fall06 final solutions

# Thomas' Calculus: Early Transcendentals

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Mathematics 104 Fall Term 2006 Solutions to Final Exam 1. Evaluate sin(ln t ) dt. Solution. We first make the substitution t = e x , for which dt = e x dx . This gives sin(ln t ) dt = 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 e x sin( x ) = - e x cos( x ) + e x cos( x ) dx. We now use u = e x , dv = cos( x ) dx du = e x dx v = sin( x ) . This gives e x sin( x ) dx = - e x cos( x ) + e x sin( x ) - e x sin( x ) dx. Moving the integral on the right to the other side of the equation and dividing by 2 gives 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 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: sin(ln t ) dt = t sin(ln t ) - t cos(ln t ) 2 + C.

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2. Evaluate dx 9 x 2 + 16 . Solution. We first make the substitution x = 4 3 tan( θ ). Note that we then have dx = 4 3 sec 2 ( θ ) and 9 x 2 + 16 = 16 sec 2 ( θ ) . We therefore have dx 9 x 2 + 16 = 4 3 sec 2 ( θ ) 16 sec 2 ( θ ) = 1 3 sec( θ ) = 1 3 ln(sec θ + tan θ ) + C. We must now rewrite the answer in terms of x . We have tan( θ ) = 3 4 x and sec( θ ) = 1 + tan 2 ( θ ) = 1 + 9 16 x 2 . Therefore we have the final answer dx 9 x 2 + 16 = 1 3 ln 3 4 x + 1 + 9 16 x 2 + C. Note that this can be rewritten as dx 9 x 2 + 16 = 1 3 ln 3 x + 16 + 9 x 2 + C.
3a. Does n =1 ln n + sin n n 3 / 2 converge or diverge? Give your reasons. Solution. We show that both A = n =1 sin n n 3 / 2 and B = 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 p -series for p = 3 / 2 that the series 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). We now show that the series B converges. Recall that for any a > 0 we have ln n < n a once n becomes sufficiently large. Taking a = 1 / 4 shows that we have the inequality ln n n 3 / 2 1 n 5 / 4 .

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