# Intergration Question (fairly basic)

• May 18th 2009, 01:11 AM
Peleus
Intergration Question (fairly basic)
$\int(e^x - 3)^2 dx$

I think it goes as following, here is my working.

$\int (e^{2x} - 6e^x + 9) dx$

$= \frac{e^{2x}}{2} - e^x + 9x$

Can anyone confirm or tell me where I've gone wrong? Thank you very much.
• May 18th 2009, 01:14 AM
Rapha
Hi.

Quote:

Originally Posted by Peleus
$\int(e^x - 3)^2 dx$

I think it goes as following, here is my working.

$\int (e^{2x} - 6e^x + 9) dx$

$= \frac{e^{2x}}{2} - e^x + 9x$

Can anyone confirm or tell me where I've gone wrong? Thank you very much.

Yea, the work seems to be correct, but you made a mistake

$\int - 6e^x dx \not= -e^x$

But I guess you know that

And maybe you should add a constant to your solution

$= \frac{e^{2x}}{2} - 6e^x + 9x + const$

Yours
Rapha
• May 18th 2009, 01:23 AM
Peleus
Yep, sorry missed that.

$\int 6e^x$

which turns into

$6 \int \frac{1}{6}e^x$

$6 * e^x$

$6e^x$

Thanks, forgot to type in the const as well.
• May 18th 2009, 01:26 AM
Rapha
Quote:

Originally Posted by Peleus
Yep, sorry missed that.

$\int 6e^x$

which turns into

$6 \int \frac{1}{6}e^x$

No!!!!

Quote:

Originally Posted by Peleus
$6 * e^x$

Yes

Quote:

Originally Posted by Peleus
$6e^x$

Yes

Actually it is $- \int 6e^x dx$ = $-6*\int e^xdx = -6*e^x$

Quote:

Originally Posted by Peleus
Thanks

You're welcome
• May 18th 2009, 01:30 AM
Peleus
I have absolutely no idea what I was thinking there, even the working doesn't work.

Yes, you're right and I understand how you got it :)

Cheers.