I am suppose to find the area of a triangle using integrals with vertices 0,0 1,2 and 3,1
This gives me
$y= 2x$
$y=\frac{1}{3}x$
$y= \frac{-1}{2}x+\frac{5}{2}$
for my slopes
I know that I can calculate the area of the first part by finding
$\int_{0}^{1}2x-\frac{1}{3}x$
and the second part by
$\int_{1}^{3}\frac{-1}{2}x+\frac{5}{2}-\frac{1}{3}x$
The anti derivative of the top is $x^2 - x^2/3$
and the other one is $-x^2/4 + 5x/2 - x^2/6$
I am not sure what I am doing wrong but I do not get the proper answer of 5/2