M408C - revfinsol - M408C Calculus 55815/20/25 Fall 2002...

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Unformatted text preview: M408C Calculus 55815/20/25 Fall 2002 Final Exam Review Problems with Solutions The Final Exam will be comprehensive, covering the material from sections 2.1, 2.32.5, 3.13.3, 3.53.7, 4.14.8, 5.15.7, 6.16.3, 7.17.5, 7.7, 8.2 of Salas and Hille's Calculus, 8th edition, with more emphasis on the last sections. Problem 1. Evaluate the limits that exist a. lim Solution. a. lim x0 tan 3x ; x0 2x2 + 5x b. lim h3 - 4h . h2 h3 - 2h2 tan 3x tan 3x 3x 3 = lim = 2x2 + 5x x0 3x x(5 + 2x) 5 b. lim h(h - 2)(h + 2) h+2 h3 - 4h = lim = lim =2 h2 h2 h3 - 2h2 h2 h2 (h - 2) h Problem 2. Differentiate a. f (x) = ex arctan x; b. f (x) = 4 - x2 + 2 arcsin(x/2); c. f (x) = arctan(ln x). Solution. a. f (x) = ex arctan x + ex (1/2)(-2x) + 4 - x2 1 1 = ex arctan x + 2 1+x 1 + x2 2(1/2) 1- (x/2)2 2-x = = 4 - x2 2-x 2+x b. f (x) = c. f (x) = (ln x) 1 = 1 + (ln x)2 x(1 + (ln x)2 ) Problem 3. Find the critical numbers and classify the extreme values for f (x) = x3 - 3x + 6 x [0, 3/2]. Solution. The function f (x) is differentiable in (0, 3/2) so the only critical points are the points where f (x) = 0. f (x) = 3x2 - 3 = 3(x2 - 1) = 3(x - 1)(x + 1) The only root of f (x) which is in (0, 3/2) is x = 1. So, x = 1 is the only critical point. Moreover, since f (x) < 0 for x < 1 close to 1, and f (x) > 0 for x > 1, we conclude that x = 1 is a point of local minimum. M408C Calculus 55815/20/25 Fall 2002 Next, near endpoints, we have f (0) < 0 and f (3/2) > 0, therefore both x = 0 and x = 3/2 are endpoint maximums. Compare now the values of f at x = 0, 1, 3/2: f (0) = 6, f (1) = 4, f (3/2) = 39/8 < 6 Hence x = 1 is point of absolute minimum, x = 0 is point of absolute maximum. Problem 4. Determine the concavity and find the points of inflection for the 2 function f (x) = e-x . Solution. We have to find the intervals where f (x) does not change the sign. f (x) = -2xe-x 2 f (x) = (-2x)2 e-x - 2e-x = (4x2 - 2)e-x Now factor f (x) f (x) = 4 x - 1/ 2 2 x + 1/ 2 e-x . 2 2 2 2 Now, since e-x is always positive, we obtain that f (x) < 0 on -1/ 2, 1/ 2 and f (x) > 0 on -, -1/ 2 and 1/ 2, . This implies that f (x) is con vex on intervals -, -1/ 2 and 1/ 2, and concave on -1/ 2, 1/ 2 . Points x = 1/ 2 are points of inflection. Problem 5. Sketch the region bounded by the curves x + y = 2y 2 and y = x3 and find the volume of the solid obtained by revolving it about x-axis. Use the shell method. 1.5 1 0.5 -0.5 -0.5 0.5 1 1.5 Solution. Represent the given curves as the graphs of x = 2y 2 - y and x = y 1/3 . Then the region will be the region between the graphs of these two functions of y, and for y between the solutions of 2y 2 - y = y 1/3 M408C Calculus 55815/20/25 Fall 2002 It's not hard to guess the solutions: y = 0, y = 1. Now, using the shell method, we obtain 1 1 V = 2 0 y(y 1/3 - (2y 2 - y))dy = 2 0 1 (y 4/3 - 2y 3 + y 2 )dy 3 1 1 - + 7 2 3 = 11 . 21 3 7/3 2 4 1 2 = 2 y - y + y 7 4 3 = 2 0 Problem 6. Sketch the region bounded by y = 3x - x2 and y = 3 - x and find the volume of the solid generated by revolving the region about the x-axis. Use the washer method. 2 1 0.5 -1 1 1.5 2 2.5 3 3.5 Solution. To find the region (call it ), we find the points of intersection of the two curves: set 3x - x2 = 3 - x (x - 1)(x - 3) = 0, hence is the region between the graphs of f (x) = 3x - x2 and g(x) = 3 - x for x [1, 3] (the shaded region on the figure.) Now, by the washer method, the volume of the solid generated by revolving around the x-axis will be 3 3 V = 1 [f (x)2 - g(x)2 ]dx = 1 (x2 (x - 3)2 - (x - 3)2 )dx. Now, to avoid lengthy computations set, u = x - 3. Then 0 0 V = -2 ((u+3)2 u2 -u2 )du = -2 (u4 +6u3 +8u2 )du = u5 3u4 8u3 + + 5 2 3 0 = -2 56 15 Problem 7. Integrate /2 a. x 10x dx; b. /6 cos x dx; 1 + sin x 2 c. 1 x3 dx 3x4 + 1 M408C Calculus 55815/20/25 Fall 2002 Solution. a. Integrate by parts, choosing u = x, dv = 10x dx. Then v = 10x / ln 10 and x 10x dx = x 1 10x - ln 10 ln 10 10x dx = x 10x 10x +C - ln 10 (ln 10)2 b. Substitute u = 1 + sin x. Then du = cos x dx. When x changes from /6 to /2, u changes from 3/2 to 2. Hence /2 /6 cos x dx = 1 + sin x 2 3/2 du = ln u u 2 3/2 = ln 2 - ln(3/2) = ln(4/3). c. Substitute u = 3x4 + 1. Then du = 12x3 dx. When x changes from 1 to 2, u changes from 4 to 49. Hence 2 1 x3 1 dx = 12 3x4 + 1 49 4 du 1 = u 6 u 49 = 4 7-2 5 = . 6 6 Problem 8. Find the derivative of f (x) = (x + 1)x . Solution. Using the logarithmic differentiation: ln f (x) = x ln(x + 1) f (x) x = ln(x + 1) + f (x) x+1 Hence f (x) = (x + 1)x ln(x + 1) + x x+1 Problem 9. Calculate g (1) by using logarithmic differentiation g(x) = (1 + x)(2 + x)x (4 + x)(2 - x) Solution. Take the logarithm of both sides ln g(x) = ln(1 + x) + ln(2 + x) + ln x - ln(4 + x) - ln(2 - x) and differentiate 1 1 1 1 1 g (x) = + + - + . g(x) 1+x 2+x x 4+x 2-x Hence at x = 1 g (1) 1 1 1 79 = + +1- +1= . g(1) 2 3 5 30 M408C Calculus 55815/20/25 Fall 2002 On the other hand 231 6 g(1) = = 51 5 and we obtain 79 6 79 g (1) = = 30 5 25 Problem 10. Use integration by parts to calculate 1/2 a. 0 x2 cos x dx; b. x ln(x + 1)dx Solution. a. 1/2 x2 cos x dx = 0 1 1/2 x2 d sin x 0 1/2 = = = = 1 2 x sin x 1 2 + 4 2 - 0 1/2 2 1/2 x sin x dx 0 x d cos x 0 1/2 1 2 + 2 x cos x 4 1 2 - 3 sin x 4 - = 0 1/2 0 2 2 1/2 cos x dx 0 1 2 - 4 3 Solution using DI chart. We are going to differentiate x2 and integrate cos x. D x2 2x 2 I cos x + - 1 sin x 1 - 2 cos x + 1 - 3 sin x Hence, 1/2 x2 cos xdx = 0 1 2 2 2 x sin x + 2 x cos x - 3 sin x 1/2 = 0 1 2 . - 4 3 b. Choose f (x) = ln(x + 1) and g (x) = x. Then we can take g(x) = x2 /2 and x ln(x + 1) dx = x2 1 ln(x + 1) - 2 2 x2 dx. x+1 M408C Calculus 55815/20/25 Fall 2002 Compute the latter integral separately: x2 x2 - 1 + 1 dx = dx = x+1 x+1 (x - 1)2 + ln(x + 1) + C. 2 Thus, x ln(x + 1) dx = = x2 1 (x - 1)2 ln(x + 1) - ln(x + 1) - +C 2 2 4 x2 - 1 (x - 1)2 ln(x + 1) - + C. 2 4 (x - 1) + 1 x+1 dx = ...
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This note was uploaded on 06/19/2008 for the course M 408 C taught by Professor Treisman during the Fall '07 term at University of Texas at Austin.

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