Evaluating $\int_{-5}^5\int_{-5}^5-\frac{3}{2}|x+y|-\frac{3}{2}|x-y|+15\,\mathrm{d}x\,\mathrm{d}y$

Question

I'm always having the wrong result from the following: $$ \int_{-5}^5\int_{-5}^5-\frac{3}{2}|x+y|-\frac{3}{2}|x-y|+15\,\mathrm{d}x\,\mathrm{d}y $$

I would really appreciate some guidance on how to go about this.

Answer 1

There are several symmetries that you can take advantage of here. First of all, we can integrate $$\int_{-5}^5 \int_{-5}^5 15 \:\mathrm{d}x \:\mathrm{d}y = 15 \cdot (10\cdot 10) = 1500.$$ The $(10\cdot 10)$ represented the area of the square $[-5,5]\times [-5,5]$.

As for what's left, $f(x,y) =|x+y|+|x-y|$ has certain symmetries. First of all $f(x,-y) = |x-y| + |x+y| = f(x,y)$, and $f(-x,y) = |-x+y| + |-x-y| = |x-y| + |x+y| = f(x,y)$. Therefore, the inner integral can be rewritten as $$(-3/2) \int_{-5}^5 |x+y| + |x-y| \:\mathrm{d}x = 2 (-3/2) \int_0^5 |x+y| + |x-y| \:\mathrm{d}x$$ using the symmetries in $x$. The outer integral can be treated the same way using the evenness in $y$.

Therefore, $$(-3/2)\int_{-5}^5 \int_{-5}^5 |x+y| + |x-y| \:\mathrm{d}x \:\mathrm{d}y = 4(-3/2)\int_{0}^5 \int_{0}^5 |x+y| + |x-y| \:\mathrm{d}x \:\mathrm{d}y.$$

Now for $x,y$ in $[0,5]$, $|x+y| = x+y$. Then we have to find a way to remove the absolute value sign on $|x-y|$. This can be handled by spliting the integration for $x$ into to integrals:

$$4(-3/2)\int_{0}^5 \int_{0}^5 |x+y| + |x-y| \:\mathrm{d}x \:\mathrm{d}y = 4(-3/2)\int_{0}^5 \int_{0}^y (x+y) + (y-x) \:\mathrm{d}x \:\mathrm{d}y + 4(-3/2)\int_{0}^5 \int_{y}^5 (x+y) + (x-y) \:\mathrm{d}x \:\mathrm{d}y$$

In the first term, $x <y$, so $|x-y|$ becomes $(y-x)$ since it must be positive. The second term we have $x>y$ so $|x-y| = x-y$.

Thus we have broken down the integral to:

$$1500 + 4(-3/2)\left(\int_{0}^5 \int_{0}^y 2y \:\mathrm{d}x \:\mathrm{d}y + \int_{0}^5 \int_{y}^5 2x \:\mathrm{d}x \:\mathrm{d}y\right).$$

This should be straight forward to integrate.

Answer 2

You are integrating over a $10 \times 10$ square centered on the origin. If you draw the diagonal from $(-5,-5)$ to $(5,5)$ you have $x-y \gt 0$ in the region above the diagonal and $x-y \lt 0$ in the region below. If you draw the other diagonal it cuts the square into a region where $x+y \gt 0$ and another where $x+y \lt 0$. If you break your integral into four pieces, one covering each quarter of the square, you know what the signs are inside the absolute value signs in each quarter. Then eliminate the absolute value signs and integrate normally.