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Question Number 57136 by mustakim420 last updated on 30/Mar/19

∫_a ^b   ((f(x))/(f(x)+f(a+b−x))) dx =

$$\underset{{a}} {\overset{{b}} {\int}}\:\:\frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left({a}+{b}−{x}\right)}\:{dx}\:= \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Mar/19

I=∫_a ^b ((f(x))/(f(x)+f(a+b−x)))dx  I=∫_a ^b ((f(a+b−x))/(f(a+b−x)+f(x)))dx  2I=∫_a ^b ((f(x)+f(a+b−x))/(f(x)+f(a+b−x)))dx  2I=∫_a ^b dx  2I=b−a  I=((b−a)/2)  note formula  ∫_a ^b f(x)dx=∫_a ^b f(a+b−x)dx

$${I}=\int_{{a}} ^{{b}} \frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left({a}+{b}−{x}\right)}{dx} \\ $$$${I}=\int_{{a}} ^{{b}} \frac{{f}\left({a}+{b}−{x}\right)}{{f}\left({a}+{b}−{x}\right)+{f}\left({x}\right)}{dx} \\ $$$$\mathrm{2}{I}=\int_{{a}} ^{{b}} \frac{{f}\left({x}\right)+{f}\left({a}+{b}−{x}\right)}{{f}\left({x}\right)+{f}\left({a}+{b}−{x}\right)}{dx} \\ $$$$\mathrm{2}{I}=\int_{{a}} ^{{b}} {dx} \\ $$$$\mathrm{2}{I}={b}−{a} \\ $$$${I}=\frac{{b}−{a}}{\mathrm{2}} \\ $$$${note}\:{formula} \\ $$$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx} \\ $$

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