If $$y = f\left( x \right)$$   makes $$+ve$$  intercept of $$2$$ and $$0$$ unit on $$x$$ and $$y$$ axes and encloses an area of $$\frac{3}{4}$$ square unit with the axes then $$\int\limits_0^2 {x\,f'\left( x \right)dx} $$   is :

Solution :
We have $$\int\limits_0^2 {f\left( x \right)dx} = \frac{3}{4}\,;$$
$$\eqalign{ & {\text{Now, }}\int\limits_0^2 {x\,f'\left( x \right)dx} \cr & = x\int\limits_0^2 {f'\left( x \right)dx} - \int\limits_0^2 {f\left( x \right)dx} \cr & = \left[ {x\,f\left( x \right)} \right]_0^2 - \frac{3}{4} \cr & = 2f\left( 2 \right) - \frac{3}{4} \cr & = 0 - \frac{3}{4}\,\,\,\,\,\,\,\left( {\because \,f\left( 2 \right) = 0} \right) \cr & = - \frac{3}{4} \cr} $$