How do you take take this integral?

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how do you take take this integral?

\int_{0}^{r} x^2 e^{-2x} dx
 
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Use integration by parts (twice)...the derivatives of x^2 are easy to find, and likewise for the antiderivative of e^{-2x}dx
 


whats the upper limit? if it's infinity answer is 1/8
 


gabbagabbahey said:
Use integration by parts (twice)...the derivatives of x^2 are easy to find, and likewise for the antiderivative of e^{-2x}dx

do you know how this integral turns into

\frac {N!}{a^{N + 1}} if I take the integral from 0 to infinity? N = 2 and a = 2
 


ice109 said:
whats the upper limit? if it's infinity answer is 1/8

answer is 1/4 if infinity
 


orthovector said:
do you know how this integral turns into

\frac {N!}{a^{N + 1}} if I take the integral from 0 to infinity? N = 2 and a = 2

If N=2 and a=2, then \frac {N!}{a^{N + 1}}=\frac {2!}{2^{2 + 1}}=\frac{1}{4} which is what you should be getting using by parts.

Are you getting something different?
 


gabbagabbahey said:
If N=2 and a=2, then \frac {N!}{a^{N + 1}}=\frac {2!}{2^{2 + 1}}=\frac{1}{4} which is what you should be getting using by parts.

Are you getting something different?

I was trying to derive the general expression

\int_{0}^{\infty} x^n e^{-ax} dx = \frac{n!}{a^{n+1}}

how is this so?
 


orthovector said:
I was trying to derive the general expression

\int_{0}^{\infty} x^n e^{-ax} dx = \frac{n!}{a^{n+1}}

how is this so?

Use integration by parts n times and remember the definition of factorial; n!=n(n-1)(n-2)\ldots (2)(1)
 
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