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
∫ top function - bottom function dx OR ∫ right function-left function dy
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:
∫ (4-x)-(2-3x) dx = 2x+x2 evaluated from -2 to 1/2
My answer to this integral was (5/4)
Then I took this integral:
∫ (4-x)-(x) = 4x-x2 evaluated from 1/2 to 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.