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July 28, 2014

July 28, 2014

Posted by **walex** on Tuesday, May 15, 2007 at 7:55pm.

integral of xe^(-2x)

There are two ways:

1) Integration by parts.

2) Differentiation w.r.t. a suitably chosen parameter.

Lets do 1) first. This is the "standard method", but it is often more tedious than 2)

You first write the integral as:

Inegral xe^(-2x) dx =

Integral -1/2 x d(e^(-2x))

Here we have used that:

d(e^(-2x)) = -2 e^(-2x)

The next is is to make use of the fact that:

d(f g) = f dg + g df --->

f dg = d(fg) - g df

This yields:

Integral -1/2 x d(e^(-2x)) =

Integral d[-1/2 x e^(-2x)] -

Integral -1/2 e^(-2x) dx =

-1/2 x e^(-2x) - 1/4 e^(-2x) + C

Method 2) is much simpler. Consider the function:

e^(ax)

It's integral is:

Integral e^(ax)dx = 1/a e^(ax)

Le's differentiate both sides w.r.t. a:

Integral x e^(ax)dx =

[ -1/a^2 + x/a] e^(ax)

And insert a = -2 to obtain the answer.

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