Minimum value of $ f(x)= \int_{0}^1 |t-x|^3dt $

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

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

Then $f$ has a minimum value for $x$ equal to:

The correct answer should be $x=\cfrac 12$

I don't know how to approach this because I haven't encountered problems like this before.

If you know where I can find similar problems to practice, let me know. Thank you very much.

Answer 1

As a general strategy: remember how you do optimization (i.e. min/max) questions from early calculus. The big idea is to find a critical number of $f$, which mostly means to take a derivative of $f$, set it to 0, and solve. (You then have to ensure that you actually found the thing you want, which in this case would be a global minimum.)

So, that's an outline of your strategy here. You should be able to convince yourself that the optimizing $x$ must be in $[0, 1]$, and some argument for why that's true would probably be required. But once you've established that, consider breaking up the integral into two pieces: $$f(x) = \int_0^1 |t - x|^3 \, \textrm d t = \int_0^x |t - x|^3 \, \textrm d t + \int_x^1 | t - x|^3 \, \textrm d t$$ Then, consider what $|t-x|$ would mean on those respective integrals (i.e. rewrite each without absolute values), and do the integration. At that point, you'll have a tidy algebraic expression for $f(x)$, and you can use the strategy I outlined above.

Answer 2

Hint: $$\int_0^x (x - t)^3 dt + \int_x^1 (t-x)^3 dt$$

Notice that for minima $x \in [0,1]$

Answer 3

If $0\le x\le 1$

$\int_0^1 |t-x|^3 dt = \int_0^x (x-t)^3 dt + \int_x^1 (t-x)^3 dt$

Otherwise $\int_0^1 |t-x|^3 dt = |\int_0^1 (t-x)^3 dt|$

You might notice a symmetry about the line $x = \frac 12,$ which would suggest that as special point to consider.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

I won't assume, at the very beginning, what is the interval of 'interest'. I prefer to find it "along the way".

Namely, \begin{align} \mrm{f}\pars{x} & \equiv \int_{0}^{1}\verts{t - x}^{3}\,\dd t = \bracks{x < 0}\int_{0}^{1}\pars{t - x}^{3}\,\dd t \\[2mm] & +\bracks{0 \leq x < 1}\bracks{\int_{0}^{x}\pars{x - t}^{3}\,\dd t + \int_{x}^{1}\pars{t - x}^{3}\,\dd t} + \bracks{x \geq 1}\int_{0}^{1}\pars{x - t}^{3}\,\dd t \\[5mm] & = \overbrace{\bracks{x < 0}% {4\verts{x}^{3} + 6\verts{x}^{2} + 4\verts{x} + 1 \over 4}} ^{\ds{\mbox{There's not any minimium !!!}}}\ +\ \overbrace{\bracks{0 \leq x < 1}{2x^{4} - 4x^{3} + 6x^{2} -4x + 1 \over 4}} ^{\ds{\mbox{Minimum value is}\ \color{red}{1 \over 8}\ \mbox{at}\ x = \color{red}{1 \over 2}}} \\[2mm] & +\ \underbrace{\bracks{x \geq 1}{4x^{3} - 6x^{2} + 4x + 1 \over 4}} _{\ds{\mbox{Minimum value is}\ 3\ \mbox{at}\ x = 1}}\ \implies\ \bbx{\mbox{Minimum value is}\ \color{red}{1 \over 8}\ \mbox{at}\ x = \color{red}{1 \over 2}} \end{align}