tapia jat4858 HW02 clark 52990 This print out should have 24 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 Consequently f x 10 0 points f x 1 7 7 2 4x 2x 4 x 2 002 Rewrite the expression 4x 1 x2 x 2 f x 3 f x 4 f x 7 1 2 x x 2 1 7 7 2 x 2x 4 x 2 1 7 7 2 2x 4x 4 x 2 5 f x 1 7 7 2 correct 4x 2x 4 x 2 Explanation We have to find A B and C so that 4x 1 x2 x 2 1 1 x 2 2 3 correct x 2 3 3 x 2 4 1 x 2 5 Ax x 2 B x 2 Cx2 x2 x 2 x 4x 1 Ax x 2 B x 2 Cx B 1 2 x 2 C 7 4 while But then on comparing coefficients of x2 we see that 7 A 4 1 x2 x 2 x3 0x2 x3 x2 x2 x2 2 x 0 2 x 2 Explanation As f x P x Q x with degP degQ we begin with long division Thus Now 2 x 2 6 A B C 2 x x x 2 A C 0 x3 2x 3 x2 x 2 find the term having denominator x 2 1 7 x2 x 2 2 f x 10 0 points In the partial fractions decomposition of the expression using partial fractions 1 f x 1 2x 3 2x 4x 3 x 2 5x 1 Thus f x x 1 5x 1 x 2 x2 On the other hand x2 x 2 x 1 x 2 tapia jat4858 HW02 clark 52990 so we look for B and C satisfying 5x 1 B C x 2 x 1 x 2 x2 Multiplying through by x 1 x 2 gives 5x 1 B x 2 C x 1 so after substituting x 1 and x 2 we see that B 2 Setting x 4 gives 12 6B i e B 2 while setting x 2 gives 6 6A i e A 1 Thus Z 1 2 I dx x 2 x 4 ln x 2 2 ln x 4 C Consequently C 3 I ln Consequently f x has partial fraction decomposition 3 2 f x x 1 x 1 x 2 keywords partial fractions 003 10 0 points Determine the indefinite integral Z x 8 I dx x 2 x 4 x 4 2 C correct 1 I ln x 2 x 4 C 2 I ln x 2 x 4 2 3 I ln C x 2 x 2 C 4 I ln x 4 2 x 2 5 I ln C x 4 2 Explanation First we have to determine the partial fraction decomposition A B x 8 x 2 x 4 x 2 x 4 Multiply through by x 2 x 4 Then x 8 A x 4 B x 2 2 x 4 2 x 2 C with C an arbitrary constant 004 10 0 points Find the unique function y satisfying the equations dy 3 dx x 1 5 x 1 y 2 y 3 y 4 y 5 y y 2 0 5 x ln 3 3 ln x 1 5 x 3 ln 3 ln 4 x 1 x 1 3 ln ln 3 5 x x 1 3 ln ln 3 correct 4 5 x x 1 1 ln ln 3 4 5 x Explanation We first find A B so that 3 A B x 1 5 x x 1 5 x by bringing the right hand side to a common denominator In this case A 5 x B x 1 3 x 1 5 x x 1 5 x tapia jat4858 HW02 clark 52990 and so To find the values of A B particular choices of 3 x are made When x 1 for instance A 4 3 while when x 5 B Thus 4 3 1 1 dy dx 4 x 1 5 x Hence after integration 3 ln x 1 ln 5 x C 4 3 x 1 ln C 4 5 x 3 1 C ln 4 3 i e C 3 ln 3 4 Consequently 3 y 4 ln 005 x 1 5 x Explanation By partial fractions A Bx C 2x2 2 2 x 1 x 1 x 1 x 1 To determine A B and C multiply through by x 1 x2 1 for then 2x2 A x2 1 x 1 Bx C A B x2 B C x A C which after comparing coefficients gives B C with C an arbitrary constant But y 2 0 1 ln 2 2 2 6 I ln 2 2 5 I A 5 x B x 1 3 y 3 ln 3 10 0 points Evaluate the integral A C B 1 Thus 1 1 x 1 I dx x 1 x2 1 0 Z 1 Z 1 1 x dx dx 2 0 x 1 0 x 1 Z 1 1 dx 2 0 x 1 Z and so h 1 I ln x 1 ln x2 1 2 i1 tan 1 x 0 I Z 0 1 I 2 I 3 I 4 I 1 2x2 x 1 x2 1 ln 8 2 1 ln 8 correct 2 2 ln 2 2 1 ln 2 2 2 dx i1 1h 2 2 1 ln x 1 x 1 2 tan x 2 0 Consequently 1 ln 8 I 2 2 006 10 0 points Determine the integral Z x2 3x 5 I dx x2 x 2 tapia jat4858 HW02 clark 52990 1 I x ln x 1 3 C correct x 2 2 I x ln x 1 3 x 2 C x 1 3 C 3 I x ln x 2 4 I x ln x 1 3 C x 2 5 I x ln x 1 3 x 2 C 6 I x ln x 1 3 C x 2 Explanation By division 4 keywords division partial 007 10 0 points At a branch of UT there are approximately 37000 students A flu epidemic is spreading among these students at a rate proportional to the product of the number of infected students and the number of students who have not been infected Initially 1000 students were infected and 10 days later 6000 were infected What percentage of the student body was infected after 20 days Hint to simplify the algebra let N t be the number of infected students in multiples of 1000 1 3 61 x2 3x 5 x2 x 2 x2 x 2 2x 7 x2 x 2 2x 7 1 2 x x 2 2 3 88 3 57 42 correct 4 55 86 But by partial fractions 2x 7 3 1 x 2 x 1 x 2 5 78 07 x2 Thus I Z 1 3 1 x 1 x 2 dx Now Z 3 dx 3 ln x 1 C1 x 1 while Z 1 dx ln x 2 C2 …
View Full Document
Unlocking...