1. ## integration #2

Hi:
Would somebody check this for me please.

Also, is this simplified correctly?

thanks

2. Originally Posted by stealthmaths
Hi:
Would somebody check this for me please.

Also, is this simplified correctly?

thanks
1- You should post this in calculus sub-forum.
2- You have a sign mistake in the last line.
3- You should not have that "x" in the second term in the last line.

3. But if this is in calculus then it is not in pre-university forum. And I am pre-university. Please advise? And how would I now move this thread to calculus?

How's this?

And is this simplified correctly?
thanks

4. Originally Posted by stealthmaths
But if this is in calculus then it is not in pre-university forum. And I am pre-university. Please advise?
Read the "sticky" post at the top of the Pre-Calculus forum. In this American-based web site, calculus is considered university-level, regardless of whether you are actually at university or not.

5. $\displaystyle \int u dv = uv- \int v du$

$\displaystyle \int 9xe^{-4x}= 9\int xe^{-4x}$

u=x, du=dx and dv=e^{-4x} , v=(-1/4)e^{-4x}

$\displaystyle 9\int xe^{-4x}= 9[-\frac{1}{4}e^{-4x}x-\int -\frac{1}{4}e^{-4x}dx]$

$\displaystyle = 9[-\frac{1}{4}e^{-4x}x +\frac{1}{4}\int e^{-4x}dx]$
$\displaystyle = 9[-\frac{1}{4}e^{-4x}x +\frac{1}{4}(e^{-4x}(-\frac{1}{4})]+C$
$\displaystyle = 9[-\frac{1}{4}e^{-4x}x -\frac{1}{8}(e^{-4x})]+C$

6. I'm sorry DBA. What is it your trying to say. The answer I have above is wrong?