# Evaluating the integral of a delta function multiplied by an exponential

How does one evaluate the integral below? I don't think we can use integration by parts here. I am basically stuck.

$$\int_{-1}^{6}{(2+e^{-t})\delta(t-2)} dt$$

$$= \int_{-1}^{6}{2\delta(t-2)dt} + \int_{-1}^{6}{e^{-t}\delta(t-2)dt}$$

• Use the fact that if $t_0 \in [a,b]$, then $\int_a^b f(t) \delta(t-t_0)dt = f(t_0)$. – muffle Nov 25 '13 at 23:30
• What if it is not in [a,b]? It evaluates to zero? – Bob Shannon Nov 25 '13 at 23:33
• Yes. To see this if $t_0 \not \in [a,b]$, then $\delta(t-t_0) = 0$ when restricted to the interval $[a,b]$, so $\int_a^b f(t) \delta(t-t_0)dt = \int 0 dt = 0$. – muffle Nov 25 '13 at 23:34

As indicated by muffle, the answer is the value of $f(t)=2 + e^{-t}$ at point $t=t_0=2$, that is $$2 + e^{-2} .$$