Question Number 261 by raj last updated on 25/Jan/15
$$\mathrm{If}\:\underset{\mathrm{0}} {\overset{{x}} {\int}}{f}\left({t}\right){dt}={x}+\underset{{x}} {\overset{\mathrm{1}} {\int}}{tf}\left({t}\right){dt},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:{f}\left(\mathrm{1}\right). \\ $$
Answered by prakash jain last updated on 17/Dec/14
![∫_0 ^x f(t)dt=x+∫_x ^1 tf(t)dt (d/dx)∫_0 ^x f(t)dt=(d/dx)[x+∫_x ^1 tf(t)dt] f(x)=1+0−xf(x) f(x)=(1/(1+x)) f(1)=(1/2)](https://www.tinkutara.com/question/Q270.png)
$$\int_{\mathrm{0}} ^{{x}} {f}\left({t}\right){dt}={x}+\int_{{x}} ^{\mathrm{1}} {tf}\left({t}\right){dt} \\ $$$$\frac{{d}}{{dx}}\int_{\mathrm{0}} ^{{x}} {f}\left({t}\right){dt}=\frac{{d}}{{dx}}\left[{x}+\int_{{x}} ^{\mathrm{1}} {tf}\left({t}\right){dt}\right] \\ $$$${f}\left({x}\right)=\mathrm{1}+\mathrm{0}−{xf}\left({x}\right) \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}} \\ $$$${f}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$