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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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