# help evaluating a double integral

• December 4th 2009, 02:13 PM
ryu991
help evaluating a double integral
heres the integral

x is between 0 and 2
y is between 0 and 5x

I need to find the exact value of the double integral 5x^2e^(xy)

I'm not really sure how to do this so any help would be greatly appreciated. Thanks in advance.
• December 4th 2009, 02:16 PM
Chris L T521
Quote:

Originally Posted by ryu991
heres the integral

x is between 0 and 2
y is between 0 and 5x

I need to find the exact value of the double integral 5x^2e^(xy)

I'm not really sure how to do this so any help would be greatly appreciated. Thanks in advance.

$\int_0^2\int_0^{5x}5x^2e^{xy}\,dy\,dx$.

Where are you stuck in the calculation? If you showed us any attempt you made at the problem, it would give us a better idea of how to best help you. :)
• December 4th 2009, 02:34 PM
qmech
$
\int_0^2\int_0^{5x}5x^2e^{xy}\,dy\,dx
$

$
=\int_0^2 5x^2 [\int_0^{5x}e^{xy}\,dy]\,dx
$

$
=\int_0^2 5x^2 [\frac{e^{xy}}{x}]|_0^{5x}\,dx
$

$

=\int_0^2 5x [e^{xy}]|_0^{5x}\,dx
$

$
=\int_0^2 5x (e^{5x^2}-1)\,dx
$

$
=\int_0^2 5x e^{5x^2}\,dx - \int_0^2 5x \,dx
$

$
=\frac{1}{2}\int_0^2 e^{5x^2}\,d{(5x^2)} - \int_0^2 5x \,dx
$

Get the idea?
• December 4th 2009, 02:59 PM
ryu991
Thank you, what I didn't realize was to factor out the 5x^2, I was staring at the whole integral and I had no idea how to approach it I was thinming integration by parts or something but it just did not seem right.

Quote:

Originally Posted by qmech
[tex]

$
=\frac{1}{2}\int_0^2 e^{5x^2}\,d{(5x^2)} - \int_0^2 5x \,dx
$

Get the idea?

just a few questions about the end?

first where does the 1/2 come from?

second how would I integrate that expression e^(5x^2)d(5x^2)? The other half of it I know, i just have a lot of trouble integrating expressions with e.

thanks again for helping me with this.
• December 4th 2009, 03:16 PM
lvleph
$1/2 \int_0^2 \! e^{5x^2}\, d(5x^2)$ is another way to write u-substitution where $u = 5x^2$.
• December 4th 2009, 03:37 PM
ryu991
Quote:

Originally Posted by lvleph
$1/2 \int_0^2 \! e^{5x^2}\, d(5x^2)$ is another way to write u-substitution where $u = 5x^2$.

so after the u substitution I would have to evaluate

(1/2)e^(5x^2) - (5/2)x^2 from 0 -2? making my answer be

[(1/2)e^20 -10] - [(1/2)e^0 - 0]

or whatever that evaluates to? or did I make a mistake along the way?
• December 4th 2009, 03:49 PM
lvleph
As long as you meant $\left[\frac{1}{2}e^{5x^2} - \frac{5}{2}x^2\right|^2_0$ then yes.
• December 4th 2009, 03:50 PM
ryu991
Quote:

Originally Posted by lvleph
As long as you meant $(1/2)e^{5x^2} - {5/2}x^2\left|^2_0\right.$ then yes.

yeah thats what I meant(I saw what the syntax error was when I quoted you) thanks for all of your help.