Sketch the region then change order of integration.

[never_lose]

Sketch the region \(\displaystyle \Omega\) and change the order of integration.

\(\displaystyle \int_{1}^{4}\int_{x}^{2x}f(x,y) dy dx\)

This is my sketch, except there should be vertical asymptotes at x = 1 and x = 4. \(\displaystyle \Omega\) would be the region between x = 1 and x = 4 and the two lines.

When I switch, y should be in terms of x... so I solved for x from these two equations: \(\displaystyle y = 2x\) and \(\displaystyle y = x\) (which are the equations for the red and blue lines)

\(\displaystyle x = \frac{y}{2}\) \(\displaystyle x = y\).

x should be expressed in terms of y so I use \(\displaystyle x = \frac{1}{2}\) \(\displaystyle x = y\).

I then get \(\displaystyle \int_{1}^{4}\int_{y}^{y/2}f(x,y) dx dy\).

But the answer should be \(\displaystyle \int_{1}^{2}\int_{1}^{y}f(x,y) dx dy \; + \; \int_{2}^{4}\int_{y/2}^{y}f(x,y) dx dy \; + \; \int_{4}^{8}\int_{y/2}^{4}f(x,y) dx dy\).

I'm not getting what is happening. How are the integrals getting split up into separate parts?

 

[galactus]

What is f(x,y)?. Is it given?. I see no reason why there should be asymptotes with two lines.

Did you inadvertently leave out f(x,y)?.

As it is, the area between y=2x and y=x from 1 to 4 would be 15/2.

I see no reason to use three separate integrals. There must be something I am missing.

\(\displaystyle \int_{1}^{4}xdx=\frac{15}{2}\)

\(\displaystyle \int_{1}^{4}\int_{x}^{2x}dydx=\frac{15}{2}\)

 

[never_lose]

f(x,y) isn't given. It's just simply some function f(x,y).

I see no reason why there should be asymptotes with two lines.

In the original function \(\displaystyle \int_{1}^{4}\int_{x}^{2x}f(x,y) dy dx\), x is going from 1 to 4, so \(\displaystyle \Omega\) would be the region between x = 1, x = 4, and the lines y = x and y = 2x I believe.

 

[pka]

f(x,y) isn't given. It's just simply some function f(x,y). In the original function \(\displaystyle \int_{1}^{4}\int_{x}^{2x}f(x,y) dy dx\), x is going from 1 to 4, so \(\displaystyle \Omega\) would be the region between x = 1, x = 4, and the lines y = x and y = 2x I believe.

Look at the three regions. As we change the order of integration, we see where the new limits come from.

 

[never_lose]

How do you know that it's separated into three regions like that? Is there some algebra work done to see this?

 

[pka]

How do you know that it's separated into three regions like that? Is there some algebra work done to see this?

You need to change from \(\displaystyle dydx\) to \(\displaystyle dxdy\).
So you have the 'outside' limits on \(\displaystyle y\) as \(\displaystyle [1,2]\cup[2,4]\cup[4,8]\)
On \(\displaystyle [1,2]~~dx:1\text{ to }y\),
on \(\displaystyle [2,4]~~dx:y/2\text{ to }y\),
on \(\displaystyle [4,8]~~dx:y/2\text{ to }4\)