Question Number 54031 by qw last updated on 28/Jan/19
$$\mathrm{If}\:\:{f}\left({a}+{b}−{x}\right)=\:{f}\left({x}\right),\:\mathrm{then}\:\underset{{a}} {\overset{{b}} {\int}}\:{x}\:{f}\left({x}\right)\:{dx}\:= \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19
![I=∫_a ^b xf(x)dx I=∫_a ^b (a+b−x)f(a+b−x)dx I=∫_a ^b (a+b−x)f(x)dx[f(a+b−x)=f(x)] 2I=∫_a ^b xf(x)dx+∫_a ^b (a+b−x)f(x)dx 2I=∫_a ^b (a+b−x+x)f(x)dx I=((a+b)/2)∫_a ^b f(x)dx others pls check...](https://www.tinkutara.com/question/Q54037.png)
$${I}=\int_{{a}} ^{{b}} {xf}\left({x}\right){dx} \\ $$$${I}=\int_{{a}} ^{{b}} \left({a}+{b}−{x}\right){f}\left({a}+{b}−{x}\right){dx} \\ $$$${I}=\int_{{a}} ^{{b}} \left({a}+{b}−{x}\right){f}\left({x}\right){dx}\left[{f}\left({a}+{b}−{x}\right)={f}\left({x}\right)\right] \\ $$$$\mathrm{2}{I}=\int_{{a}} ^{{b}} {xf}\left({x}\right){dx}+\int_{{a}} ^{{b}} \left({a}+{b}−{x}\right){f}\left({x}\right){dx} \\ $$$$\mathrm{2}{I}=\int_{{a}} ^{{b}} \left({a}+{b}−{x}+{x}\right){f}\left({x}\right){dx} \\ $$$${I}=\frac{{a}+{b}}{\mathrm{2}}\int_{{a}} ^{{b}} {f}\left({x}\right){dx} \\ $$$${others}\:{pls}\:{check}… \\ $$
Commented by maxmathsup by imad last updated on 28/Jan/19
$${correct}\:{sir}\:{Tanmay}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19
$${thank}\:{you}\:{sir}… \\ $$