Answer
$$f\left( 4 \right) = 24$$
Work Step by Step
$$\eqalign{
& {\text{We have this information:}} \cr
& \int_1^2 {f'\left( {2x} \right)dx} = 10{\text{ and }}f\left( 2 \right) = 4 \cr
& {\text{Let }}t = 2x,{\text{ }}dt = 2dx,{\text{ }}dx = \frac{1}{2}dt \cr
& {\text{The new limits of integration are:}} \cr
& x = 1 \Rightarrow t = 2\left( 1 \right) = 2 \cr
& x = 2 \Rightarrow t = 2\left( 2 \right) = 4 \cr
& {\text{Apply the substitution}} \cr
& \int_1^2 {\underbrace {f'\left( {2x} \right)}_{f'\left( t \right)}\underbrace {dx}_{\frac{1}{2}dt}} = \int_2^4 {f'\left( t \right)\left( {\frac{1}{2}} \right)dt} \cr
& = \frac{1}{2}\int_2^4 {f'\left( t \right)dt} \cr
& {\text{Recall that }}\int_a^b {f'\left( x \right)dx = f\left( b \right) - f\left( a \right),{\text{ then}}} \cr
& \frac{1}{2}\int_2^4 {f'\left( t \right)dt} = \frac{1}{2}\left( {f\left( 4 \right) - f\left( 2 \right)} \right) \cr
& \int_2^4 {f'\left( t \right)dt} = f\left( 4 \right) - f\left( 2 \right) \cr
& {\text{Where }}\int_2^4 {f'\left( t \right)dt} = 2\int_1^2 {f'\left( {2x} \right)dx} = 20 \cr
& 20 = f\left( 4 \right) - f\left( 2 \right) \cr
& {\text{Solve for }}f\left( 4 \right) \cr
& f\left( 4 \right) = 20 + f\left( 2 \right) \cr
& {\text{Substitute }}f\left( 2 \right) = 4 \cr
& f\left( 4 \right) = 20 + 4 \cr
& f\left( 4 \right) = 24 \cr} $$
