Question Number 39342 by rahul 19 last updated on 05/Jul/18
$$\mathrm{Let}\:\mathrm{f}\left({x}\right)\:=\:\int_{\mathrm{0}\:} ^{\mathrm{2}} \:\mid{x}−{t}\mid\:\mathrm{dt}\:\left({x}>\mathrm{0}\right)\:,\:\mathrm{then} \\ $$$$\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left({x}\right)\:\mathrm{is}\:? \\ $$
Answered by MrW3 last updated on 05/Jul/18
![for 0<x≤2: f(x) = ∫_(0 ) ^2 ∣x−t∣ dt =∫_(0 ) ^x (x−t)dt+∫_x ^2 (x−t)dt =[xt−(t^2 /2)]_0 ^x −[xt−(t^2 /2)]_x ^2 =(x^2 −(x^2 /2))−(2x−2−x^2 +(x^2 /2)) =x^2 −2x+2 =(x−1)^2 +1≥1 for 2≤x: f(x) = ∫_(0 ) ^2 ∣x−t∣ dt = ∫_(0 ) ^2 (x−t) dt =[xt−(t^2 /2)]_0 ^2 =2(x−1)≥2 min. f(x)=1 at x=1.](https://www.tinkutara.com/question/Q39363.png)
$${for}\:\mathrm{0}<{x}\leqslant\mathrm{2}: \\ $$$$\mathrm{f}\left({x}\right)\:=\:\int_{\mathrm{0}\:} ^{\mathrm{2}} \:\mid{x}−{t}\mid\:\mathrm{dt} \\ $$$$\:=\int_{\mathrm{0}\:} ^{{x}} \:\left({x}−{t}\right)\mathrm{dt}+\int_{{x}} ^{\mathrm{2}} \left({x}−{t}\right){dt} \\ $$$$=\left[{xt}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{{x}} −\left[{xt}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{{x}} ^{\mathrm{2}} \\ $$$$=\left({x}^{\mathrm{2}} −\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)−\left(\mathrm{2}{x}−\mathrm{2}−{x}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2} \\ $$$$=\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\geqslant\mathrm{1} \\ $$$$ \\ $$$${for}\:\mathrm{2}\leqslant{x}: \\ $$$$\mathrm{f}\left({x}\right)\:=\:\int_{\mathrm{0}\:} ^{\mathrm{2}} \:\mid{x}−{t}\mid\:\mathrm{dt} \\ $$$$=\:\int_{\mathrm{0}\:} ^{\mathrm{2}} \:\left({x}−{t}\right)\:\mathrm{dt} \\ $$$$=\left[{xt}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=\mathrm{2}\left({x}−\mathrm{1}\right)\geqslant\mathrm{2} \\ $$$$ \\ $$$${min}.\:{f}\left({x}\right)=\mathrm{1}\:{at}\:{x}=\mathrm{1}. \\ $$
Commented by rahul 19 last updated on 05/Jul/18
Thank you sir ! ����