I don't know anything about that. You posed the problem, and you have gotten a couple of strategies for solving it. If you don't know how to do integration by parts, say so, and we'll help you out.kingwinner said:This integral actually occurs in the middle of a statistics problem. If I know the expectation of a gamma distribution, can I possibly avoid integrating by parts in the above integral? If so, how?
The general formula for integrating x^5 exp(x^2) is ∫x^n exp(ax)dx = (x^(n+1)/a)*exp(ax) - n/(a^2)*∫x^(n-1)exp(ax)dx, where n is a positive integer and a is a constant.
The technique for solving integrals of the form x^5 exp(x^2) is to use integration by parts, where we choose u = x^5 and dv = exp(x^2)dx. This will allow us to reduce the power of x in u and simplify the integral.
Yes, we can also use substitution to solve this integral. We can let u = x^2, du = 2x dx and rewrite the integral as ∫x^3 exp(u)du. This can then be solved using the power rule for integration.
The final answer after integrating x^5 exp(x^2) is (x^6/2)*exp(x^2) - 15/4*∫x^3 exp(x^2)dx + C, where C is the constant of integration.
Yes, we can check our answer by taking the derivative of our solution. If the derivative matches the original integrand, then our solution is correct. In this case, the derivative of (x^6/2)*exp(x^2) - 15/4*∫x^3 exp(x^2)dx + C is indeed x^5 exp(x^2), so our solution is verified.