- #1

MissEuropa

- 20

- 0

## Homework Statement

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

## Homework Equations

∫ 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:

1/2

∫ (4-x)-(2-3x) dx = 2x+x

^{2}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-x

^{2}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.