How do you take take this integral?

[orthovector]

how do you take take this integral?

$$\int_{0}^{r} x^2 e^{-2x} dx$$

 

[gabbagabbahey]

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

 

[ice109]

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

 

[orthovector]

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

 

[orthovector]

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

answer is 1/4 if infinity

 

[gabbagabbahey]

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?

 

[orthovector]

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?

 

[gabbagabbahey]

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; $n!=n(n-1)(n-2)\ldots (2)(1)$