# 18117 - 7.10, #12. d dx 1 x ln t dt = - d dx x 1 ln t dt =...

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Unformatted text preview: 7.10, #12. d dx 1 x ln t dt = - d dx x 1 ln t dt = - ln x. 2 7.10, #19. The definition of erf(x) implies erf (x) = (2/ )e-x . Thus 2 (x erf(x)) = erf(x) + (2/ )xe-x . 7.10, #21. Substitute t = F (x) = 2u, dt = 2du in the integral defining F (x). Then x 0 2 e-t 2 /2 2 dt = x/ 2 0 2 e-u du = erf(x/ 2). 7.Rev, #2. The area of the triangle is b 0 hx h dx = x2 b 2b b = 0 hb . 2 7.Rev, #3. The area of the circle is 4 times the area of the sector in the first quadrant. Thus the area is r 4 0 r2 - x2 dx = 2x r2 - x2 + 2r2 arcsin x r r 0 = 2r2 = r2 . 2 The antiderivative is given by formulas 30 and 28 on page 367. 7.Rev, #4. Use the substitution w = 2x, dw = 2 dx. Then dx 1 = 2 1 - 4x2 In particular, /8 (1/ 1 0 dw 1 1 = arcsin w + C = arcsin 2x + C. 2 2 1 - w2 - 4x2 dx = (0.5) arcsin(/4) = 0.4517. 7.Rev, #6. 2 1 2 f (t) dt = 0 0 t dt + 1 2 (2 - t) dt = 1 3 t 3 1 0 + (2t - (t2 /2)) 2 1 1 5 = + (2 - (3/2)) = . 3 6 7.Rev, #11. Use the substitution w = 1 - x, dw = -dx. Then x 1 - x dx = = = -(1 - w) w dw (w3/2 - w1/2 ) dw = 2 5/2 2 3/2 w - w +C 5 3 2 2 (1 - x)5/2 - (1 - x)3/2 + C. 5 3 The integral can also be done by parts with u = x v = 1 - x. 7.Rev, #16. Use integration by parts with u = (ln x)2 and v = 1. Then (ln x)2 dx = x(ln x)2 - x(2 ln x)(1/x) dx = x(ln x)2 - 2 ln x dx. The last integral can also be done by parts or by the tables: (ln x)2 dx = x(ln x)2 - 2x ln x + 2x + C. 7.Rev, #17. ln(x2 ) dx = 2 ln x dx = 2x ln x - 2x + C. ln x dx = x ln x - x. Thus ...
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## This note was uploaded on 07/06/2011 for the course MATH 180 taught by Professor Tan during the Fall '08 term at Ill. Chicago.

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18117 - 7.10, #12. d dx 1 x ln t dt = - d dx x 1 ln t dt =...

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