-2

\begin{align} \int_{-1}^1\int_{-1}^1 \max(x,y)\,dx \,dy \end{align} How can I calculate this double integral?

Help me please.

Thomas
  • 4,363
Edward Puente De La Vega
  • 15
  • 2
    Any ideas so far? – Thomas Nov 11 '14 at 13:46
  • 4
    i see at least two typos in your question, one is in the title, one in the question. Also, you posted no ideas or work you did on your own, strongly indicating this is some sort of "solve my homework for me" question. – 5xum Nov 11 '14 at 13:48

2 Answers2

1

Define \begin{align} I=\int_{-1}^1\int_{-1}^1 \max(x,y)\,dx \,dy \end{align} and we want to calculate $I$.

Note that we can split the integral in the following way: \begin{align} I=\int_{-1}^1\int_{-1}^1 \max(x,y)\,dx \,dy =\int_{-1}^1\Bigl(\int_{-1}^y \max(x,y)\,dx +\int_{y}^1 \max(x,y)\,dx\Bigr)\,dy \end{align} Now for the first part, we note, that $x$ runs from $-1$ to $y$ and is therefore always smaller than $y$. For the second part, $x$ runs from $y$ to $1$ and is therefore always greater than $y$. We can then rewrite the expression as \begin{align} I&=\int_{-1}^1\int_{-1}^1 \max(x,y)\,dx \,dy =\int_{-1}^1\Bigl(\int_{-1}^y \max(x,y)\,dx +\int_{y}^1 \max(x,y)\,dx\Bigr)\,dy \\ &=\int_{-1}^1\Bigl(\int_{-1}^y y\,dx +\int_{y}^1 x\,dx\Bigr)\,dy =\int_{-1}^1 y\cdot x|_{-1}^y+\frac{x^2}{2}|_{y}^1 \,dy =\int_{-1}^1 y^2+y+\frac12-\frac{y^2}{2} \,dy \\ &=\int_{-1}^1 \frac{y^2}{2} +y+\frac12\,dy = \Bigl(\frac{y^3}{6}+\frac{y^2}{2}+\frac{y}{2}\Bigr)|_{-1}^1=\frac{1}{6}+\frac12 +\frac12+\frac16-\frac12+\frac12\\ &=\frac43 \end{align}

However, I strongly encourage you to understand the splitting of the integral and replacing the $\max$ function with the corresponding value $x$ or $y$.

Thomas
  • 4,363
-1

Hint

$$\max\{x,y\}=\frac{|x+y|+|x-y|}{2}$$

Then $$\int_{-1}^1\int_{-1}^1\max\{x,y\}dxdy=\frac{1}{2}\int_{-1}^1\int_{-1}^1|x+y|dxdy+\frac{1}{2}\int_{-1}^1\int_{-1}^1|x-y|dxdy$$

idm
  • 11,824
  • I doubt this is helpful, since you face similar problems integrating the $|x\pm y|$ as when you integrate $\max(x,y)$. – Thomas Nov 11 '14 at 14:02
  • of course not, it's not complicate to integrate $$|x\pm y|.$$ you can use $$\iint |x-y|dxdy=\iint \sqrt{(x\pm y)}^2dxdy,$$ and it's easy to conclude :-) – idm Nov 11 '14 at 19:44