Question

If f (a + b − x) = f (x), then $\underset{a}{\overset{b}{\int }}$ x f (x) dx is equal to

(a) $\frac{a+b}{2}\underset{a}{\overset{b}{\int }}f\left(b-x\right)dx$

(b) $\frac{a+b}{2}\underset{a}{\overset{b}{\int }}f\left(b+x\right)dx$

(c) $\frac{b-a}{2}\underset{a}{\overset{b}{\int }}f\left(x\right)dx$

(d) $\frac{a+b}{2}\underset{a}{\overset{b}{\int }}f\left(x\right)dx$

Answer 1

(d) $\frac{a+b}{2}\underset{a}{\overset{b}{\int }}f\left(x\right)dx$


$$\begin{gathered} \mathrm{Let},I={\int }_a^{b}xf\left(x\right)dx...\left(\mathrm{i}\right) \\ ={\int }_a^b\left(a+b-x\right)f\left(a+b-x\right)dx \\ ={\int }_a^b\left(a+b-x\right)f\left(x\right)dx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ 2I={\int }_a^b\left(x+a+b-x\right)f\left(x\right)dx \\ =\left(a+b\right){\int }_a^{b}f\left(x\right)dx \\ \mathrm{Hence}I=\frac{a+b}{2}{\int }_a^{b}f\left(x\right)dx \end{gathered}$$