Double integral of a piecewise function over a rectangle?

Question

Let $f$ be defined on the rectangle $R=[1,2] \times [2,4]$ as follows:

$$ f(x,y) = \begin{cases} (x+y)^{-2}, & \text{if }x\leq y \leq 2x; > \\\\ 0, & \text{otherwise. } \end{cases} $$

Compute the value of the double integral $\int\int_Rf$.

I think since $f=0$ in the second case, I should just ignore it.

However, what confuses me is the condition in the first case. The interval of $y$ is not $[x,2x]$ for all values of $x$. Since $2\leq y \leq 4$, this condition implies that $x=2$ for this interval. But does that mean I ignore $x\in [1,2)$? I'm just not sure how to begin this problem.

score 1 · Accepted Answer · answered Apr 28 '14 at 15:44

1

The region you want is the common interior (intersection) of the following six halfplanes:

$x\ge 1$
$x\le 2$
$y\ge 2$
$y\le 4$
$y\ge x$
$y\le 2x$

Try sketching all six and see what shape you get.

answered Apr 28 '14 at 15:44

vadim123

82,796

A right triangle? Hmm... do I integrate the first case over the interval $1\leq x \leq 2$ and $2 \leq y \leq 2x$? – Bobby Lee Apr 28 '14 at 15:53
You should set up your integral to be exactly over the shape you get. – vadim123 Apr 28 '14 at 16:14

Double integral of a piecewise function over a rectangle?

1 Answers1