$\int\limits_0^{10}e^{-0.04t}\cdot e^{-0.001t^2}dt$

Question

I need to find the following integral

$$\int\limits_0^{10}e^{-0.04t ~-0.001t^2}dt$$ This integral seems to "scream" for the error function, but I have never worked with the error function yet, so I have no idea how to do this. Can anyone please show me how this definite integral can be determined?

Update: After following the hints given, I have the following:

\begin{align}\int\limits_0^{10}e^{\frac{-1}{1000}(t^2+40t)}dt &= e^{\frac{400}{1000}}\int\limits_0^{10} e^{\frac{-1}{1000}(t+20)^2}dt\end{align}

Now, let $u = t+20$, then our integral changes to \begin{align*}e^{\frac{400}{1000}}\int\limits_{20}^{30}e^{\frac{-1}{1000}u^2}du\end{align*} I am stuck here though, since I only know how to use the error function of $\int_0^xe^{-at^2}dt$, which is different from what I have, since the lower limit is not zero? Please help!

Answer 1

Hint: complete the square in the exponent so you get $e^{(a -(b+t)^2)}$ for constants $a,b$. Pull out the $e^a$ as a multplicative constant. Do a $u$ substitution to make it $e^{-u^2}$ to match the definition of the error function.