How would I split the two absolute values of this double integral?
$\int_{-1}^1\int_{-1}^1(|x|+|y|)dxdy$
The answer key shows the integral = 2 with work:
$\int_{-1}^1(|x|+|y|)dx$
split into
$\int_{-1}^0((-x)-y)dx$ + $\int_{0}^1((x)-y)dx$
but why does y become -y instead of staying |y| inside the dx integral? Could I not carry |y| over and split it inside the dy integral?
Best Answer
Note that \begin{eqnarray*} \int_{-1}^{1}\int_{-1}^{1} |x|+|y|dxdy&=&\int_{-1}^{1} \left(\int_{-1}^{1}|x|dx+\int_{-1}^{1}|y|dx \right)dy\\ &=&\int_{-1}^{1} \left(2\int_{0}^{1}|x|dx+\int_{-1}^{1}|y|dx \right)dy\\ &=&\int_{-1}^{1}1+2|y|dy\\ &=&\boxed{4} \end{eqnarray*}
Explication:
We need to calculate $$\int_{-1}^{1}1+2|y|dy$$ so we can see by linearity, $$\int_{-1}^{1}1+2|y|dy=\int_{-1}^{1}1dy+\int_{-1}^{1}|y|dy$$ Now, we can see that $|y|$ is an even function and interval $[-1,1]$ ($f$ is even par function if $f(-x)=f(x)$ for all $x$) and is symmetric about $0$, so $$\int_{-1}^{1}|y|dy=2\int_{0}^{1}|y|dy$$ and since that $y$ is always positive when $0<x<1$, so we can to make $|y|=y$, and also $$2\int_{0}^{1}|y|dy=2\int_{0}^{1}ydy=2\left.\left(\frac{y^{2}}{2} \right)\right|_{0}^{1}=2$$so, we can see that $$\int_{-1}^{1}1+2|y|dy=2+\int_{-1}^{1}dy=4$$
Similarly we can see that $$\int_{-1}^{1}|x|+|y|dx=1+2|y|$$