This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Mathematics 104 Fall Term 2006 Solutions to Final Exam 1. Evaluate Z sin(ln t ) dt. Solution. We first make the substitution t = e x , for which dt = e x dx . This gives Z sin(ln t ) dt = Z e x sin( x ) dx. To evaluate the integral on the right we integrate by parts twice. First we use u = e x , dv = sin( x ) dx du = e x dx v = cos( x ) . This gives Z e x sin( x ) = e x cos( x ) + Z e x cos( x ) dx. We now use u = e x , dv = cos( x ) dx du = e x dx v = sin( x ) . This gives Z e x sin( x ) dx = e x cos( x ) + e x sin( x ) Z e x sin( x ) dx. Moving the integral on the right to the other side of the equation and dividing by 2 gives Z e x sin( x ) dx = e x sin( x ) e x cos( x ) 2 . Remember, we now have to add on a + C since this is an indefinite integral. Going back to the original integral, we now have Z sin(ln t ) dt = e x sin( x ) e x cos( x ) 2 + C where t = e x . Rewriting in terms of t gives the final answer: Z sin(ln t ) dt = t sin(ln t ) t cos(ln t ) 2 + C. 2. Evaluate Z dx 9 x 2 + 16 . Solution. We first make the substitution x = 4 3 tan( ). Note that we then have dx = 4 3 sec 2 ( ) d and 9 x 2 + 16 = 16 sec 2 ( ) . We therefore have Z dx 9 x 2 + 16 = Z 4 3 sec 2 ( ) p 16 sec 2 ( ) d = 1 3 Z sec( ) d = 1 3 ln(sec + tan ) + C. We must now rewrite the answer in terms of x . We have tan( ) = 3 4 x and sec( ) = q 1 + tan 2 ( ) = q 1 + 9 16 x 2 . Therefore we have the final answer Z dx 9 x 2 + 16 = 1 3 ln 3 4 x + q 1 + 9 16 x 2 + C. Note that this can be rewritten as Z dx 9 x 2 + 16 = 1 3 ln 3 x + 16 + 9 x 2 + C. 3a. Does X n =1 ln n + sin n n 3 / 2 converge or diverge? Give your reasons. Solution. We show that both A = X n =1 sin n n 3 / 2 and B = X n =1 ln n n 3 / 2 converge. This will imply that the original series converges. First note that sin n n 3 / 2 1 n 3 / 2 . It follows by direct comparison with the pseries for p = 3 / 2 that the series X n =1 sin n n 3 / 2 converges. This shows that the series A converges (since the series obtained by taking the absolute value of each of its terms converges)....
View
Full
Document
This homework help was uploaded on 02/12/2008 for the course MATH 104 taught by Professor Nelson during the Fall '07 term at Princeton.
 Fall '07
 Nelson
 Math

Click to edit the document details