Math 132 Exam III Spring 2008 1 This exam contains 15 multiple choice questions and 2 hand graded questions The multiple choice questions are worth 5 points each and the hand graded questions are worth a total of 25 points The latter questions will be evaluated not only for having the correct solutions but also for clarity Points may be taken for confusing and disorganized writing even when the answer is correct 1 Evaluate the integral Z 4 5 2x 1 dx x2 5x 6 A 3 ln 3 2 B 5 ln 2 C ln 2 3 5 ln 3 D 3 ln 3 4 ln 2 E 5 ln 2 3 ln 5 F ln 2 5 G 3 ln 2 3 5 ln 2 H 3 5 ln 2 I 3 ln 5 J 4 ln 2 3 7 ln 2 First note that x2 5x 6 x 2 x 3 We can expand the integrand into partial fractions by solving for A and B 2x 1 A B x2 5x 6 x 2 x 3 Writing both sides over a common denominator and equating the numerators gives 2x 1 A x 3 B x 2 This is easily solved and gives A 3 B 5 Therefore Z 5 Z 5 2x 1 3 5 dx dx 2 x 2 x 3 4 x 5x 6 4 3 ln x 2 5 ln x 3 54 3 ln 2 3 5 ln 2 Math 132 Exam III Spring 2008 2 Evaluate the integral Z 1 A ln 2 5 B ln 1 2 C ln 3 2 D ln 3 2 E ln 7 3 F ln G ln ln 5 2 ln 1 5 2 5 ln 5 5 H ln 7 2 I ln 2 7 ln 2 dx dx x x2 1 2 3 2 J ln 3 The partial fractions decomposition has the form A Bx C 1 2 x x2 1 x x 1 Equating the numerators after setting both sides over the same denominator A x2 1 Bx C x 1 Equivalently A B x2 Cx A 1 This gives B A C 0 and A 1 Therefore Z 2 Z 2 1 1 x dx dx 2 x x2 1 1 x x 1 1 h i2 p ln x ln 1 x2 1 ln 2 5 ln 2 2 Math 132 Exam III Spring 2008 3 3 Determine whether the following improper integrals converge or diverge Z Z 5 Z Z 1 dx dx dx dx a b c d 19 20 20 19 3 2 x x x 4 1 0 3 0 x Below c stands for converges and d for diverges A c c c c B d c d d C d d c d D c c c d E c d c c F d d d d G c c d c H c c d d I d c c d J c d d d Ra Recall that if a is a finite positive number the integral 0 xdxp converges for p R and diverges for p 1 Therefore both b and d diverge The integral a xdxp converges for p 1 and diverges for p 1 Therefore a diverges The integral R c is equal to 1 udu 3 2 this is seen by doing a substitution u x 4 which converges Math 132 Exam III Spring 2008 4 4 Determine whether the integral 1 Z I x ln x dx 0 is convergent or not If it is convergent evaluate it A convergent I 1 8 B convergent I 1 2 C convergent I 1 D convergent I 1 2 E convergent I 1 F convergent I 2 G convergent I 1 4 H divergent I convergent I 4 J convergent I 4 I lima 0 R1 a x ln x dx Integration by parts gives 1 Z 1 x 2 a 2 2 1 x x ln x 2 4 a x ln x dx a x2 ln x 2 Z 1 a2 ln a a2 4 2 4 It can be shown using L Ho spital s rule that lima 0 a ln a 0 From this limit it immediately follows that lima 0 a2 ln a 0 Therefore the limit as a 0 exists and is equal to 1 4 Math 132 Exam III Spring 2008 5 5 Calculate the arc length of the graph of y x3 2 over the interval 1 2 A 2e 7 h 5 2 i 7 5 5 2 B 27 34 2 h i 8 13 3 2 11 3 2 C 27 2 4 h i 5 2 11 5 2 D 83 11 2 4 h i 1 2 1 2 E 83 92 47 h 1 2 i 3 2 15 F 14 13 2 4 h 1 2 i 3 2 G 13 72 17 4 h i 3 2 15 3 11 3 2 H 19 2 4 8 13 3 2 I 9 4 8 11 3 2 J 27 2 The arclength I is given by 2 Z I Z p 1 f 0 x 2 dx 1 1 2 r 1 9 x dx 4 The change of variables u 1 9x 4 gives 4 I 9 Z 1 9 2 1 9 4 u1 2 du i 8 h 8 h 3 2 i11 2 u 11 2 3 2 13 4 3 2 27 27 13 4 Math 132 Exam III Spring 2008 6 6 Compute the surface area of revolution defined by the function y x 1 over the interval 0 1 A 2 2 B 5 2 C 7 2 D 3 5 E 7 5 F 2 5 G 8 2 H 3 2 I 3 7 J 5 3 The surface area is given by Z 1 Z 1 p p x 1 1 12 dx 2 f x 1 f 0 x 2 dx 2 0 0 Z 1 2 2 x 1 dx 0 1 2 x x 2 2 2 0 2 2 3 2 3 2 Math 132 Exam III Spring 2008 7 7 Which of the following integrals correctly represents the surface area of revolution obtained by rotating the graph of y sin x about the x axis over the interval 0 R A 0 sin x 1 cos2 x dx R B 0 sin x 1 cos2 x dx p R C 2 0 cos x 1 sin2 x dx R D 2 0 sin x 1 cos2 x dx R E 2 0 1 cos2 x dx R F 0 1 cos2 x dx R G 0 1 cos2 x dx R H 2 0 cos x 1 cos2 x dx p R I 2 0 sin x 1 sin2 x dx p R J 4 0 cos x 2 sin2 x dx TheR general p integral expression for the area of a surface of revolution is I b 2 a f x 1 f 0 x 2 dx Therefore if f x sin x we have …
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