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

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

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

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    $\begingroup$ Use the fact that if $t_0 \in [a,b]$, then $\int_a^b f(t) \delta(t-t_0)dt = f(t_0)$. $\endgroup$ – muffle Nov 25 '13 at 23:30
  • $\begingroup$ What if it is not in [a,b]? It evaluates to zero? $\endgroup$ – Bob Shannon Nov 25 '13 at 23:33
  • $\begingroup$ 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 $\int_a^b f(t) \delta(t-t_0)dt = \int 0 dt = 0$. $\endgroup$ – muffle Nov 25 '13 at 23:34
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As indicated by muffle, the answer is the value of $f(t)=2 + e^{-t}$ at point $t=t_0=2$, that is $$ 2 + e^{-2} . $$

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