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# p1sol - Problem 1a Find the derivative of G(x if G(x = d dx...

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Problem 1a: Find the derivative of G ( x ) if G ( x ) = R 1 x e t 2 dt . d dx R 1 x e t 2 dt = - d dx R x 1 e t 2 dt = - e x 2 where the second equality is by the fundamental theorem of calculus. Problem 1b: Suppose R 2 0 f ( t ) dt = 3. Calculate R 1 0 f (2 t ) dt . By substitution, u = 2 t , du 2 = dt , R 1 0 f (2 t ) dt = 1 2 R 2 0 f ( u ) du = 3 2 . Problem2a: Integration by parts, with U = log x , V = x 2 2 . Answer: log x x 2 2 - x 2 4 + C . Problem 2b: R e x dx Let w = x , then dw = 1 2 x dx = 1 2 w dx , so that 2 w dw = dx . So Z e x dx = 2 Z we w dw Now we do integration by parts: Let u = w and dv = e w dw . So du = dw and v = e w . Now we have: 2 ± we w - Z e w dw ² = 2 we w - 2 e w + C = 2 xe x - 2 e x + C Problem 2c: R π 2 0 sin 2 xdx We use a trig identity to get the following: Z π 2 0 sin 2 xdx = Z π 2 0 ± 1 2 - cos 2 x 2 ² dx = Z π 2 0 1 2 dx - 1 2 Z π 2 0 cos 2 xdx We use direct subsitution for the second integral in the above sum: Let u = 2 x , so du = 2 dx , or 1 2 du = dx . To change our bounds, notice that when x = 0, then u = 0, and when x = π 2 , then u = π . 1

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2 So Z π 2 0 sin 2 xdx = 1 2 x | π 2 0 - 1 4 Z π 0 cos udu ; = π 4 - 1 4 sin u | π 0 = π 4 - 1 4 (sin π - sin 0) = π 4 - 1 4 (0 - 0) = π 4 Problem 2d: Evaluate R dx x 3 - x . 1 x 3 - x = A x + B x - 1 + C x +1 = A ( x 2 - 1)+ B ( x 2 + x )+ C ( x 2 - x ) x 3 - x = ( A + B + C ) x 2 +( B - C ) x - A x 3 - x This yields the system of linear equations: A + B + C = 0 B - C = 0 - A = 1 which results in the solution
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p1sol - Problem 1a Find the derivative of G(x if G(x = d dx...

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