Answer to Question #159826 in Calculus for Sean Sith

Question

Set up a sum of integrals that will give the area of the quadrilateral with vertices at (0,0), (1,1), (0,2), and (2,1), connected in the stated order.

Answer 1

$1. BC\implies y-2=\frac{2-1}{0-2}(x-0)\implies y-2=\frac{1}{2}(x)$

$BC\implies 2y+4=x\implies x=4-2y$

$2. CA\implies y-1=\frac{1-0}{2-0}(x-2)\implies y-1=\frac{1}{2}(x-2)$

$CA\implies 2y-2=x-2\implies x=2y$

$3. BD\implies y-2=\frac{2-1}{0-1}(x-0)\implies y-2=\frac{1}{-1}(x)$

$BD\implies -y+2=x\implies x=2-y$

$4. DA\implies y-1=\frac{1-0}{1-0}(x-1)\implies y-1=\frac{1}{1}(x-1)$

$DA\implies y-1=x-1\implies x=y$

Area of shaded region =$\int_1^2x_{BC}dy-\int_1^2x_{BD}dy+\int_0^1x_{CA}dy-\int_0^1x_{DA}dy$

$\int_1^2(4-2y)dy-\int_1^2(2-y)dy+\int_0^12ydy-\int_0^1ydy$

$[4y-y^2]_1^2-[2y-0.5y^2]_1^2+[y^2]_0^1-[0.5y^2]_0^1$

$1-0.5+0.5+0$

$1$