How do you sovle the equation: $\int_{-3}^{2} ( e^{-t+1} + sin (\frac{2\pi}{3}t) ) \delta(t- \frac{3}{2}) dt$
Because of the $\delta(t- \frac{3}{2})$ this is only non-zero at $t=\frac{3}{2}$
But I don't get why I can't do this: $\int_{-3}^{2} ( e^{-\frac{3}{2}+1} + sin (\frac{2\pi}{3} \times \frac{3}{2}) ) dt$
Why you try to solve the indefinite integral, it supposes to be: $\int( e^{-t+1} + sin (\frac{2\pi}{3}t) ) \delta(t- \frac{3}{2}) dt = \frac{u(2t-3)-1}{\sqrt{e}} + C $. In which $u$ is the unit step function. But I don't know how to solve this indefinite integral step-by-step.
$$\int_a^b f(x) \delta(x-c)dx = \begin{cases} f(c) & \text{ if }c \in (a,b)\\ 0 & \text{ otherwise}\end{cases}$$