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how do you evaluate the integral of xe^(-x^2)dx on the integral (0,infinity)?
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sub u=-x^2 so du=-2xdx
1/2\[Limit as R->\infty of 1/2 \int\limits_{0}^{R} \]
ah.. lol
I would do sub: u=x^2/2, so du=u and integral ( e^u du)... should be easy from this point
are you good with the improper integral part of it?
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Let me know what your answer is
give me a min
Okay. Medal plz
Im getting: integral (e^(-u)du)=-e(-u) now: e^0=1 lim e(-u) when u is going to infinity = lim (1/ e^u) =0 my answer would be 1
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