#### Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 16 maths

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

Given: $\text { If } \mathrm{f}(\mathrm{a}-\mathrm{x})=\mathrm{f}(\mathrm{x}) \text { and } \int_{0}^{a} f(x) d x=K \int_{0}^{a / 2} f(x) d x$

Solution:

$\mathrm{f}(\mathrm{a}-\mathrm{x})=\mathrm{f}(\mathrm{x})$                                ................(1)

We have, $\int_{0}^{a} f(x) d x=K \int_{0}^{a / 2} f(x) d x$

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

$\left[\because \int_{0}^{2 a} f(x) d x=\int_{0}^{a} f(x) d x+\int_{0}^{a} f(2 a-x) d x\right]$

\begin{aligned} &\Rightarrow \int_{0}^{a / 2} f(x) d x+\int_{0}^{a / 2} f(x) d x=K \int_{0}^{a / 2} f(x) d x \\\\ &\Rightarrow 2 \int_{0}^{a / 2} f(x) d x=K \int_{0}^{a / 2} f(x) d x \end{aligned}

On comparing, we get K = 2
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