Question

Let f(x) be a continuous function of x defined on [0, a] such that f(a - x) = f(x). Then, ${\int }_0^{a}xf\left(x\right)dx$ is equal to

• A. $\frac{a}{2}{\int }_0^{a}f\left(x\right)\mathrm{dx}$
• B. $a{\int }_0^{a}f\left(x\right)\mathrm{dx}$
• C. $\frac{a}{2}{\int }_0^{a}f(a-x)\mathrm{dx}$
• D. $a{\int }_0^{a}f(a-x)\mathrm{dx}$

Answer 1

Answer: (A), (C)

$\frac{a}{2}{\int }_0^{a}f(a-x)\mathrm{dx}$

$\mathrm{Let}I={\int }_0^{a}xf\left(x\right)\mathrm{dx}\Rightarrow I={\int }_0^a\left(a-x\right)f(a-x)\mathrm{dx}\Rightarrow I={\int }_0^a\left(a-x\right)f\left(x\right)\mathrm{dx}\left(\mathrm{As}f(a-x)=f\left(x\right)\right]\Rightarrow I=a{\int }_0^{a}f\left(x\right)\mathrm{dx}-{\int }_0^{a}xf\left(x\right)\mathrm{dx}\Rightarrow I=a{\int }_0^{a}f\left(x\right)\mathrm{dx}-I\Rightarrow 2I=a{\int }_0^{a}f\left(x\right)\mathrm{dx}\Rightarrow I=\frac{a}{2}{\int }_0^{a}f\left(x\right)\mathrm{dx}=\frac{a}{2}{\int }_0^{a}f(a-x)\mathrm{dx}$