We have $f:RR->RR;f(a)=int_0^1abs(x-a)dx$ .What is minimum value of the function $f$ ?

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Answer 1 · 2017-05-13T21:16:47

Please see the other correct answer by Cesareo.

Answer 2 · 2017-05-13T21:25:43

$f(1/2)=1/4$

With a change of variable $y = x-a$

$int_0^1 abs(x-a)dx equiv int_(-a)^(1-a) abs y dy = 1/2((1-a)^2"sign"(1-a)-a^2"sign"(-a)) = 1/2((1-a)^2"sign"(1-a)+a^2 "sign"(a))$

The minimum is located at $a=1/2$ and $f(1/2)=1/4$

Attached a plot of $f(a)$

enter image source here

We have f:RR->RR;f(a)=int_0^1abs(x-a)dxf:RR->RR;f(a)=int_0^1abs(x-a)dx.What is minimum value of the function ff?