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I know that you can define the absolute value of a as: |a| = a when a>0 and |a| = -a when a<0

At first sight I thought this function would always evaluate to a positive value; however, after analysing it correctly, there is a small interval of values for which the result will be negative, so the question is, how do I define the proper interval for which I have to integrate negative values...? It's not clear as in some other functions I've worked with, in which by just looking at them I can tell in which value the things will start to change...

I think I have to divide this integral in 3 sub integrals...

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up vote 4 down vote accepted

$$3x^2-x=3x\left(x-\frac13\right)\ge 0\iff x\le 0\;\;or\;\;x\ge\frac13\implies$$

$$\int\limits_{-1}^1|3x^2-x|dx=\int\limits_{-1}^0(3x^2-x)dx+\int_0^{1/3}(-3x^2+x)dx+\int\limits_{1/3}^1(3x^2-x)dx\;\;\ldots$$

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Recall the formula for absolute value. It is $$|x|=\begin{cases}x, &x\ge 0\\ -x, & x\lt 0\\ \end{cases}$$ Therefore, you will need to find when $3x^2-x\lt 0$ and $3x^2-x\ge 0$. In your case, it will give you $3x^2-x\lt 0$ when $x\in (0,\frac13)$ and $3x^2-x\ge 0$ when $x\in [-1,0]$ and $x\in [\frac13, 1].$ Therefore your integral becomes $$\int_{-1}^1 |3x^2-x|dx = \int_{-1}^0 (3x^2-x)dx -\int_0^{\frac13} (3x^2-x)dx + \int_{\frac13}^1 (3x^2-x)dx$$

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\color{#00f}{\large\int_{-1}^{1}\verts{3x^{2} - x}\,\dd x}= \left. x\verts{3x^{2} - x}\vphantom{\huge A}\right\vert_{-1}^{1} -\int_{-1}^{1}x\sgn\pars{3x^{2} - x}\pars{6x - 1}\,\dd x \\[3mm]&=6-\int_{x = -1}^{x = 1}\sgn\pars{3x^{2} - x}\, \dd\pars{2x^{3} - {x^{2} \over 2}} \\[3mm]&=6 - \left.\sgn\pars{3x^{2} - x}\pars{2x^{3} - {x^{2} \over 2}} \right\vert_{-1}^{1} + \int_{-1}^{1}\pars{2x^{3} - {x^{2} \over 2}}2\delta\pars{3x^{2} - x}\pars{6x - 1} \,\dd x \\[3mm]&=2 + \int_{-1}^{1}\pars{4x^{3} - x^{2}}\pars{6x - 1} \bracks{\delta\pars{x} + \delta\pars{x - {1 \over 3}}}\,\dd x =2 + \left.\pars{4x^{3} - x^{2}}\pars{6x - 1}\right\vert_{x\ =\ 1/3} \\[3mm]&=2 + {1 \over 27} = \color{#00f}{\large{55 \over 27}} \end{align}

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