If $f: R \in R$ is continuous and differentiable function such that $\int\limits_{-1}^x f(t) d t+f'''(3) \int\limits_x^0 d t=\int\limits_1^x t^3 d t-f'(1) \int\limits_0^x t^2 d t+f''(2) \int\limits_x^3 t d t$ then the value of $f'(4)$, is

$48-8 f'(1)-f''(2)$

We have,

$\int\limits_{-1}^x f(t) d t+f'''(3) \int\limits_x^0 d t$

$= \int\limits_1^x t^3 d t-f'(1) \int\limits_0^x t^2 d t+f''(2) \int\limits_x^3 t d t$

$\Rightarrow \int\limits_{-1}^x f(t) d t-x f'''(3)=\left(\frac{x^4}{4}-\frac{1}{4}\right) - f'(1) \frac{x^3}{3}+f''(2)\left(\frac{9}{2}-\frac{x^2}{2}\right)$

Differentiating w.r. to $x$, we get

$f(x)-f'''(3)=x^3-x^2 f'(1)-x f''$

Differentiating w.r. to $x$, we get

$f'(x) =3 x^2-2 x f'(1)-f''$

$\Rightarrow f'(4) =48-8 f'(1)-f''(2)$