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Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 46 maths textbook solution.

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Answer:-  Proved

Hints:-  You must know the continuity of integral functions.

Given:- f(x) is a continuous on \left [ 0,2a \right ]

Solution : I=\int_{0}^{2 a} f(x) d x

By additive property

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

Consider the integral \int_{0}^{2 a} f(x) d x

\begin{aligned} &\text { Let } x=2 a-t, \text { then } d x=-d t \\ &\text { When } x=a, t=a \quad x=2 a, t=0 \end{aligned}


\begin{aligned} &\int_{a}^{2 a} f(x) d x=-\int_{a}^{0} f(2 a-t) d t \\ &\int_{0}^{a} f(2 a-t) d t=\int_{0}^{a} f(2 a-x) d x \end{aligned}


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

Hence proved.

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