#### Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 43 maths textbook solution.

Hints:-  You must know the rules of continuous integral functions.

Given:-  $f(2 a-x)=f(x)$

Prove : $\int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x$

Solution : $\int_{a}^{c} f(x) d x=\int_{a}^{b} f(x) d x+\int_{b}^{c} f(x) d x$

So L.H.S,

$\int_{a}^{c} f(x) d x=\int_{a}^{b} f(x) d x+\int_{b}^{c} f(x) d x$                                   ....(1)

Let    $x=2 a-t \; \; \; \; \; \; \; \; \; \quad \text { when, } x=2 a, t=0$

$dx=-dt \; \; \; \; \; \; \; \; \; \; \; \; \; \; x = a, t=a$

Subscription

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

Using the property of integration                                      $\left[\int_{0}^{b} f(x) d x=-\int_{b}^{a} f(x) d t\right]$

Substituting in eq (1)                                                        $\left[\int_{0}^{2 a} f(x) d x=\int_{0}^{a} f(2 a-t) d t\right]$

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

Now using the property

\begin{aligned} &\int_{a}^{b} f(x) d x+\int_{a}^{b} g(x) d x=\int_{a}^{b} f(x)+g(x) \cdot d x \\ &\int_{0}^{2 a} f(x) d x=\int_{0}^{a} f(x)+f(2 a-x) \cdot d x \end{aligned}

Since $f(2a-x)=f(x)$

\begin{aligned} &\therefore \int_{0}^{2 a} f(x) d x=\int_{0}^{a} f(x)+f(x) \cdot dx \\ &\therefore \int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x \end{aligned}

Hence proved