If f(0)=1,f(2)=3,f'(2)=5, then find the  value of ∫01 xf′′(2x)dx

I1=∫01 xf′′(2x)dx. Putting t=2x, i.e. dx=dt2, we get  I1=14∫02 tf′′(t)dt =14tf′(t)02−∫02 f′(t)dt (Integrating by parts)    =14tf′(t)02−f(t)02=142f′(2)−f(2)+f(0)=14(10−3+1)=2