So i know the basic form for integrating delta functions with the bounds of integration being 0 to infinity, but I've been given a problem with different bounds and i'm a little lost on how to approach the problem. Here it is:
$$ \int_{-1}^6 (5+e^{-2t})\, \delta(t-2) \ dt$$
$f(t)$ would be defined as $5+e^{-2t}\,$ and normally I think the answer would be $f(a)$ which is $f(2)$ in this case, but i think the bounds change that.
$$\int_{-1}^6 (5+e^{-2t})\, \delta(t-2) \ dt= 5+e^{-2(2)}=5+e^{-4}$$
as $2 \in (-1,6)$.