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Unformatted text preview: Math 1B. Sample Answers to Final Exam 1. (24 points) Find the following integrals: (a). Z / 4 sin 2 x cos 6 x dx Convert into tan x and sec x , and substitute u = tan x , du = sec 2 xdx : Z / 4 sin 2 x cos 6 x dx = Z / 4 tan 2 x sec 4 xdx = Z / 4 tan 2 x (tan 2 x + 1)sec 2 xdx = Z 1 u 2 ( u 2 + 1) du = Z 1 ( u 4 + u 2 ) du = u 5 5 + u 3 3 1 = 1 5 + 1 3 = 8 15 . (b). Z dx x 2 + 2 x + 2 Completing the square gives x 2 +2 x +2 = ( x +1) 2 +1, so we substitute u = x +1 and then u = tan , and use the provided formula for R sec xdx : Z dx x 2 + 2 x + 2 = Z du u 2 + 1 = Z sec 2 d sec = Z sec d = ln  sec + tan  + C = ln  p u 2 + 1 + u  + C = ln  p x 2 + 2 x + 2 + x + 1  + C . (It would also have been good to draw a figure of a right triangle, with the side opposite labeled u , the adjacent side labeled 1, and the hypotenuse labeled u 2 + 1. Its not so easy on a computer, though.) 1 2 2. (16 points) Find the integral Z dx x 3 + x 2 . Use partial fractions. The denominator factors as x 2 ( x + 1), so we look for A , B , and C that satisfy 1 x 2 ( x + 1) = A x + B x 2 + C x + 1 . Multiplying this equation by x 2 ( x + 1) gives 1 = Ax ( x + 1) + B ( x + 1) + Cx 2 . Setting x = 0, x = 1, and x = 1, give the equations B = 1 C = 1 2 A + 2 B + C = 1 . These have solutions B = C = 1, A = 1, so we have Z dx x 3 + x 2 = Z 1 x + 1 x 2 + 1 x + 1 dx = ln  x   1 x + ln  1 + x  + C . 3. (16 points) The curve y = x 2 4 ln x 2 , 1 x 2 is rotated about the yaxis. Find the area of the resulting surface. 1 2 We have y = x/ 2 1 / 2 x , so ds = p ( y ) 2 + 1 dx = r x 2 4 1 2 + 1 4 x 2 + 1 dx = r x 2 4 + 1 2 + 1 4 x 2 dx = x 2 + 1 2 x dx . Therefore the area is 2 Z 2 1 xds = 2 Z 2 1 x x 2 + 1 2 x dx = Z 2 1 ( x 2 + 1) dx = x 3 3 + x 2 1 = 8 3 + 2 1 3 1 = 10 3 . 3 4. (24 points) Determine whether the series is absolutely convergent, conditionally con vergent, or divergent. (a). X n =1 2 n n !...
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 Fall '08
 Reshetiken
 Integrals

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