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Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 20 maths

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Answer: \frac{b-a}{2}

Hint: you must know the rule of integration for function of x


Given: \int_{a}^{b} \frac{f(x)}{f(x)+f(a+b-x)} d x

Solution:  \mathrm{I}=\int_{a}^{b} \frac{f(x)}{f(x)+f(a+b-x)} d x            ...............(i)

Using property \int_{a}^{b} f(x)=\int_{a}^{b} f(a+b-x) d x

 

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

\mathrm{I}=\int_{a}^{b} \frac{f(a+b-x)}{f(a+b-x)+f(x)} d x        ................(ii)

Add (i) and (ii)

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

\begin{aligned} &2 \mathrm{l}=[x]_{a}^{b} \\\\ &2 \mathrm{l}=\mathrm{b}-\mathrm{a} \\\\ &\mathrm{I}=\frac{b-a}{2} \end{aligned}

 

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