Math 132 Exam 2 Fall 2016 15 multiple choice questions worth 5 points each 2 hand graded questions worth 12 and 13 points each Exam covers sections 6 3 6 5 7 1 7 5 7 8 8 1 8 2 No calculators For the multiple choice questions mark your answer on the answer card Show all your work for the written problems Your ability to make your solution clear will be part of the grade Useful Formulas Pn i 1 Pn i 1 i3 Pn n n 1 2 i n n 1 2 i 1 2 i2 n n 1 2n 1 6 sin2 cos2 1 1 tan2 sec2 1 cot2 csc2 sin A B sin A cos B sin B cos A cos A B cos A cos B sin A sin B tan A B cos A cos B R 1 2 tan A tan B 1 tan A tan B cos A B cos A B sin A sin B 1 2 cos A B cos A B sin A cos B 1 2 sin A B cos A B sin2 x 12 1 cos 2x cos2 x 21 1 cos 2x sin 2 2 sin cos cos 2 cos2 sin2 csc x dx ln csc x cot x C R sec x dx ln sec x tan x C Math 132 Exam 2 Z Page 2 of 21 3 f x dx 12 and the average value of f x on 3 5 is 6 1 Suppose you know that Z 5 f x dx Find 0 0 A 0 B 2 C 3 D 6 E 8 F 9 G 12 Solution Since the average value on 3 5 is 6 that means Z 5 1 f x dx 6 5 3 3 R5 which gives us that 3 f x dx 12 Then using properties of integrals Z 5 Z Z 0 5 f x dx 12 12 0 f x dx f x dx 0 3 3 2 Find the average value of f x sin x e1 cos x on the interval 0 A e2 e 1 e e 1 B 2 C e2 2 D e 1 Math 132 Exam 2 Page 3 of 21 e2 1 2 2 F e 1 e 1 G E Solution The average value of f x sin x e1 cos x on the interval 0 is Z 1 sin x e1 cos x dx 0 To compute this do the substitution u 1 cos x so du sin x dx and the bounds become u 0 and u 2 Then Z Z 1 2 u 1 1 cos x sin x e dx e du 0 0 1 2 eu 0 e2 1 Math 132 Exam 2 Z 12 3 Evaluate the definite integral sin2 x cos2 x dx 0 A 48 B E 48 3 32 1 64 96 643 96 F 24 C D 96 G 1 64 3 64 1 12 3 64 24 H Solution Z 12 2 2 Z 12 1 1 1 cos 2x 1 cos 2x dx 2 2 0 Z 12 1 1 cos2 2x dx 4 0 Z 1 12 1 1 1 cos 4x dx 4 0 2 Z 12 1 1 1 cos 4x dx 4 0 2 12 1 1 x sin 4x 8 4 0 1 1 sin 3 8 12 4 3 96 64 sin x cos x dx 0 This can also be done using the trig identities sin2 x cos2 x 14 2 sin x cos x 2 14 sin2 2x 81 1 cos 4x Page 4 of 21 Math 132 Exam 2 Page 5 of 21 4 If you are going evaluate the integral Z x3 dx x2 6x 13 using trig substitution which substitution should you use A x 3 sin 2 B x 23 sec C x 3 tan 6 D x 2 tan 3 E x 9 sec 3 F x 2 sin 3 G x 23 tan H x 3 sec 6 Solution To do this you need to complete the square x2 6x 13 x2 6x 9 9 13 x 3 2 4 Thus Z x3 dx x2 6x 13 This integral is of the form x3 Z p x 3 2 4 dx u2 a2 You could do this in two substitutions starting with u x 3 and then followed by u 2 tan Putting these together gives x 2 tan 3 Math 132 Exam 2 Page 6 of 21 5 Identify the form of the partial fraction decomposition of the rational function 3x2 10 x2 x4 25 A x A B x A C x A D x A E Bx C x4 25 Bx C Dx E 2 2 x 5 x 5 B C Dx E F x 2 2 2 x x 5 x 5 B D E Fx C 2 2 x x 5 x 5 x 5 A B C D E Fx 2 2 x x x 5 x 5 x 5 Cx D Ex A B 2 2 2 x x x 5 x 5 A B C D E 2 G 2 x x x 5 x 5 x 5 F Solution First note that x2 5 is not irreducible and it factors x Facting the denominator gives 5 x x2 x4 25 x2 x2 5 x2 5 x2 x 5 x 5 x2 5 This factorization gives the correct form for the partial fraction decomposition 6 Use polynomial long division to write 2x4 x3 3x2 4x 9 p x Q x 2 2 x x x x where Q x and p x are polynomials What is p 2 A 0 B 1 2 C 1 D 2 5 Math 132 Exam 2 E 3 F 4 G 6 Solution 2x2 x 2 x2 x 2x4 x3 3x2 4x 9 2x4 2x3 x3 3x2 x3 x2 2x2 4x 2x2 2x And thus 2x4 x3 3x2 4x 9 6x 9 2x2 x 2 2 2 x x x x so p x 6x 9 and p 2 3 Page 7 of 21 Math 132 Exam 2 Page 8 of 21 Z 1 7 Evaluate the integral dx 2 x 16 x 4 ln x x2 16 C A arcsin 4 x B ln 4 x2 16 C x C ln 4 D arcsec x 4 x2 16 C 4 4 ln x x2 16 C 1 E ln x C x2 16 x2 16 x C F 4 4 3 2 2 G x 16 2 C 3 Solution Use the trig substitution x 4 sec so dx 4 sec tan d and 2 x 16 4 tan Later we will use the fact that we can rewrite these two equations 2 as sec x4 and tan x 4 16 Applying our substitution the integral becomes Z Z 4 sec tan d 1 dx 4 tan x2 16 Z sec d ln sec tan C Now we use our earlier formulas for tan and sec in terms of x to obtain Z 1 x x2 16 dx ln C 4 …
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