Question Number 155488 by SANOGO last updated on 01/Oct/21
Answered by puissant last updated on 01/Oct/21
$${Q}=\int_{{a}} ^{{b}} \frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left({a}+{b}−{x}\right)}{dx}\:\:\left(\mathrm{1}\right) \\ $$$${u}={a}+{b}−{x}\:\rightarrow\:{x}={a}+{b}−{u}\:\rightarrow\:{dx}=−{du} \\ $$$$\Rightarrow\:{Q}=\int_{{b}} ^{{a}} \frac{{f}\left({a}+{b}−{u}\right)}{{f}\left({a}+{b}−{u}\right)+{f}\left({u}\right)}\left(−{du}\right) \\ $$$$=\int_{{a}} ^{{b}} \frac{{f}\left({a}+{b}−{u}\right)}{{f}\left({u}\right)+{f}\left({a}+{b}−{u}\right)}{du}\:\:\left(\mathrm{2}\right) \\ $$$$\frac{\left(\mathrm{1}\right)+\left(\mathrm{2}\right)}{\mathrm{2}}\:\Rightarrow\:{Q}=\frac{\mathrm{1}}{\mathrm{2}}\int_{{a}} ^{{b}} \frac{{f}\left({x}\right)+{f}\left({a}+{b}−{x}\right)}{{f}\left({x}\right)+{f}\left({a}+{b}−{x}\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{{a}} ^{{b}} {dx}=\frac{{b}−{a}}{\mathrm{2}} \\ $$$${Divers}\::\:\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({u}\right){du}=….. \\ $$$$\:\because\therefore{Q}=\int_{{a}} ^{{b}} \frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left({a}+{b}−{x}\right)}{dx}\:=\:\frac{{b}−{a}}{\mathrm{2}} \\ $$
Commented by peter frank last updated on 01/Oct/21
$$\mathrm{good} \\ $$
Commented by SANOGO last updated on 01/Oct/21
$${merci}\:{le}\:{general} \\ $$