Mathematics - Definite Integration Question with Solution | TestHub

${\int }_1^4{f}^{\prime }(x)dx={\left[x{f}^{\prime }(x)\right]}_1^4-{\int }_1^{4}x{f}^{\prime \prime }(x)dx$

$I={\int }_1^{4}x{f}^{\prime \prime }(x)dx={\int }_1^4(5-x)f^{\prime \prime }(5-x)dx$

$=5{\int }_1^4{\mathrm{f}}^{\prime \prime }(\mathrm{x})\mathrm{dx}-\mathrm{I}$

$\mathrm{I}=\frac{5}{2}\left[{\mathrm{f}}^{\prime }(4)-{\mathrm{f}}^{\prime }(1)\right]$

So, ${\int }_1^4{\mathrm{f}}^{\prime }(\mathrm{x})\mathrm{dx}=\frac{3}{2}\left[{\mathrm{f}}^{\prime }(4)+{\mathrm{f}}^{\prime }(1)\right]$

Now, ${\mathrm{f}}^{\prime \prime }(\mathrm{x})={\mathrm{f}}^{\prime \prime }(5-\mathrm{x})\Rightarrow {\mathrm{f}}^{\prime }(\mathrm{x})=-{\mathrm{f}}^{\prime }(5-\mathrm{x})+\mathrm{c}$

${\mathrm{f}}^{\prime }(0)+{\mathrm{f}}^{\prime }(5)=\mathrm{c}\Rightarrow \mathrm{c}=8$

so $f^{\prime }(x)+f^{\prime }(5-x)=8\Rightarrow f^{\prime }(4)+f^{\prime }(1)=8$