Answer
$G\left( t \right) = 7{e^t} - {e^3}t + C$
Work Step by Step
$$\eqalign{
& g\left( t \right) = 7{e^t} - {e^3} \cr
& {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr
& {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}{e^x} \to {\text{Particular antiderivative: }}{e^x} \cr
& {\text{Function: }}f\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + h\left( x \right) \cr
& {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr
& G\left( t \right) = 7{e^t} - {e^3}t + C \cr
& \cr
& {\text{Checking the answer by differentiation}} \cr
& G'\left( t \right) = \frac{d}{{dt}}\left[ {7{e^t}} \right] - \frac{d}{{dt}}\left[ {{e^3}t} \right] + \frac{d}{{dt}}\left[ C \right] \cr
& G'\left( t \right) = 7{e^t} - {e^3} + 0 \cr
& G'\left( t \right) = 7{e^t} - {e^3} \cr} $$
