Mathematics - Definite Integration Question with Solution | TestHub

${\int }_1^3{x}^2{F}^{\prime }\left(x\right)dx=-12 and {\int }_1^3{x}^3 F^{\prime \prime }\left(x\right)dx=40$
${\int }_1^3{x}^3{F}^{\prime \prime }\left(x\right)=x^3{F}^{\prime }\left(x\right){]}_1^3-{\int }_1^{3}3x^{2}F\prime \left(x\right)$
$40=27F^{\prime }\left(3\right)-F^{\prime }\left(1\right)+36$
$4=9 \left(f\left(3\right)+4\right)-f\left(1\right)$
$\Rightarrow 9f^{\prime }\left(3\right)-f^{\prime }\left(1\right)+32=0$
Also
${\int }_1^3{x}^{2}F\prime \left(x\right)dx$
$=x^{2}F\left(x\right){]}_1^3-{\int }_1^{3}2xF\left(x\right)dx=-12$
$9F\left(3\right)-F\left(1\right)-2{\int }_1^{3}xF\left(x\right)dx=-12$
$-36-0-2{\int }_1^{3}f\left(x\right)dx=-12$
$-36+12=2{\int }_1^{3}f\left(x\right)dx$
$-12={\int }_1^{3}f\left(x\right)dx$