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### 2007a_x1a_sols

Course: MATH 152, Fall 2008
School: Texas A&M
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Word Count: 1141

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2007 Spring Math 152 Exam 1A: Problems and Solutions c 2007, Art Belmonte Mon, 19/Feb 2 2 3. (b) Find the value of 0 x 3 e x d x. 2 1. (e) Compute 0 5x + 7 d x. x 2 + 4x + 3 Use integration by parts. Compute an antiderivative, 2 then apply the FTC. Let u = x 2 and dv = e x x d x. 1 2 Then du = 2x d x and v = 2 e x . Hence x 3ex d x = 1 x 2ex - 2 2 3 x2 0 x e 2 2 ex x d x = 2 2 1 2 x 2 - 1 ex . 2...

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2007 Spring Math 152 Exam 1A: Problems and Solutions c 2007, Art Belmonte Mon, 19/Feb 2 2 3. (b) Find the value of 0 x 3 e x d x. 2 1. (e) Compute 0 5x + 7 d x. x 2 + 4x + 3 Use integration by parts. Compute an antiderivative, 2 then apply the FTC. Let u = x 2 and dv = e x x d x. 1 2 Then du = 2x d x and v = 2 e x . Hence x 3ex d x = 1 x 2ex - 2 2 3 x2 0 x e 2 2 ex x d x = 2 2 1 2 x 2 - 1 ex . 2 Split the rational integrand into a sum of partial fractions. 5x + 7 = (x + 1) (x + 3) 5x + 7 = B A + x +1 x +3 A (x + 3) + B (x + 1) Thus or 3e4 + 1 . 2 dx = 1 2 x 2 - 1 ex 2 0 3 = 2 e4 - - 1 2 5x + 7 = (A + B) x + (3 A + B) 4. (c) Find the average value of f (x) = interval [0, 3]. The average value is given by f ave = = = 1 b-a 1 3-0 1 3 1 2 b x x 2 + 16 on the Equate coefficients of like terms. 3A + B A+B = 5 7 = f (x) d x a 3 0 Subtracting the second equation from the first gives -2 A = -2 or A = 1. Thus B = 5 - A = 4. Integrate term-by-term. 2 0 x 2 + 16 -1/2 x dx 4 1 + dx x +1 x +3 (2) x 2 + 16 1/2 3 0 = (ln |x + 1| + 4 ln |x + 3|) 0 = (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 - x 2 is revolved about the y-axis. X1A/P2 X1A/P2: 3-D 2 = 1 5 4 - = . 3 3 3 5. (d) The region in the first quadrant bounded by y = sin x 2 and y = cos x 2 , 0 x 1 , is revolved about the 2 y-axis. Find the volume of the resulting solid. X1A/P5 X1A/P5: 3-D 1 1 0.5 0.5 0 0 0.5 x 1 0.5 0 -0.5 z 0.5 0 -0.5 x y 0 y 4 3 y 4 3 2 1 0 2 1 2 2 1 0 0 1 x 2 y Via cylindrical shells, the volume swept out by the revolving the stated region about the y-axis is given by 0 z 0 x 1 -1 -2 -2 -1 V = = 2r h d x /2 0 Via disks, the volume is V = = = = 0 2x cos x 2 - sin x 2 /2 0 dx r 2 d y = 4 x2 dy = = sin x 2 + cos x 2 2 - 4 - y dy 4 (4 - y)2 0 2 0 - (-8) = 8. - 6. (e) The base of a solid is bounded by y = cos x, - 1 x 1 , and the x-axis. Each cross-section 2 2 perpendicular to the x-axis is a square region whose bottom is sitting on this base. Find the volume of the solid. 1 X1A/P6 1 y Via Hooke's Law, we have F (x) = kx or 5 = 2k, whence k = 5 . 2 The work done is W = 4 0 5 2x b a F (x) d x or = 20 - 0 = 20 ft-lb. 0 -2 -1 0 x 1 2 5 dx = 4 x2 4 0 X1A/P6: 3-D boundaries and cross-sections 10. (d) Find the area of the region bounded by the line y = x + 4 and the parabola y = x 2 - 2. X1A/P10 8 6 y -1 0 1 x X1A/P6: 3-D picture of solid; surface patches 1 0.5 0 0 0.5 y 1 1 z 0 x -1 z 1 0.5 0 0 0.5 y 4 2 0 -2 -2 0 2 x The volume by slicing is V = = = = y2 d x /2 When the curves intersect, their y-coordinates are equal. This yields x +4 = 0 = x2 - 2 x2 - x - 6 cos x d x sin x -/2 /2 -/2 1 - (-1) = 2. 0 = x = (x + 2) (x - 3) -2, 3. 7. (c) A 10-lb object hangs over a ledge at the end of a 20-ft 1 chain that weighs 2 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 20 At x = 0, an interior point of [-2, 3], the line's y-coordinate is 4, whereas that of the parabola is Thus -2. the line lies above the parabola. Hence the area is given by 3 A = = -2 3 -2 (x + 4) - x 2 - 2 d x 6 + x - x2 dx 3 -2 Wrope = 0 1 2x 1 dx = 4 x2 20 0 = 100 ft-lb. = = = = 11. Evaluate x2 x +1 +4 1 1 6x + 2 x 2 - 3 x 3 The total amount of work is 200 + 100 = 300 ft-lb. e 18 + 27 2 9 2 - 9 - -12 + 2 + 8 3 8. (d) Compute 1 ln x d x. x2 - - 22 3 Use integration by parts. Compute an antiderivative, then apply the FTC. Let u = ln x and dv = x -2 d x. 1 Then du = x d x and v = -x -1 . Accordingly, ln x ln x 1 + ln x dx = - + x -2 d x = - . 2 x x x e ln x 2 1 + ln x e = - - (-1) or Thus dx = - 2 1 x e 1 x 2 1- . e 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? 2 81 + 44 125 = . 6 6 3/2 d x. Use trigonometric substitution. Let x = 2 tan . Then d x = 2 sec2 d . Changing variables, the integral becomes 2 tan + 1 2 sec2 d 8 sec3 1 2 = = sin + 1 4 1 4 cos d sin + C. [continued...] 1 - 2 cos + Finally, we express the antiderivative in terms of the original variable x. X1A/P11 14. Compute cos t cos 4t dt. Use a trigonometric product formula cos A cos B = 1 2 cos (A - B) + cos (A + B) (x2 + 4)1/2 x 2 Integrate the transformed integral as follows. cos t cos 4t dt = = = or 1 2 1 2 1 2 1 6 plus the fact the the cosine function is even: cos (- ) = cos . (cos (-3t) + cos 5t) dt cos 3t + cos 5t dt 1 3 1 - 2 = or x 1 + +C x2 + 4 4 x2 + 4 1 x - +C 2+4 2+4 4 x x x -4 +C 4 x2 + 4 2 sin 3t + 1 10 1 5 sin 5t + C sin 3t + sin 5t + C 12. Find the volume of the solid generated by revolving the region in the first quadrant bounded by y = x 3 , the line x = 2, and the x-axis, about the line y = 8. X1A/P12 15. A tank full of water has the depicted shape of a solid of revolution obtained by rotating about the y-axis the regio...

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