1. ## Integration question

I've obtained the anti-derivative, but I'm having problems solving for the constant a from this step. Just to let you know, the expected value of t is 6, so I already have the anti-derivative of that equation but I'm having problems going forward. Also, the lower limit can be considered as starting from 0. Thanks

2. Sorry, but as written the integral evaluates to infinity, not 1: the value of $e^{-a(t-2)}$ for $t \to -\infty$ is $\infty$.

3. Originally Posted by bidii
I've obtained the anti-derivative, but I'm having problems solving for the constant a from this step. Just to let you know, the expected value of t is 6, so I already have the anti-derivative of that equation but I'm having problems going forward. Also, the lower limit can be considered as starting from 0. Thanks

For $a\neq 0$ , $\int\limits_{-\infty}^\infty ae^{-a(t-2)}\,dt=-[e^{-a(t-2)}]^\infty_{-\infty}=-\lim_{b\to\infty}e^{-a(b-2)}+\lim_{h\to -\infty}e^{-a(h-2)}$ . Let's now check each limit separatedly:

$\lim_{b\to\infty}e^{-a(b-2)}=\lim_{b\to\infty}\frac{1}{e^{a(b-2)}}$ $=\left\{\begin{array}{lc}0&,\,if\,\,a>0\\\infty&,\ ,if\,\,a<0\end{array}\right.$ , and likewise $\lim_{h\to -\infty}e^{-a(h-2)}=\lim_{h\to -\infty}\frac{1}{e^{a(h-2)}}$ $=\left\{\begin{array}{lc}\infty&,\,if\,\,a>0\\0&,\ ,if\,\,a<0\end{array}\right.$

According to the above the value of the improper integral can't be 1 at all, since the integral diverges for any value of $a$.

I don't understand what you mean by "the expected value of t is 6"...

Tonio