Question

If $\underset{0}{\overset{1}{\int }}f\left(x\right)dx=1,\underset{0}{\overset{1}{\int }}xf\left(x\right)dx=a,\underset{0}{\overset{1}{\int }}{x}^{2}f\left(x\right)dx=a^2,\mathrm{then}\underset{0}{\overset{1}{\int }}{\left(a-x\right)}^{2}f\left(x\right)dx$ equals

(a) 4a²
(b) 0
(c) 2a²
(d) none of these

Answer 1

(b) 0

$$\begin{gathered} {\int }_0^1{\left(a-x\right)}^{2}f\left(x\right)dx \\ =a^2{\int }_0^{1}f\left(x\right)dx+{\int }_0^1{x}^{2}f\left(x\right)dx-2a{\int }_0^{1}xf\left(x\right)dx \\ =a^2\times 1+a^2-2aa\left(\mathrm{As}\mathrm{per}\mathrm{given}\mathrm{values}\right) \\ =2a^2-2a^2 \\ =0 \end{gathered}$$