Find the area between three curves

1. Sep 26, 2012

MissEuropa

1. The problem statement, all variables and given/known data
Sketch the region enclosed by the curves and compute its area as an integral along the x or y axis.
y+x=4 y-x=0 y+3x=2

2. Relevant equations

top function - bottom function dx OR right function-left function dy

3. The attempt at a solution
I originally had chosed to integrate with respect to y, I solved all equations for y as follows:
y=4-x y=x y=2-3x

I set all equations equal to each other to find three intersection x values: These are
x=2, x=-2, and x=1/2

This makes since considering the way I sketched the graph: see attachment.

Now the Area is shaded in orange and labeled A: to set up the integral I did:
1/2
(4-x)-(2-3x) dx = 2x+x2 evaluated from -2 to 1/2
-2

My answer to this integral was (5/4)

Then I took this integral:
2
(4-x)-(x) = 4x-x2 evaluated from 1/2 to 2
1/2

My answer to this integral was (7/4)

It was olvious to me at this point that something went wrong because (5/4)-(7/4)=-(1/2) and areas cannot be negative.

This is the first time I've attempted to find the area between three curves. I am not sure how to arrange the integrals and which one is considered "top" or "bottom" because the graph that I drew shows the lower functions switching, which I tried to reflect in my calculations.

I'm really working hard to wrap my head around this stuff, so any help would be greatly appreciated.

2. Sep 26, 2012

SammyS

Staff Emeritus
The lines y=4-x , and y=2-3x intersect at x = -1 not at x = -2.

By the way, the lines y=4-x, and y=x are perpendicular to each other, so it's easy to check the answer.

3. Sep 26, 2012

MissEuropa

Thanks, I wouldn't of caught that.
So I've edited my limits of integration:
1/2
((4-x)-(2-3x)) = 4.25
-1

2
((4-x)-(x))= (7/4)
1/2

4.25-(7/4) = 2.5

Alas, I am still incorrect. Is there something else I am missing?

Thanks again.

4. Sep 26, 2012

SammyS

Staff Emeritus
Be careful with your signs !

You integrations are correct in the previous post.

You are messing up when plugging in the limits of integration.