20b_04f_1s

# 20b_04f_1s - Math 20B 1(a d dx 2x First Midterm Solutions...

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Unformatted text preview: Math 20B 1. (a) d dx 2x First Midterm Solutions u3 + 1 du = −x October 18, 2004 d dx −x d dx 2x 0 u3 + 1 du − 1 − x3 . u3 + 1 du 0 = 2 8x3 + 1 + (b) Integrate by parts with u = sin−1 x and dv = dx and then let 1 − x2 = t: sin−1 x dx = x sin−1 x − x dx 1 √ = x sin−1 x + 2 2 1−x t−1/2 dt = x sin−1 x + t1/2 + C = x sin−1 x + π /2 π /2 1 − x2 + C. sin3 x 3 π /2 (c) 0 cos3 x dx = 0 (cos x − sin2 x cos x) dx = sin x − = 0 2 . 3 (d) Use substitution with t = 3x − 1: x=1 x=0 t=2 (3x − 1)4 dx = t=−1 t4 dt t5 = 3 3×5 25 +1 3×5 . 2 −1 = 11 25 + 1 = . 3×5 5 You can leave your answer as, e.g., 2. The easiest way to prove it is to use the Fundamental Theorem of Calculus: Verify that the derivative of x sin(ln x) + C is the integrand. 3. The curve intersects the x-axis at x = −2, x = 0 and x = 1. The function is positive for −2 < x < 0 and negative for 0 < x < 1. The area is 1 −2 0 1 |x3 + x2 − 2x| dx = = −2 (x3 + x2 − 2x) dx + 3 0 0 4 −(x3 + x2 − 2x) dx 1 0 x x x3 x + − x2 + − x2 − 4 3 4 3 −2 11 16 8 − −4 − + −1 =− 4 3 43 71 37 = − +1 = . 34 12 4 4. The curves intersect at (0, 0) and (1, 1). Thus the integral is 1 π 0 (x1/2 + 2)2 − (x2 + 2)2 dx. ...
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