#### Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 23

Answer: $\frac{a}{2}$

Hint: Using $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$

Given: $\text { If } \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a}-\mathrm{x}) \text { and } \int_{0}^{a} x f(x) d x=K \int_{0}^{a} f(x) d x$

Solution:  $f(x)=f(a-x)$                    .............(*)

We have  $\int_{0}^{a} x f(x) d x=K \int_{0}^{a} f(x) d x$

Taking LHS,

\begin{aligned} &\text { Let } I=\int_{0}^{a} x \end{aligned}                    .............(1)

$\Rightarrow \mathrm{I}=\int_{0}^{a}(a-x) f(a-x) d x$

From (*),

$\mathrm{I}=\int_{0}^{a}(a-x) f(x) d x$            ...........(2)

$2I=\int_{0}^{a} x\; {\mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\; +} \int_{0}^{a}(a-x) f(x) d x$

$2I=\int_{0}^{a}(x+a-x) f(x) d x$

$\mathrm{I}=\frac{a}{2} \int_{0}^{a} f(x) d x$

From the Given, $K\int_{0}^{a} f(x) d x =\frac{a}{2} \int_{0}^{a} f(x) d x$

On comparing, we get $\mathrm{K}=\frac{a}{2}$