x sin(x2)dx x2 e-x dx Solution d) Let u = x^2. Then du = 2x dx => du/2 = x dx We

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d) Let u = x^2. Then du = 2x dx => du/2 = x dx We change the bounds in terms of u, so the integral is from 0 to pi^2. Thus we have the integral from 0 to pi^2 of (1/2)sin(u). The integral of that is -(1/2)cos(u). We now evaluate from 0 to pi^2. The answer is -(1/2)(cos(pi^2) - 1) e) You must use integration by parts. Let u = x^2 and dv = e^-x dx. Then du = 2x dx and v = -e^-x So it's the integral of -(x^2)e^-x - integral of (2x - e^-x) dx You must use integration by parts once more. Let u = 2x and dv = - e^-x. Then du = 2dx and v = e^-x So we have -(x^2)e^-x - (2xe^-x - integral 2e^-x) Thus the answer is -(x^2)e^-x - 2xe^-x - 2e^-x

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