How do you break two absolute values inside double integral?

Question

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?

Answer 1

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*}

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Explication:

1. 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 $$\int_{-1}^{1}1+2|y|dy=2+\int_{-1}^{1}dy=4$$

2. Similarly we can see that $$\int_{-1}^{1}|x|+|y|dx=1+2|y|$$

Answer 2

The area of integration is a square of side length $2$ centred at the origin and with sides parallel to the axes. $$|x|+|y|=\begin{cases}x+y,&0\le x,y\le1\text{ (Ist quadrant)}\\-x+y,&0\le -x,y\le1\text{ (IInd quadrant)}\\-x-y,&0\le-x,-y\le1\text{ (IIIrd quadrant)}\\x-y,&0\le x,-y\le1\text{ (IVth quadrant)}\end{cases}$$and hence the integral is split into four integrals, one over each quadrant.

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Note that what you propose is also correct. You can chose to retain $|y|$ as is and split it up in the $dy$ integral. $$\int_{-1}^1(|x|+|y|)dx=2|y|+\int_{-1}^1|x|dx=2|y|+1$$so$$I=\int_{-1}^1(2|y|+1)dy=4$$

Answer 3

You could use the fact that $$ \int |x|\,dx=\frac{1}{2}x|x|+c $$

Then

\begin{eqnarray} \int_{-1}^{1}\int_{-1}^{1} |x|+|y|\,dx\,dy&=&\int_{-1}^{1}\left[\frac{1}{2}x|x|+x|y|\right]_{-1}^1\,dx\,dy\\ &=&\int_{-1}^{1}1+2|y|\,dy\\ &=&\left[\,y+y|y|\,\right]_{-1}^1\\ &=&4 \end{eqnarray}

Answer 4

In most cases with some function which changes definition as we vary the domain(*), we integrate it by splitting the domain into fragments where a certain definition of the function pertains and integrate over those partitions.

You could actually carry the |y| and split it after that, I demonstrate as follows:

$$ \int _{-1}^1 \int_{-1}^1 |x| dx dy + \int _{-1}^1 \int_{-1}^1 |y| dx dy$$

Now actually, there is a trick, both the integrals are exactly the same but with different variables, so we actually only have to think of one and twice the answer to it. Let's take twice the one on the left,

$$ 2 \int _{-1}^1 \int_{-1}^1 |x| dx dy $$

Now, The easy part is to integrate with $y$ first:

$$ 2 \int_{-1}^1 \int_{-1}^1 |x| dy dx$$

Since the bounds aren't dependent on $x$ , i.e: we are integrating over a rectangle, $|x|$ is simply a constant and hence:

$$ 2 \int_{-1}^1 [y]_{-1}^1 |x| dx$$

Or,

$$ 4 \int_{-1}^1 |x| dx$$

Now this is integral is where the first part of my answer comes in:

$$ 4 [ \int_0^1 |x| dx + \int_{-1}^0 |x| dx]$$

Now, for the first integral, we can simply remove that modulus, and the second we negate because the domain is negative:

$$ 4 [ \int_0^1 x dx + \int_{-1}^0 -x dx]$$

Or,

$$ 4 [ [\frac{x^2}{2}]_{0}^1 - [\frac{x^2}{2}]_{-1}^0]$$

This evaluates to :

$$ 4$$

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*:Usually, functions are consistent rules like $x^3$ where we cube the quantity given or more generally do some algorithm of calculations on the number to get our answer.

However, there is certain function who has their rule/ algorithm to evaluate changing as we move across domains. For example, the modulus function has a rule that we should take negative for $x<0$ and positive for $x>0$. Ultimately these two rules form one rule which has two specific cases.

Anyways the point I'm trying to convey is that split the domain such that we can easily evaluate the quantity for each subrule