Answer
$f\left( t \right) = \frac{{{t^4}}}{{12}} - 2{t^2} + {C_1}t + {C_2}$
Work Step by Step
$$\eqalign{
& f''\left( t \right) = {t^2} - 4 \cr
& {\text{Find the general antiderivative }}f'\left( t \right) \cr
& {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right) \cr
& {\text{Function: }}cf\left( t \right) \to {\text{Particular antiderivative: }}cF\left( t \right) \cr
& {\text{Function:}}{t^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}\frac{{{t^{n + 1}}}}{{n + 1}} \cr
& {\text{we obtain}} \cr
& f'\left( t \right) = \frac{{{t^{2 + 1}}}}{{2 + 1}} - 4t + {C_1} \cr
& f'\left( t \right) = \frac{{{t^3}}}{3} - 4t + {C_1} \cr
& {\text{Find the general antiderivative }}f\left( t \right) \cr
& {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right) \cr
& {\text{Function: }}cf\left( t \right) \to {\text{Particular antiderivative: }}cF\left( t \right) \cr
& {\text{Function:}}{t^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}\frac{{{t^{n + 1}}}}{{n + 1}} \cr
& {\text{we obtain}} \cr
& f\left( t \right) = \frac{{{t^4}}}{{3\left( 4 \right)}} - 4\left( {\frac{{{t^2}}}{2}} \right) + {C_1}t + {C_2} \cr
& f\left( t \right) = \frac{{{t^4}}}{{12}} - 2{t^2} + {C_1}t + {C_2} \cr} $$
