Correct Answer - Option 1 : 2
\(x\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1,}&{\left| t \right| \le 2}\\ {0,}&{\left| t \right| > 2} \end{array}} \right.\)
\(\mathop \smallint \limits_0^5 2x\left( {t - 3} \right)\delta \left( {t - 4} \right)dt\)
From the property of unit impulse function,
\(\smallint x\left( t \right)\delta \left( {t - {t_0}} \right)dt = x\left( {{t_0}} \right)\)
\(= \mathop \smallint \limits_0^5 2x\left( {t - 3} \right)\delta \left( {t - 4} \right)dt\)
= 2x(4 – 3) = 2x(1)
As, x(1) = 1, the integral value = 2