 12 x2 − 13 x3 parenrightBigvextendsinglevextendsingle vextendsingle3−2 = parenleftBig 18 + 92 − 9 parenrightBig − parenleftBig −12 + 2 + 83 parenrightBig = 272 − parenleftBig −223 parenrightBig = 81 + 446 = 1256 . 11. Evaluate integraldisplay x + 1 parenleftbigx2 + 4parenrightbig3/2 dx. • Use trigonometric substitution. Let x = 2 tan θ. Then dx = 2 sec2 θ dθ. • Changing variables, the integral becomes integraldisplay 2 tan θ + 1 8 sec3 θ · 2 sec 2 θ dθ = integraldisplay 1 2 sin θ + 1 4 cos θ dθ = −12 cos θ + 14 sin θ + C. [continued...] 2 • Finally, we express the antiderivative in terms of the original variable x. x (x2 + 4)1/2 2 θ X1A/P11 −12 2radicalbig x2 + 4 + 14 xradicalbig x2 + 4 + C = x 4 radicalbig x2 + 4 − 1radicalbig x2 + 4 + C or x − 4 4 radicalbig x2 + 4 + C 12. Find the volume of the solid generated by revolving the region in the first quadrant bounded by y = x3, the line x = 2, and the x-axis, about the line y = 8. 0 1 2 0 2 4 6 8 x y X1A/P12 0 1 2 0 8 160 8 16 z X1A/P12: 3−D x y • Via washers, the volume is V = integraldisplay pir2o − pir2i dx = pi integraldisplay 2 0 82 − parenleftBig 8 − x3 parenrightBig2 dx = pi integraldisplay 2 0 16x3 − x6 dx = pi parenleftBig 4x4 − 17 x7 parenrightBigvextendsinglevextendsingle vextendsingle20 = parenleftBig 64 − 1287 parenrightBig pi − 0 = 3207 pi 13. Find integraldisplay cos3 3θ sin−2 3θ dθ. • Use the trigonometric identity sin2 x + cos2 x = 1. integraldisplay cos 3θ parenleftBig 1 − sin2 3θ parenrightBig sin−2 3θ dθ = integraldisplay (sin 3θ)−2 cos 3θ − cos 3θ dθ = −13 (sin 3θ)−1 − 13 sin 3θ + C or −13 (csc 3θ + sin 3θ)+ C 14. Compute integraldisplay cos t cos 4t dt. • Use a trigonometric product formula cos A cos B = 12parenleftbigcos (A − B)+ cos (A + B)parenrightbig plus the fact the the cosine function is even: cos (−θ) = cos θ. • Integrate the transformed integral as follows. integraltext cos t cos 4t dt = integraltext 1 2 (cos (−3t)+ cos 5t) dt = 12 integraltext cos 3t + cos 5t dt = 12 parenleftBig1 3 sin 3t + 1 5 sin 5t parenrightBig + C or 16 sin 3t + 110 sin 5t + C 15. A tank full of water has the depicted shape of a solid of revolution obtained by rotating about the y-axis the region in the first quadrant bounded by x = (4y)1/3 and the lines y = 2 and x = 0. Find the work (in joules) required to pump the water out of the tank. Lengths are in meters. The density of water is ρ = 1000 kg/m3 and g = 9.8 m/s2. Front view of tank filled with water with a typical water layer • From the diagram on the exam depicting a typical layer of water, we follow the “march of the differentials” for volume, mass, force, and work. dV = pir2 dy = pix2 dy = pi (4y)2/3 dy dm = ρ dV = ρpi (4y)2/3 dy d F = (dm) g = gρpi (4y)2/3 dy dW = (d F)(D) = gρpi (4y)2/3 (2 − y) dy • Set up the and dispatch the work integral. (This begs for calculator firepower. Perhaps in future terms... ) W = integraldisplay dW = gρpi integraldisplay 2 0 2 (4)2/3 y2/3 − 42/3 y5/3 dy = gρpi parenleftBig 2 (4)2/3 parenleftBig3 5 parenrightBig y5/3 − 42/3 parenleftBig3 8 parenrightBig y8/3 parenrightBigvextendsinglevextendsingle vextendsingle20 = gρpi parenleftBig48 5 − 48 8 parenrightBig − 0 = 48gρpi parenleftBig8−5 40 parenrightBig = 185 gρpi = 185 parenleftBig98 10 parenrightBig (1000)pi = 35,280pi joules 3 x1a_sols.dvi

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