How do you evaluate this integral? Evaluate the integral $\int_{T} (x-y)\text{d}x + (x+y)\text{d}y$ where T is counterclockwise around the triangle with vertices $(0,0), (1,0)$, and $(0,1)$.
I'm completely lost on how to solve this one. And if it does involve finding a parameterization, $r(t)$, could you please explain to me how we found it? 
 A: It is a line integral over a curve made of three line segments.
There is a standard way to parametrize a line segment from the point ${\bf r_1}=(x_1,y_1)$ to the point ${\bf r_2}=(x_2,y_2)$ which is 
$${\bf r}(t) = (x(t),y(t)) = {\bf r_1} + t({\bf r_2}-{\bf r_1}) ~~~~~ 0\leq t\leq1$$
Once you parametrize the curves, you can set up three integrals, each of which will look like $$\int_0^1(x(t)-y(t))x^\prime(t)dt + (x(t)+y(t))y^\prime(t)dt$$
A: You don't have to do any integrals explicitly if you resort to Greens Theorem to evaluate the line integral:
$$\int_T (x-y)dy + (x+y)ydx = \int_{A_T}\frac{\partial (x+y)}{\partial x} - \frac{\partial (x-y)}{\partial y}dxdy = 2\int_{A_T}dxdy = 1$$
since $\int_{A_T}dxdy = \frac{1}{2}$ is the area of the triange.
A: You should do the integration piecewise on the 3 line segments separately, and then sum them together.
The parameterization $r_1(t)$ of the line segment joining $(0,0)$ and $(1,0)$ is easy.
$$r_1(t) = (t, 0) \quad t \in [0,1]$$
$r_2(t)$ of the line segment joining $(1,0)$ and $(0,1)$ is given by
$$r_2(t) = (1 - t, t) \quad t \in [0,1]$$
$r_3(t)$ of the line segment joining $(0,1)$ and $(0,0)$ is also easy.
$$r_3(t) = (0, 1 - t) \quad t \in [0,1]$$
