Question

If $\delta \left(t\right)$ is an unit impulse function, a signal $g\left(t\right)$ is represented as follows; $g\left(t\right)=6\delta (2t+1)$
Then the value of ${\int }_{-1}^{0}g\left(t\right)dt$ is_______

• A. 3

Answer 1

The correct option is A 3
Given,

$g\left(t\right)=6\delta (2t+1)$

${\int }_{-1}^{0}g\left(t\right)dt={\int }_{-1}^{0}6\delta (2t+1)dt$

$=\frac{6}{2}{\int }_{-1}^0\delta (t+\frac{1}{2})dt=3$

$\because {\int }_{t1}^{t2}\delta (t-t_0)dt=1;t_1<t_0<t_2$