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Question Number 42410 by soufiane zarik last updated on 25/Aug/18

∫_( 0) ^(2a)   ((f(x))/(f(x)+f(2a−x))) dx =

$$\underset{\:\mathrm{0}} {\overset{\mathrm{2}{a}} {\int}}\:\:\frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left(\mathrm{2}{a}−{x}\right)}\:{dx}\:= \\ $$

Commented by maxmathsup by imad last updated on 25/Aug/18

let A = ∫_0 ^(2a)     ((f(x))/(f(x) +f(2a−x)))dx   changement 2a−x =t give  A =∫_0 ^(2a)       ((f(2a−t))/(f(2a−t) +f(t)))dt  ⇒ 2A = ∫_0 ^(2a)     ((f(x))/(f(x)+f(2a−x)))dx +∫_0 ^(2π)    ((f(2a−x))/(f(2a−x) +f(x)))dx  =∫_0 ^(2a)    ((f(x)+f(2a−x))/(f(2a−x) +f(x)))dx =∫_0 ^(2a)  dx =2a ⇒ A =a.

$${let}\:{A}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} \:\:\:\:\frac{{f}\left({x}\right)}{{f}\left({x}\right)\:+{f}\left(\mathrm{2}{a}−{x}\right)}{dx}\:\:\:{changement}\:\mathrm{2}{a}−{x}\:={t}\:{give} \\ $$$${A}\:=\int_{\mathrm{0}} ^{\mathrm{2}{a}} \:\:\:\:\:\:\frac{{f}\left(\mathrm{2}{a}−{t}\right)}{{f}\left(\mathrm{2}{a}−{t}\right)\:+{f}\left({t}\right)}{dt}\:\:\Rightarrow\:\mathrm{2}{A}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} \:\:\:\:\frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left(\mathrm{2}{a}−{x}\right)}{dx}\:+\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{f}\left(\mathrm{2}{a}−{x}\right)}{{f}\left(\mathrm{2}{a}−{x}\right)\:+{f}\left({x}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}{a}} \:\:\:\frac{{f}\left({x}\right)+{f}\left(\mathrm{2}{a}−{x}\right)}{{f}\left(\mathrm{2}{a}−{x}\right)\:+{f}\left({x}\right)}{dx}\:=\int_{\mathrm{0}} ^{\mathrm{2}{a}} \:{dx}\:=\mathrm{2}{a}\:\Rightarrow\:{A}\:={a}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Aug/18

∫_a ^b f(x)dx=∫_a ^b f(a+b−x)dx formula  I=∫_0 ^(2a) ((f(x))/(f(x)+f(2a−x)))dx  =∫_0 ^(2a) ((f(2a−x))/(f(2a−x)+f(x)))dx  adding  2I=∫_0 ^(2a) 1.dx  2I=2a  I=a

$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx}\:{formula} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{2}{a}} \frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left(\mathrm{2}{a}−{x}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}{a}} \frac{{f}\left(\mathrm{2}{a}−{x}\right)}{{f}\left(\mathrm{2}{a}−{x}\right)+{f}\left({x}\right)}{dx} \\ $$$${adding}\:\:\mathrm{2}{I}=\int_{\mathrm{0}} ^{\mathrm{2}{a}} \mathrm{1}.{dx} \\ $$$$\mathrm{2}{I}=\mathrm{2}{a} \\ $$$${I}={a} \\ $$$$ \\ $$

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