Question Number 259 by a@b.c last updated on 25/Jan/15
![∫_a ^b ((f(x))/(f(x)+f(a+b−x)))dx=](https://www.tinkutara.com/question/Q259.png)
$$\int_{{a}} ^{{b}} \:\frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left({a}+{b}−{x}\right)}{dx}= \\ $$
Answered by prakash jain last updated on 17/Dec/14
![Substitue x=a+b−y ⇒dx=−dy The given integral I I=∫_b ^a ((f(a+b−y))/(f(a+b−y)+f(y)))(−dy)=∫_b ^a ((f(a+b−y))/(f(a+b−y)+f(y)))dy =∫_a ^b ((f(a+b−y))/(f(a+b−y)+f(y)))dy=∫_a ^b ((f(a+b−x))/(f(a+b−x)+f(x)))dx 2I=∫_a ^b ((f(a+b−x)+f(x))/(f(a+b−x)+f(x)))dx=∫_a ^b 1∙dx=b−a I=((b−a)/2)](https://www.tinkutara.com/question/Q262.png)
$$\mathrm{Substitue}\:{x}={a}+{b}−{y}\:\Rightarrow{dx}=−{dy} \\ $$$$\mathrm{The}\:\mathrm{given}\:\mathrm{integral}\:{I} \\ $$$${I}=\int_{{b}} ^{{a}} \:\frac{{f}\left({a}+{b}−{y}\right)}{{f}\left({a}+{b}−{y}\right)+{f}\left({y}\right)}\left(−{dy}\right)=\int_{{b}} ^{{a}} \:\frac{{f}\left({a}+{b}−{y}\right)}{{f}\left({a}+{b}−{y}\right)+{f}\left({y}\right)}{dy} \\ $$$$=\int_{{a}} ^{{b}} \:\frac{{f}\left({a}+{b}−{y}\right)}{{f}\left({a}+{b}−{y}\right)+{f}\left({y}\right)}{dy}=\int_{{a}} ^{{b}} \:\frac{{f}\left({a}+{b}−{x}\right)}{{f}\left({a}+{b}−{x}\right)+{f}\left({x}\right)}{dx} \\ $$$$\mathrm{2}{I}=\int_{{a}} ^{{b}} \:\frac{{f}\left({a}+{b}−{x}\right)+{f}\left({x}\right)}{{f}\left({a}+{b}−{x}\right)+{f}\left({x}\right)}{dx}=\int_{{a}} ^{{b}} \mathrm{1}\centerdot{dx}={b}−{a} \\ $$$${I}=\frac{{b}−{a}}{\mathrm{2}} \\ $$