Minimum of parameter integral

Question

Find the minimum value of the function $f: R \to R, f(x)= \int_{0}^{1} {|x-t|}^3dt$

I computed the function analysing 3 cases: $x \leq 1, x \in (0,1), x \geq 1 $ And then i studied the extrem with derivatives obtaining that f has $1/2$ as a minimum point. I am interested if there is a shorter solution without too much computation.

score 1 · Answer 1 · 2019-03-15T21:08:55.253

1

Hint:

$f'(x)=\int_0^1 \frac{d{|x-t|^3}}{dx}dt=\int_0^13(x-t)^2\text{sign}(x-t)dt=\\ =-(x-t)^3\text{sign}(x-t)\Big|^1_0=x^3\text{sign}(x)-(x-1)^3\text{sign}(x-1)$

edited Mar 15 '19 at 21:08

answered Mar 15 '19 at 20:05

John Wayland Bales · Answer 2 · 2019-03-15T20:56:26.393

1

Since $f(x)= \int_{0}^{1} {|x-t|}^3dt$ it follows that

\begin{eqnarray} f^\prime(x)&=&\int_0^1\frac{d}{dx}(|x-t|^3)\,dt\\ &=&\int_0^13(x-t)|x-t|\,dt\\ &=&-\left[|x-t|^3\right]_0^1\\ &=&-\left[(x-t)^2|x-t|\right]_0^1\\ &=&x^2|x|-(x-1)^2|x-1| \end{eqnarray}

which is an increasing function and has no minimum.

edited Mar 15 '19 at 20:56

answered Mar 15 '19 at 20:07

John Wayland Bales

21,814

The theorem is the Leibnitz Integral Rule. And yes, the derivative for absolute value is undefined where it vanishes. – John Wayland Bales Mar 16 '19 at 18:13
Actually that modulus is derivable everywhere because it is cubed. If it was simple modulus it wouldn;t be derivable . My error. – amarius8312 Mar 16 '19 at 21:23
@amarius8312 That is correct. $|x|^\prime=\dfrac{x}{|x|}$ and we have $(|x|^n)^\prime=nx|x|^{n-2}$. – John Wayland Bales Mar 16 '19 at 22:04

Minimum of parameter integral

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