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Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 39

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Answer:

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

Hint:

To solve this we will split (a+b-x)

Given:

\int(a+b-x)=f(x)

Solution:

\int(a+b-x)=f(x)

Let

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

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

I=\int_{a}^{b}(a+b) \cdot f(x) d x-\int_{a}^{b} x f(x) d x \ldots(i i)

Adding equation (i) and (ii)

I+I=\int_{a}^{b} x f(x) d x+\int_{a}^{b}(a+b) \cdot f(x) d x-\int_{a}^{b} x f(x) d x

\begin{aligned} &2 I=\int_{a}^{b}(a+b) f(x) d x \\\\ &I=\frac{1}{2} \int_{a}^{b}(a+b) f(x) d x \\\\ &I=\frac{a+b}{2} \int_{a}^{b}(a+b) f(x) d x \end{aligned}

 

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