Expert: Paul Klarreich - 9/19/2009

Question
I need help solving these two equations that I'm stuck on. Please try to show all the steps you go through. Thanks.



1) Derive a reduction formula for... the integral of x^n e^-x2 dx.

2) Use this reduction formula to evaluate...
the integral of x^5 e^-x2 dx.


Answer
Questioner: Paul Dunner
Country: United States
Category: Calculus
Private: No
Subject: Reduction Formulas
Question: I need help solving these two equations that I'm stuck on. Please try to show all the steps you go through. Thanks.

1) Derive a reduction formula for... the integral of x^n e^-x2 dx.

2) Use this reduction formula to evaluate...
the integral of x^5 e^-x2 dx.
................................................................
Have you tried integration by parts?  Assuming you mean:

{
| x^n e^-x^2 dx
}


The usual scheme is to make the u = x^n, BECAUSE  diff'ing x^n REDUCES the exponent.

However, there is no way to integrate  e^-x^2 without an 'x' to go with it.
AH! That is the clue: take one of the x's from  x^n -- we have plenty of them.

u =  x^n-1,    du = (n-1) x^n-2  dx

dv = x e^-x^2 dx,    v = e^-x^2/(-2) = -1/2 e^-x^2

Now do the IBP.  The uv part is:

x^n-1 -1/2 e^-x^2 = -1/2 x^n-1 e^-x^2

The - INT(vdu) part, is:


INT  - (n-1) x^n-2  dx [-1/2 e^-x^2] =

(n-1)/2 INT  x^n-2  e^-x^2 dx =

So your whole thing is:

-1/2 x^n-1 e^-x^2 + (n-1)/2 INT  x^n-2  e^-x^2 dx
--------------------> REDUCED INTEGRAL PART

.........................................
To apply it to

INT x^5 e^-x2 dx, take n = 5 and do:

-1/2 x^4 e^-x^2 + (4)/2 INT x^3 e^-x^2 dx =

-1/2 x^4 e^-x^2 + 2 INT x^3 e^-x^2 dx

Then apply it again to the second term, this time with n = 3:
That will reduce the integral part to INT x e^-x^2, which you will just integrate without any need for a reduction formula.

You thought this would be simple?