Hi everyone, I'm having some trouble doing this question that involves reversing the order of equations (so that I guess you do dxdy first vs doing the initial dydx).

Here is the question:

Can anyone tell me what the graph would look like too? What areas would be shaded? I came up with something below...

2. Originally Posted by finalfantasy
Hi everyone, I'm having some trouble doing this question that involves reversing the order of equations (so that I guess you do dxdy first vs doing the initial dydx).

Here is the question:

Can anyone tell me what the graph would look like too? What areas would be shaded? I came up with something below...

That is the same as
$\displaystyle 3\left[\int_{x= -1}^\frac{1}{2}(x+ y) dydx- \int_{x= \frac{1}{2}}^1 (x+ y) dy dx\right]$

Notice the "-" between the two integrals. That is because the two boundaries cross as x= 1/2.

3. Originally Posted by HallsofIvy
That is the same as
$\displaystyle 3\left[\int_{x= -1}^\frac{1}{2}(x+ y) dydx- \int_{x= \frac{1}{2}}^1 (x+ y) dy dx\right]$

Notice the "-" between the two integrals. That is because the two boundaries cross as x= 1/2.

Hi, thank you for your response!

I was told by my friend you need to subtract a triangle part from another and this seems to be what he was talking about.

So how would I solve that integral? Is my idea of dividing the bigger triangle in half to solve for its area, solving the 2 parts of the smaller triangle and subtracting them both a good idea?

Also, for your integrals I have to do reverse the order of integration, so I have to use horizontal sections first instead of vertical.

Thank you!