Question Number 138415 by tugu last updated on 13/Apr/21
$${A},{B}\:\in{R},\:\:{f}\left(\mathrm{1}\right)=\mathrm{0}\:,\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left({f}\left({x}\right)\right)^{\mathrm{2}} {dx}\:={A}\:{and}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{xf}\left({x}\right){dx}={B}\: \\ $$$${what}\:{is}\:{the}\:{integral}\:{value}\:{of}\:\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{xf}\left({x}\right)\left({f}\:'\left({x}\right)−\mathrm{1}\right){dx}\:{by}\:{using}\:{trrms}\:{of}\:{A}\:{and}\:{B}\:?\: \\ $$
Answered by Ar Brandon last updated on 13/Apr/21
![I=∫_0 ^1 xf(x)(f ′(x)−1)dx =∫_0 ^1 xf(x)f ′(x)dx−∫_0 ^1 xf(x)dx =[((x(f(x))^2 )/2)−(1/2)∫(f(x))^2 dx]_0 ^1 −B =(((f(1))^2 )/2)−(1/2)∫_0 ^1 (f(x))^2 dx−B =−(A/2)−B](https://www.tinkutara.com/question/Q138439.png)
$$\mathcal{I}=\int_{\mathrm{0}} ^{\mathrm{1}} {xf}\left({x}\right)\left({f}\:'\left({x}\right)−\mathrm{1}\right){dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} {xf}\left({x}\right){f}\:'\left({x}\right){dx}−\int_{\mathrm{0}} ^{\mathrm{1}} {xf}\left({x}\right){dx} \\ $$$$\:\:\:=\left[\frac{{x}\left({f}\left({x}\right)\right)^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int\left({f}\left({x}\right)\right)^{\mathrm{2}} {dx}\right]_{\mathrm{0}} ^{\mathrm{1}} −{B} \\ $$$$\:\:\:=\frac{\left({f}\left(\mathrm{1}\right)\right)^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({f}\left({x}\right)\right)^{\mathrm{2}} {dx}−{B} \\ $$$$\:\:\:=−\frac{{A}}{\mathrm{2}}−{B} \\ $$