`(2,0), (0,2), (-1,1)` Use calculus to find the area of the triangle with the given vertices.

Given the coordinates (2, 0), (0, 2), and (-1, 1).

Let A=(2, 0), B(0, 2), and C(-1, 1).

Find the equation of line AB using A(2, 0) and B(0, 2).

The slope of line AB is `(2-0)/(0-2)=-1`

The equation of line AB is `y=-1x+2`

Find the equation of line BC using B(0, 2) and C(-1 1).

The slope of line BC is `(1-2)/(-1-0)=(-1)/-1=1`

The equation of line BC is `y=1x+2`

Find the equation of line AC using A(2,...

Given the coordinates (2, 0), (0, 2), and (-1, 1).

Let A=(2, 0), B(0, 2), and C(-1, 1).

Find the equation of line AB using A(2, 0) and B(0, 2).

The slope of line AB is `(2-0)/(0-2)=-1`

The equation of line AB is `y=-1x+2`

Find the equation of line BC using B(0, 2) and C(-1 1).

The slope of line BC is `(1-2)/(-1-0)=(-1)/-1=1`

The equation of line BC is `y=1x+2`

Find the equation of line AC using A(2, 0) and C(-1, 1).

The slope of line AC is `(1-0)/(-1-2)=-1/3`

The equation of line AC is `y-0=(-1/3)(x-2)=>y=-1/3x+2/3`

Set up the intervals for integration.

`int_-1^0(x+2)-(-1/3x+2/3)dx+int_0^2(-x+2)-(-1/3x+2/3)dx`  

`=int_-1^0(4/3x+4/3)dx+int_0^2(-2/3x+4/3)dx`

`=[4/3*x^2/2+4/3x]` from x=-1 to x=0 + `[-2/3*x^2/2+4/3x]`  from x=0 to x=2

`=[2/3x^2+4/3x]`  from x=-1 to x=0 + `[-1/3x^2+4/3x]`  from x=0 to x=2

`=[0-(2/3-4/3)]+[(-4/3+8/3)-0]`

`=(-2/3+4/3)+(-4/3+8/3)`

`=2/3+4/3`

`=6/3`

`=2`

The area of the triangle with vertices (2, 0), (0, 2), and (-1, 1) is 2 units squared.