0

I'm given a function that's piece-wise defined for $x \leq y^2$ and "otherwise".

I.e.

$$f(x,y) = \begin{cases} x+y^3 & x \leq y^2 \\ 2xy^2 +1 & \text{otherwise} \end{cases}$$

How do I form the integration bounds using this information?

mavavilj
  • 7,270
  • Can you describe more precisely the function ? What is $x$ what is $y$ ? Is it a funtion $\mathbb{R}^2 \rightarrow \mathbb{R}$ ? – Zubzub Feb 03 '17 at 09:17
  • This is a function whose domain is $\Bbb{R}^2$. Under which region do you want to integrate it? The question is meaningless if you don't specify it (anyway, the integral over $\Bbb{R}^2$ is clearly divergent since this function is not bounded above). – Crostul Feb 03 '17 at 09:31

1 Answers1

1

Assume $$f(x)=\begin{cases}g(x),& \text{if}\ x\leq y^2\\h(x),& \text{if}\ x> y^2\end{cases}$$ Then you can compute \begin{align}\iint_Df(x,y)dxdy=\iint_{D\cap\{x\leq y^2\}}f(x,y)dxdy+\iint_{D\cap\{x> y^2\}}f(x,y)dxdy\\=\iint_{D\cap\{x\leq y^2\}}g(x,y)dxdy+\iint_{D\cap\{x>y^2\}}h(x,y)dxdy. \end{align} Of course this might not be an easy task, but at least you are now dealing with a non-piecewise defined function.

b00n heT
  • 16,360
  • 1
  • 36
  • 46
  • And what are the bounds of the double integrals precisely? I.e. what are the bounds of $x$ and $y$? – mavavilj Feb 03 '17 at 09:23
  • It depends on the domain $D$ you want to integrate over. If you want to integrate over the whole plane $\mathbb{R}^2$ then for the first integral you have $$\int_{-\infty}^{\infty}\int_{-\infty}^{y^2}g(x,y)dxdy$$ and for the second $$\int_{-\infty}^{\infty}\int_{y^2}^{\infty}h(x,y)dxdy$$ – b00n heT Feb 03 '17 at 09:25
  • So if one considers $[0,1] \times [0,1]$, then the bounds would be $\int_0^1 \int_0^{y^2}$ and so on? – mavavilj Feb 03 '17 at 09:27
  • That's it! Drawing a picture might be of help for you, to understand what is going on :) – b00n heT Feb 03 '17 at 09:29