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  • Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 48 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 48 maths textbook solution.

Answers (1)

Answer:- Proved

Hints:-  You must know the continuity rules of integration.

Given:-  f(x) is a continuous function on [-a,a]

Solution:-   \int_{-a}^{a} f(x) d x=\int_{0}^{a} f(x)+f(-x) d x

            \begin{aligned} &{\left[\begin{array}{ll} f(-x)=f(x) & f(x) \rightarrow \text { even } \\ f(-x)=-f(x) & f(x) \rightarrow \text { odd } \end{array}\right]} \\ &\int_{-a}^{a} f(x) d x=\left\{\begin{array}{lr} \int_{0}^{a} f(x)+f(-x) d x & f(x) \rightarrow \text { even } \\ \int_{0}^{a} f(x)+f(-x) d x & f(x) \rightarrow \text { odd } \end{array}\right] \end{aligned}

         \int_{-a}^{a} f(x) d x=\left\{\begin{array}{ll} \int_{0}^{a} 2 f(x) d x & f(x) \rightarrow \text { even } \\ 0 & f(x) \rightarrow \text { odd } \end{array}\right]

Hence \left.\int_{-a}^{a} f(x) d x=\int_{0}^{a} f(x)+f(-x)\right] d x is true because this function is not zero

         

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