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Here is the question.

I put the equation in the following double integral:

$$\int_{0}^{1} \int_{-1}^{1} |x + y|\,\mathrm dy\,\mathrm dx$$

I know you can break up the absolute function into the following. I'm not really sure what to do next. I looked at this link but I don't know when x + y is nonnegative and when it's negative as we need the values of both x and y and when calculating the inner integral we're only doing it with respect to y. So I'm not exactly sure how to determine when |x + y| is nonnegative and when it is negative. $$ |\,x + y\,| = \left\{ \begin{array}{lr} x + y & : x + y >0 \Leftrightarrow y > -x \\ -(x + y) & : x + y < 0 \Leftrightarrow y < -x \end{array} \right. $$

However, I came up with a "solution" which is quite silly but I thought it might work.

$$\int_{0}^{1} | \int_{-1}^{1} x \, dy \, + \int_{-1}^{1} y \,dy\, | = \int_{0}^{1} |\, 2x\, | = 1 $$

I'm lost and any help would be greately appreciated.

Community
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user2635911
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2 Answers2

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Hint, applying your formula for $|x+y|$ gives: $$\int_{0}^{1} \int_{-1}^{1} |x + y|\,dy\,dx = \int_{0}^{1} \int_{-1}^{-x} [-(x + y)]\,dy\,dx + \int_{0}^{1} \int_{-x}^{1} (x + y)\,dy\,dx $$

p.s.
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1\,\right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{{\cal I} \equiv \int_{0}^{1}\int_{-1}^{1}\verts{x + y}\,\dd y\,\dd x = \int_{0}^{1}{\cal F}\pars{x}\,\dd x\quad\mbox{where}\quad {\cal F}\pars{x} \equiv \int_{-1}^{1}\verts{x + y}\,\dd y}$

\begin{align} {\cal F}\pars{x} &= \left.\vphantom{\LARGE A}\verts{x + y}\,y\,\right\vert_{y\ =\ -1}^{y\ =\ 1} - \int_{-1}^{1}y\sgn\pars{x + y}\,\dd y \\[3mm]&= \verts{x + 1} + \verts{x - 1} -\bracks{% \left.\vphantom{\LARGE A}\sgn\pars{x + y}\,{y^{2} \over 2} \right\vert_{y\ =\ -1}^{y\ =\ 1}} + \int_{-1}^{1}{y^{2} \over 2}\bracks{2\delta\pars{x + y}}\,\dd y \\[3mm]&= \verts{x + 1} + \verts{x - 1} - {1 \over 2}\,\sgn\pars{x + 1} + {1 \over 2}\,\sgn\pars{x - 1} + x^{2}\Theta\pars{1 - \verts{x}} \end{align}

$$ {\cal F}\pars{x} = \verts{x + 1} + \verts{x - 1} - {1 \over 2}\,\sgn\pars{x + 1} + {1 \over 2}\,\sgn\pars{x - 1} + x^{2}\Theta\pars{1 - \verts{x}} $$

\begin{align} {\cal I}&= \int_{1}^{2}{1 \over 2}\,\verts{x}\,\dd x + \int_{-1}^{0}{1 \over 2}\,\verts{x}\,\dd x -\int_{1}^{2}\sgn\pars{x}\,\dd x + \int_{-1}^{0}\sgn\pars{x}\,\dd x + \int_{0}^{1}x^{2}\,\dd x \\[3mm]&= \left.{x^{2} \over 4}\right\vert_{1}^{2} - \left.{x^{2} \over 4}\right\vert_{-1}^{0} - \left.x\right\vert_{1}^{2} + \left.\pars{-x}\right\vert_{-1}^{0} + \left.{x^{3} \over 3}\right\vert_{1}^{2} \\[3mm]&= \pars{1 - {1 \over 4}} - \pars{-\,{1 \over 4}} - 1 - 1 + \pars{{8 \over 3} - {1 \over 3}} = {4 \over 3} \end{align}

$$\color{#0000ff}{\large% \int_{0}^{1}\int_{-1}^{1}\verts{x + y}\,\dd y\,\dd x = {4 \over 3} } $$

Felix Marin
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