Question

If $f:R\to R$ is continuous and differentiable function such that ${\int }_{-1}^{x}f\left(t\right)dt+f^{\prime \prime \prime }\left(3\right){\int }_x^{0}dt={\int }_1^x{t}^{3}dt-f^{\prime }\left(t\right){\int }_0^x{t}^{2}dt+f^{\prime \prime }\left(2\right){\int }_x^{3}tdt$, then $f^{\prime }\left(4\right)=48-a{f}^{\prime }\left(1\right)-b{f}^{\prime \prime }\left(2\right)$, where $a+b=$

Answer 1

From the given equation, we have
${\int }_{-1}^{x}f\left(t\right)dt-x{f}^{\prime \prime \prime }\left(3\right)$
$=(\frac{x^4}{4}-\frac{1}{4})-f^{\prime }\left(1\right)\frac{x^3}{3}+f^{\prime \prime }\left(2\right)(\frac{9}{2}-\frac{x^2}{2})$

Differentiating w.r.t. $x$, we get
$f\left(x\right)-f^{\prime \prime \prime }\left(3\right)=x^3-x^2{f}^{\prime }\left(1\right)-x{f}^{\prime \prime }\left(2\right)$

Differentiating again, we get
$f^{\prime }\left(x\right)=3x^2-2x{f}^{\prime }\left(1\right)-f^{\prime \prime }\left(2\right)$

$\therefore f^{\prime }\left(4\right)=48-8f^{\prime }\left(1\right)-f^{\prime \prime }\left(2\right)$