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

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

Answers (1)

Answer:-  Proved

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

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    \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

 

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