5 points Z arctan x dx64 MIDTERM SOLUTIONS 111 Integration by parts u arctan x

# 5 points z arctan x dx64 midterm solutions 111

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(5 points) Z arctan x dx
6.4. MIDTERM SOLUTIONS 111 Integration by parts! u = arctan x v = x du = 1 1 + x 2 dx dv = dx Z arctan x dx = x arctan x - Z x 1 + x 2 dx u = 1 + x 2 du = 2 xdx Z arctan x dx = x arctan x - 1 2 Z 1 u du = x arctan x - 1 2 ln(1 + x 2 ) + C (5 points) Z 1 x + x 2 dx Partial Fractions: Z 1 x (1 + x ) dx = Z 1 x - 1 x + 1 dx = ln | x | - ln | x + 1 | + C. 5. (10 points) Does the following integral converge or diverge? If it con- verges, compute its value. Z 0 e - x sin x dx. Converge! If you can’t find the antiderivative, then you can use the bound | sin x | ≤ 1 to see that Z 0 e - x sin x dx Z 0 e - x dx = lim b →∞ ( - e - b + 1) = 1 .
112 CHAPTER 6. QUIZZES AND MIDTERMS However, we can find this using integration by parts. Let I = R e - x sin x dx . u = e - x v = - cos x du = - e - x dv = sin xdx I = - e - x cos x - Z e - x cos x dx u = e - x v = sin x du = - e - x dv = cos xdx I = - e - x cos x - ( e - x sin x - Z - e - x sin x dx = - e - x (cos x + sin x ) - I 2 I = - e - x (cos x + sin x ) I = - e - x 2 (cos x + sin x ) Now that we have the antiderivative, we can compute the value of our improper integral. Z 0 e - x sin x dx = lim b →∞ ( - e - x 2 (cos x + sin x ) | b 0 = lim b →∞ ( - e - b 2 (cos b + sin b ) - ( - e 0 2 (cos 0 + sin 0) = 1 2 . 6. (Extra Credit, +5 points) Show that k =1 1 k 2 is less than 2 by com- paring it to a series or integral that you know how to compute exactly. Fun fact, it’s exactly π 2 / 6. X k =1 1 k 2 < 1 + X k =2 1 k ( k - 1) = 1 + X k =2 1 k - 1 - 1 k = 1 + 1 = 2 . Or, we can notice that making a Riemann sum for 1 x 2 with right end- points and Δ x k = 1 gives us X k =1 1 k 2 = 1 + X k =2 1 k 2 < 1 + lim N →∞ Z N 1 1 x 2 dx = 1 + 1 = 2 .
6.4. MIDTERM SOLUTIONS 113 6.4.3 Practice Midterm 2 1. Find the second-degree Taylor approximation to the function f ( x ) = sec x centered at x = π/ 4, a.k.a. with a = π/ 4. The second-degree Taylor approximation of f ( x ) at a is f ( a ) + f 0 ( a )( x - a ) + f 00 ( a ) 2 ( x - a ) 2 . We compute that f 0 ( x ) = sec x tan x and f 00 ( x ) = sec x tan 2 x + sec 3 x , so f 0 ( π/ 4) = 2 and f 00 ( π/ 4) = 3 2. Thus, we our answer is sec x 2 + 2( x - π/ 4) + 3 2 2 ( x - π/ 4) 2 . 2. (a) Estimate the integral Z 2 0 sin( x ) dx using the Trapezoid Method with n = 5. One way to write the Trapezoid method is Z b a f ( x ) dx b - a 2 n ( f ( x 0 ) + 2 f ( x 1 ) + · · · + 2 f ( x n - 1 ) + f ( x n )) where the x i are evenly spaced in the interval with x 0 = a and x n = b . For this problem, we have b - a = 2 and n = 5, so we get Z 2 0 sin( x ) dx 1 5 (sin(0)+2 sin(2 / 5)+2 sin(4 / 5)+2 sin(6 / 5)+2 sin(8 / 5)+sin(2)) . (b) Give an upper bound on the error of this approximation. For the Trapezoid method, our error estimate is Z b a f ( x ) dx - T n K ( b - a ) 3 12 n 2 where K is the maximum value of the magnitude of the second deriva- tive of the function. In our case, K = max 0 x 2 | sin 00 ( x ) | = max 0 x 2 sin( x ) = 1 .
114 CHAPTER 6. QUIZZES AND MIDTERMS So, we have Z 2 0 sin( x ) dx - T 5 8 12(25) = 2 75 = . 02 ¯ 6 . (c) Compute the integral exactly. What does your estimate say about the value of cos 2? This is a simple integral. R 2 0 sin xdx = - cos x | 2 0 = 1 - cos 2. So, we know that cos 2 = 1 - Z 2 0 sin x dx 1 - 1 5 (sin(0) + 2 sin(2 / 5) + 2 sin(4 / 5) + 2 sin(6 / 5) + 2 sin(8 / 5) + sin(2)) where the approximation is accurate within 2 75 . Not terribly useful in this case.
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