tapia jat4858 HW01 clark 52990 This print out should have 22 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points Evaluate the integral Z 6 ln x I dx x2 e 1 I 2 I 3 I 4 I 5 I 6 I 6 1 ln 6 1 e 2 6 1 ln 6 1 e 2 2 1 ln 6 1 e 6 2 1 ln 6 1 correct e 6 6 1 ln 6 1 e 2 2 1 ln 6 1 e 6 Explanation After integration by parts Z Z 1 2 1 2 2 x ln x dx x ln x x2 ln x dx 2 x Z 1 x2 ln x 2 x ln x dx 2 Thus Consequently 10 0 points 1 C I 2x2 ln x 2 ln x 2 Determine the integral Z I 4x ln x 2 dx 1 I 4x2 ln x 2 ln x 2 1 1 x2 ln x x2 C 2 4 1 1 1 x2 ln x 2 x2 ln x x2 C 2 2 4 2 1 ln 6 1 I e 6 1 But after integration by parts once again Z Z 1 1 1 2 x2 dx x ln x dx x ln x 2 2 x Z 1 1 x dx x2 ln x 2 2 Thus Z x ln x 2 dx h1 1 i6 I ln x x x e Consequently 1 2 I 2x2 ln x 2 ln x C 2 1 C 3 I 4x2 ln x 2 ln x 2 1 C 4 I 4x2 ln x 2 ln x 2 1 2 2 5 I 2x ln x ln x C 2 correct 1 C 6 I 2x2 ln x 2 ln x 2 Explanation After integration by parts h1 i6 Z 6 1 I ln x dx 2 x e e x 002 1 keywords integration by parts log function C 003 10 0 points tapia jat4858 HW01 clark 52990 Evaluate the definite integral I Z 1 I 1 3e4 x dx 0 1 I 2 I 3 I 4 I 5 I 6 I 3 4 3e 1 2 3 4 3e 1 correct 8 3 4 3e 1 8 3 4 4e 1 2 3 4 4e 1 2 3 4 4e 1 8 Explanation First we use the substitution u x to eliminate the square root in the integrand For then 1 dx du 2 x so dx 2 xdu 2u du Thus I 6 Z 3 I 4 I 3 5 I 3 4 I 3e 1 8 10 0 points Evaluate the integral 4 0 1 3 2 2 Explanation After integration by parts i 4 1h 4 x 5 sin 2x 0 2 Z 1 4 d sin 2x 4 x 5 dx 2 0 dx Z 4 5 1 sin 2x dx 2 2 2 0 I 5 1 4 cos 2x 0 2 2 Consequently u e4u du and so I 7 1 correct 2 2 1 This last integral we evaluate using integration by parts Hence h3 i1 3 Z 1 4u I ue e4u du 2 2 0 0 h3 3 4u i1 4u ue e 0 2 8 Z 1 3 2 2 I 7 I 0 004 2 4 x 5 cos 2x dx 005 7 1 2 2 10 0 points Evaluate the integral Z I e x cos x dx 0 1 I e 1 1 2 I e 1 2 3 I 1 1 e 2 4 I 1 e tapia jat4858 HW01 clark 52990 3 5 I 1 e 1 correct 2 4 I 3 tan x ln sec x x C 6 I 1 e 1 2 1 5 I 3 x tan x ln sec x x2 C 2 1 6 I 4 x tan x ln sec x x C 2 7 I e 1 8 I e 1 Explanation After integration by parts h i Z x I e cos x e x sin x dx 0 e 1 Z 0 e x Explanation As tan2 x sec2 x 1 the integrand can be rewritten as x 2 sec2 x sec2 x 1 x 3 sec2 x 1 Thus sin x dx I 0 To evaluate this last integral we integrate by parts once again For then Z h i e x sin x dx e x sin x Z x 3 sec2 x 1 dx 3 Z 1 x sec2 x dx x2 2 On the other hand 0 0 Z d tan x sec2 x dx e x cos x dx 0 I 0 so integration by parts is suggested for evaluating this last integral since then Z Z 2 3 x sec x dx 3 x tan x tan x dx in which case I e 1 I Consequently But I 006 1 e 1 2 10 0 points Determine the integral Z I x 2 sec2 x tan2 x dx 1 I 4 tan x ln sec x x C 2 I 4 tan x ln sec x x2 C 1 3 I 3 x tan x ln sec x x2 C 2 correct Z tan x dx ln sec x C Consequently 1 I 3 x tan x ln sec x x2 C 2 with C an arbitrary constant keywords integration by parts trig function 007 10 0 points The base of a solid S is the region enclosed by the graphs of p y ln x x e y 0 tapia jat4858 HW01 clark 52990 If the cross sections of S perpendicular to the x axis are squares determine the volume V of S 1 V 2 cu units 3 1 I 3 cos 4 1 cos3 C 3 2 I 3 cos 2 V 1 cu unit correct 3 I 3 cos 3 V 2 e3 1 cu units 4 V 1 cu units 2 5 V 1 3 e 1 cu units 3 1 3 sin C 3 4 I 3 cos 1 cos3 C 3 5 I 3 cos 1 3 sin C 3 6 I 3 cos Explanation The base of the solid S is given by 1 cos3 C correct 3 1 3 sin C 3 Explanation After simplification we see that 2 sin sin3 sin 2 sin2 1 On the other hand sin2 1 cos2 1 e 2 Since the square cross section has side length y its area is given by A y y 2 ln x Thus Z e V A y dy 1 Z 1 e 2 y dx Z e ln x dx 1 After Integration by Parts therefore h ie V x ln x x 1 Consequently Thus the integrand can be rewritten as sin 2 1 cos2 sin 3 cos2 As this is now of the form sin f cos the subsitution x cos is suggested For then dx sin d in which case Z 1 3 2 I 3 x dx 3x x C 3 Consequently volume e e 1 1 cu unit 008 …
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