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Question Number 39342 by rahul 19 last updated on 05/Jul/18

$$\mathrm{Let}\mathrm{f}\left(x\right)={\int }_0^2\mid x-t\mid \mathrm{dt}(x>0),\mathrm{then}$$$$\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{f}\left(x\right)\mathrm{is}?$$

Answered by MrW3 last updated on 05/Jul/18

$$for0<x⩽2:$$$$\mathrm{f}\left(x\right)={\int }_0^2\mid x-t\mid \mathrm{dt}$$$$={\int }_0^x(x-t)\mathrm{dt}+{\int }_x^2(x-t)dt$$$$={[xt-\frac{t^2}{2}]}_0^x-{[xt-\frac{t^2}{2}]}_x^2$$$$=(x^2-\frac{x^2}{2})-(2x-2-x^2+\frac{x^2}{2})$$$$=x^2-2x+2$$$$={(x-1)}^2+1⩾1$$$$$$$$for2⩽x:$$$$\mathrm{f}\left(x\right)={\int }_0^2\mid x-t\mid \mathrm{dt}$$$$={\int }_0^2(x-t)\mathrm{dt}$$$$={[xt-\frac{t^2}{2}]}_0^2$$$$=2(x-1)⩾2$$$$$$$$min.f\left(x\right)=1atx=1.$$

Commented by rahul 19 last updated on 05/Jul/18

Thank you sir ! ����

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