Is there an easier way to solve this integral?

**Ares_D1** · August 22nd 2009, 09:07 AM

Sorry, I'm terrible at typing this out using the math code so I'll paste an image:

I've been trying to solve this using parts, but I'm getting nowhere fast. Should I throw some substitution in there also? If someone could show me how to do this one it would be greatly appreciated.

**Rapha** · August 22nd 2009, 09:10 AM

Hello Ares_D1!

Originally Posted by Ares_D1

Sorry, I'm terrible at typing this out using the math code so I'll paste an image:

I've been trying to solve this using parts, but I'm getting nowhere fast. If someone could show me how to do this one it would be greatly appreciated.

Here is my suggestion:

substitute z:=x^2

=> z' := 2x and dx = dz/z'

Therefore

$\int -2x^3*e^{-x^2}\pi dx = \int -2x^3*e^{-z}\pi \frac{dz}{2x}$

$=\int -x^2*e^{-z}\pi dz = \int -z*e^{-z}\pi dz$

Yours
Rapha

**o_O** · August 22nd 2009, 09:12 AM

Imagine it like this: $\pi\int {\color{blue}-2x} \cdot x^2 \cdot e^{{\color{red}-x^2}} \ {\color{blue}dx}$

So now: ${\color{red}u = -x^2} \ \Rightarrow \ {\color{blue}du = -2x \ dx}$

So your integral becomes: $-\pi\int u e^u du$

which is an easy one by parts.

**Ares_D1** · August 22nd 2009, 09:29 AM

Thanks guys, very helpful!

Math Help - Is there an easier way to solve this integral?

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