Exam 1 Solutions (Form A) Spring 2007- Detailed.pdf

Spring 2007 Math 152 Exam 1A: Problems and Solutions Mon, 19/Feb cİ2007, 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 ?. Then dx = 2 sec2 ? d?. ? Changing variables, the integral becomes integraldisplay 2 tan ? + 1 8 sec3 ? · 2 sec 2 ? d? = integraldisplay 1 2 sin ? + 1 4 cos ? d? = ?12 cos ? + 14 sin ? + C. [continued...] 2 ? Finally, we express the antiderivative in terms of the original variable x. x (x2 + 4)1/2 2 ? 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 3? sin?2 3? d?. ? Use the trigonometric identity sin2 x + cos2 x = 1. integraldisplay cos 3? parenleftBig 1 ? sin2 3? parenrightBig sin?2 3? d? = integraldisplay (sin 3?)?2 cos 3? ? cos 3? d? = ?13 (sin 3?)?1 ? 13 sin 3? + C or ?13 (csc 3? + sin 3?)+ 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 (??) = cos ?. ? 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 ? = 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 = ? dV = ?pi (4y)2/3 dy d F = (dm) g = g?pi (4y)2/3 dy dW = (d F)(D) = g?pi (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 = g?pi integraldisplay 2 0 2 (4)2/3 y2/3 ? 42/3 y5/3 dy = g?pi parenleftBig 2 (4)2/3 parenleftBig3 5 parenrightBig y5/3 ? 42/3 parenleftBig3 8 parenrightBig y8/3 parenrightBigvextendsinglevextendsingle vextendsingle20 = g?pi parenleftBig48 5 ? 48 8 parenrightBig ? 0 = 48g?pi parenleftBig8?5 40 parenrightBig = 185 g?pi = 185 parenleftBig98 10 parenrightBig (1000)pi = 35,280pi joules 3 x1a_sols.dvi