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Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 23

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Answer: \frac{a}{2}

Hint: Using \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x


Given: \text { If } \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a}-\mathrm{x}) \text { and } \int_{0}^{a} x f(x) d x=K \int_{0}^{a} f(x) d x


Solution:  f(x)=f(a-x)                    .............(*)

We have  \int_{0}^{a} x f(x) d x=K \int_{0}^{a} f(x) d x

Taking LHS,

        \begin{aligned} &\text { Let } I=\int_{0}^{a} x \end{aligned}                    .............(1)

        \Rightarrow \mathrm{I}=\int_{0}^{a}(a-x) f(a-x) d x

From (*),

        \mathrm{I}=\int_{0}^{a}(a-x) f(x) d x            ...........(2)

Adding (1) and (2),

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


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

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

From the Given, K\int_{0}^{a} f(x) d x =\frac{a}{2} \int_{0}^{a} f(x) d x

On comparing, we get \mathrm{K}=\frac{a}{2}


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