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Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 16 maths textbook solution.

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Answer : \int_{a}^{b} x f(x) d x=\frac{a+b}{2} \int_{a}^{b} f(x) d x

Given : f(a+b-x)=f(x) and prove that

\int_{a}^{b} x f(x)=\frac{a+b}{2} \int_{a}^{b} f(x) d x

Hint : You must know the formula of \int_{a}^{b} f(x) d x

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

\begin{aligned} &I=\int_{a}^{b}(a+b-x) f(a+b-x) d x\left(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right) \\ &I=\int_{a}^{b}(a+b-x) f(x) d x[f(a+b-x)=f(x)] \\ &I=\int_{a}^{b}(a+b) f(x) d x-\int_{a}^{b} x f(x) d x \end{aligned}

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

\int_{a}^{b} x f(x) d x=\frac{a+b}{2} \int_{a}^{b} f(x) d x \text { (proved) }

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