Question

The function $f$ is a continuous on $R$ such that $f\left(4\right)=8$ and $f^{\prime }\left(4\right)=10$. If $g\left(x\right)=\underset{0}{\overset{x}{\int }}(x-t{)}^{2}f\left(t\right)dt$, then the value of $\frac{g^{\prime \prime \prime }\left(4\right)}{g^{\prime \prime \prime \prime }\left(4\right)}$ is equal to

Answer 1

$g\left(x\right)=\underset{0}{\overset{x}{\int }}(x-t{)}^{2}f\left(t\right)dt\Rightarrow g\left(x\right)=x^2\underset{0}{\overset{x}{\int }}f\left(t\right)dt-2x\underset{0}{\overset{x}{\int }}tf\left(t\right)dt+\underset{0}{\overset{x}{\int }}{t}^{2}f\left(t\right)dt$
$\Rightarrow g^{\prime }\left(x\right)=\underset{0}{\overset{x}{\int }}2(x-t)f\left(t\right)dt\Rightarrow g^{\prime \prime }\left(x\right)=2\underset{0}{\overset{x}{\int }}f\left(t\right)dt-0\Rightarrow g^{\prime \prime \prime }\left(x\right)=2f\left(x\right)\Rightarrow g^{\prime \prime \prime }\left(4\right)=2\times f\left(4\right)=16\Rightarrow g^{\prime \prime \prime \prime }\left(x\right)=2f^{\prime }\left(x\right)\Rightarrow g^{\prime \prime \prime \prime }\left(4\right)=20$
$\frac{g^{\prime \prime \prime }\left(4\right)}{g^{\prime \prime \prime \prime }\left(4\right)}=\frac{8}{10}=0.80$