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# Review Sheet for MATH 116 with Professor Mermin at KU

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This 5 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Kansas taught by a professor in Fall. Since its upload, it has received 25 views.

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Date Created: 02/06/15

Math 116 Je Mermin7s sections Test 1 Review 1 TrueFalse questions There will be ten truefalse questions taken verbatim from the quizzes See the quizzes and solutions 2 Indicate the rst step you would take in evaluating each of the integrals below Or if they re easy enough to require no intermediate steps eval uate themi Otherwise do not actually evaluate the integrals The rst ve are done as examples Several of these have more appropriate answers in addition to those given here a f 2x3dxi This is x4 C 01 Factor out the 2 b f Use algebra to simplify the expression C f xe dx Use integration by parts with u x and dv exdxi d f 295952 420dxi Substitute u x2 4 e 952 e dxi Break it up into a sum of two integrals f 7 95 955195 i 3 a Thzs zs gym 7 952 x10 C g f 11517 2 Reunite the integrand as 2 7 W or 7 7 1 Use integration by parts with u 7 4x and dv 7 males or Substitute u 2x 1 h f aim Substitute u 7x 1 fz2x3 9dx Substitute u x3 9 j f xmdx Substitute u 2x2 1 k ln x2dx Use integration by parts with u In as and dv dag l f xgexdx Use integration by parts with u x3 and dv e dx m f xgexgdx Substitute u x2 11 f 1quotde Substitute u In x 0 f 5195 Use algebra to simplify the expression 3 Evaluate the inde nite integrals a f We have 9 dz x2 71m x37xCi b 6x2e e dx We have 6x2e e dx f6x2e dx fe xclxi To evaluate the second integral set u 795 so clx idu and fe xclx f feudu 7e C 7e Cl To evaluate the rst integral set u 12 s0 clx ght and f6x2e dx f4euclu 4e C 4e Cl Thus we have 6x2e e xdx 6x2e dx e xdx 4e ie xCi C f x In xdx Set u lnx and clv xdx s0 clu idx and v x2 Then we have xlnxdxudv uv7vdu Inacx ziw lt x2gtlt dxgt lenx7 xdac 1 2 1 2 5x lnx71x Ci 4 Express the area bounded by the given curves in terms of a de nite integral or integrals Do not actually compute a numerical answer a b acly952ancly807ac3 The curves y x2 and y 807 x3 meet at x2 807 x3 ie ac 4 Each of these curves meets ac 1 when ac 1 Thus the region is bounded on the left at ac l and on the right by ac 4 For ac between 1 and 4 we take ac 2 as a sample point 22 4 lt 72 80 7 23 so y 80 7 x3 is the top boundary and y x2 is the bottom boundary Thus the area is given by 14 c1 80 7 x3 7 dac i yac473ac2 andy6x2i The curves intersect at x4 7 3x2 6x2 ie at ac 0 and ac i3 Thus there are two regions to consider the rst bounded on the left by ac 73 and on the right by ac 0 and the second bounded on the left by ac 0 and on the right by ac 3 For the rst region we use ac 7l as a sample point and compute 7l4737l2 72 lt 6 67l2 Thus the top boundary is given by y 6x2 and the bottom boundary is given by y x4 7 3952 Thus 10 the area of the rst region is 6952 7 x4 7 3952 dac 173 For the second region we use as 1 as a sample point and compute 14 7 312 72 lt 6 61 Thus the top boundary is given by y 6x2 and the bottom boundary is given by y x4 7 3952 Thus 13 the area of the rst region is 6952 7 x4 7 3952 dost 10 The entire area is the sum of these two areas or 523 1 6952 7 x4 7 3952 doc 6952 7 x4 7 3952 doc l 73 520 13 Or equivalently 6952 7 x4 7 3952 doc 73

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