For the exponential distribution, $f(x)=(1/\theta) e^{-x/\theta}$ for $x>0,$ and $f(x)=0$ for $x \leq0$
$(i)$ Determine the exact value for the probability $P(0<X<3\theta).$
I need help with the integration. It has been a while since I have done this. Can anyone show me the working. Here is something that I did.
Here is what I did:
$E(X)=\int_0^{3\theta}xf(x)dx=1/\theta \int_0^{3\theta}xe^{-x/\theta}dx$
$u=x$ and $dv=e^{-x/\theta}dx$
$du=dx$ and $v=\int e^{-x/\theta}dx=-\theta e^{-x/\theta}$
$\int_0^{3\theta}xe^{-x/\theta}dx=uv |_0^{3\theta}-\int_0^{3\theta}vdu=(-\theta xe^{-x/\theta})|_0^{3\theta}+\theta \int_0^{3\theta}e^{x/\theta}dx$