Let f x and g x be two functions satisfying f x 2 + g ( 4 - x ) = 4 x 3 and g 4 - x + g x = 0 , then the…

$I={\int }_{-4}^{4}f\left(x^2\right)dx=2{\int }_0^{4}f\left(x^2\right)dx$ {Even function}

$=2{\int }_0^4\left(4x^3-g\left(4-x\right)\right)dx$

$=2\left(\frac{4x^4}{4}{|}_0^4-{\int }_0^{4}g\left(4-x\right)dx\right)=512-2I_1$

$I_1={\int }_0^{4}g\left(4-x\right)dx={\int }_0^{2}g\left(4-x\right)dx+{\int }_2^{4}g\left(4-x\right)dx$

Let $I_2={\int }_2^{4}g\left(4-x\right)dx$

If $4-x=t$ then 

$I_2=-{\int }_2^{0}g\left(t\right)dt={\int }_0^{2}g\left(t\right)dt={\int }_0^{2}g\left(x\right)dx$

So, $I_1={\int }_0^{2}g\left(4-x\right)dx+{\int }_0^{2}g\left(x\right)dx=0$

Hence, $I=512$