• May 29th 2007, 04:48 AM
owenji81
Please see the attached file !!!!

Thanks a lot!
• May 29th 2007, 05:20 AM
CaptainBlack
Quote:

Originally Posted by owenji81
Please see the attached file !!!!

Thanks a lot!

Are those definite or indefinite integrals?

RonL
• May 29th 2007, 05:39 AM
owenji81
indefinite integrals
• May 29th 2007, 08:02 AM
ThePerfectHacker
The problem is we cannot find anti-derivatives on closed intervals, we require that $\displaystyle |x|<.5$ in that case $\displaystyle f(x) = 1 \mbox{ on }-.5<x<.5$ and hence the anti-derivative is $\displaystyle x$ on $\displaystyle (-5,5)$ and a constant everywhere.

So,
$\displaystyle f(x) = \left\{ \begin{array}{c}C_1 \mbox{ for }x<-.5\\ x+C_2 \mbox{ for }-.5<x<.5 \\ C_3 \mbox{ for }x>.5 \end{array} \right\}$

But $\displaystyle f$ needs to be continous differenciable at $\displaystyle x=\pm .5$ as well. So $\displaystyle C_1 = C_2 - .5$ and $\displaystyle C_3 = C_2+.5$ where $\displaystyle C_2$ is any arbitraty konstant.
• May 29th 2007, 09:03 AM
CaptainBlack
Quote:

Originally Posted by owenji81
Please see the attached file !!!!

Thanks a lot!

Well I still don't like it so I will do the following instead:

$\displaystyle c_K(x) = \int_{-\infty}^x K^2(u) du$

Then:

$\displaystyle c_K(x) = \left\{ \begin{array}{cc} 0 &\mbox{ for }x<-0.5 \\ x + 0.5& \mbox{ for }-0.5\le x \le 0.5 \\ 1& \mbox{ for }x>0.5 \end{array}\right.$

and if you add an arbitary constant onto this $\displaystyle c_K(x)$ you will still have an anti-derivative of $\displaystyle K(x)$.

If everything has worked out right with the extra arbitary constant this should be the same as the soution given by ImPerfectHacker.

RonL