$\text{ If }f\left(0\right)=1,f\left(2\right)=3,f^{\text{'}}\left(2\right)=5,\text{ then find the value of }{\int }_0^1 x{f}^{\prime \prime }\left(2x\right)dx$

$I_1={\int }_0^1 x{f}^{\prime \prime }\left(2x\right)dx.\text{ Putting }t=2x,\text{ i.e. }$

$$\begin{gathered} dx=\frac{dt}{2},\text{ we get } \\ {I}_1=\frac{1}{4}{\int }_0^2 t{f}^{\prime \prime }\left(t\right)dt \end{gathered}$$

$\begin{array}{rcl} & =& \frac{1}{4}\left[{\left.t{f}^{\mathrm{\prime }}\left(t\right)\right|}_0^2-{\int }_0^2 f^{\mathrm{\prime }}\left(t\right)dt\right]\text{ (Integrating by parts) }\\ & & \end{array}$

$=\frac{1}{4}\left[{\left.t{f}^{\mathrm{\prime }}\left(t\right)\right|}_0^2-{\left.f\left(t\right)\right|}_0^2\right]$

$=\frac{1}{4}\left(2f^{\mathrm{\prime }}\left(2\right)-f\left(2\right)+f\left(0\right)\right)=\frac{1}{4}(10-3+1)=2$