midterm2_16b_solu

midterm2_16b_solu - Math 16B Section 1 Sarason MIDTERM...

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Spring 2006 Sarason MIDTERM EXAMINATION 2 – SOLUTIONS 1. Perform the integrations. (a) I 1 = Z π 0 x sin( x 2 ) dx (b) I 2 = Z π 0 x 2 sin x dx Solution. (a) We make the substitution u = x 2 , du = 2 x dx : I 1 = 1 2 Z π 0 sin u du = - 1 2 cos u ± ± ± ± π 0 = - 1 2 ( - 1) + 1 2 (1) = 1 . (b) We integrate by parts twice: I 2 = - Z π 0 x 2 d (cos x ) = - x 2 cos x ± ± π 0 + Z π 0 cos x d ( x 2 ) = π 2 + 2 Z π 0 x cos x dx = π 2 + 2 Z π 0 x d (sin x ) = π 2 + 2 x sin x ± ± π 0 - 2 Z π 0 sin x dx = π 2 + 0 + 2 cos x ± ± π 0 = π 2 + 2( - 1) - 2(1) = π 2 - 4 2. For the diﬀerential equation y 0 = - ( y + 1) 2 ( t + 1): (a) What are the constant solutions, if any? (b) What is the general solution? (c) What is the solution satisfying the initial condition y (0) = 0? Solution. (a) y = - 1 (b) The equation is separable. We can rewrite it as - y 0 ( y + 1) 2 = t + 1 , or, in diﬀerential notation, as - 1 ( y + 1) 2 dy = ( t + 1) dt. Integration gives

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This note was uploaded on 09/06/2010 for the course CHEM 10894 taught by Professor Pederson during the Fall '10 term at Berkeley.

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midterm2_16b_solu - Math 16B Section 1 Sarason MIDTERM...

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