Answer
$$f\left( x \right) = 12{x^3}$$
Work Step by Step
$$\eqalign{
& 3{x^4} - 48 = \int_2^x {f\left( t \right)} dt \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {3{x^4} - 48} \right] = \frac{d}{{dx}}\left[ {\int_2^x {f\left( t \right)} dt} \right] \cr
& {\text{By the fundamental theorem of calculus }}\frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)} dt} \right] = f\left( x \right) \cr
& \frac{d}{{dx}}\left[ {3{x^4} - 48} \right] = f\left( x \right) \cr
& f\left( x \right) = \frac{d}{{dx}}\left[ {3{x^4} - 48} \right] \cr
& f\left( x \right) = 12{x^3} \cr
& \cr
& {\text{Checking by substitution}} \cr
& \int_2^x {f\left( t \right)} dt = \int_2^x {12{t^3}} dt \cr
& \int_2^x {f\left( t \right)} dt = \left[ {\frac{{12{t^4}}}{4}} \right]_2^x \cr
& \int_2^x {f\left( t \right)} dt = \frac{{12{{\left( x \right)}^4}}}{4} - \frac{{12{{\left( 2 \right)}^4}}}{4} \cr
& \int_2^x {f\left( t \right)} dt = 3{x^4} - 48 \cr} $$
