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

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