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### Integration by Parts

```
Date: 15 Feb 1995 17:08:26 -0500
From: Courtois
Subject: Help!!!

Since most people did calculus in high school, I assume this question you
folks will be able to answer.  I've been working all day on it, and no luck.

I need the integral from 0 to 1 of X^4 * exp(-x/2).  Can you help?

Stuck in Seattle
```

```
Date: 15 Feb 1995 22:10:07 -0500
From: Dr. Ken
Subject: Re: Help!!!

Hello there!

This is a classic example of a problem that can be solved by using the
technique of Integration by Parts.

But first, let's make the problem a little bit nicer by getting rid of the
-x/2.  Let's let y = -x/2 (i.e. x = -2y), and then we have dx = -2dy.  So we
have the integral from 0 to -1/2 of 16y^4 * e^y dy.  We can pull the 16
outside of the integral.

So now we've got 16 times the integral from 0 to -1/2 of y^4 * e^y dy.
Using Integration by Parts, let u = y^4 and let dv = e^y dy.  So then du =
4y^3 dy, and v = e^y.

Remember the formula for integration by parts?  It's
Integral(u dv) = uv - Integral(v du).  So we get

16 * Integral(y^4 e^y dy) = 16*y^4 e^y - 16*Integral(4y^3 e^y dy)
= 16/(16Sqrt{e}) - 64*Integral(y^3 e^y dy))
= 1/Sqrt{e} - 64*Integral(y^3 e^y dy).

***note: all integrals are between 0 and -1/2***

Notice that we've knocked the power down on the y^4 so that it became y^3.
You can apply this method again and knock the power down another notch, and
another, until you get the integral of e^y.  Which you can do, because it's
e^y.  I hope this makes sense to you.

-Ken "Dr." Math
```
Associated Topics:
High School Calculus

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