# How do you find the area between y=x, y=2-x, y=0?

Jun 11, 2017

Let $A$ be the area bounded by $y = x , y = 2 - x ,$ and $y = 0$

$\implies$
$A = {\int}_{0}^{1} {\int}_{y}^{2 - y} \mathrm{dx} \mathrm{dy} = 1$

#### Explanation:

First, a good thing to do would be sketch the graph. So, we are looking for the area of that triangle.

Take note of the points of intersection.
Here we are dealing with $y = 2 - x$ and $y = x$
So we set both equations equal and get

$x = 2 - x$

$\iff$ add $x$ to both sides

$2 x = 2$

$\iff$ divide both sides by $2$

$x = 1$

So, both functions intersect at $\left(1 , 1\right)$

Since we are only dealing with values greater than $0$ we will have some ${\int}_{0}^{1} f \mathrm{dy}$

And since our area is bounded on the left by $y = x$ and on the right by $y = 2 - x$, If we let $A$ be our area we get

$A = {\int}_{0}^{1} {\int}_{y}^{2 - y} \mathrm{dx} \mathrm{dy}$
$A = {\int}_{0}^{1} {\int}_{y}^{2 - y} \mathrm{dx} \mathrm{dy} = {\int}_{0}^{1} \left(2 - y - y\right) \mathrm{dy} = \int 2 - 2 y \mathrm{dy}$
=2int1-ydy=2(y-y^2/2)]_0^1=2[(1-1/2)-(0-0/2)]=2[1/2]=1