Evaluate the following 2nd kind line integral

Question

$$\int _{\gamma }\:\left(x+y\right)dx-\left(x-y\right)dy\:$$ $$\:\gamma \::\:y=1-\left|1-x\right|$$ $$\:x\in \left[0,2\right]$$

Is this way of solving it correct? I have trouble figuring out how to parameterize the path, so I am not sure if it's alright this way

where O(0,0) ; A(1,1) ; B(2,0)

Answer 1

The first leg can simply have $x=t, y=t, 0 \leq t \leq 1$. The second leg can be $x = t, y = 2-t, 1 \leq t \leq 2$

Answer 2

I would like to point out a way to 'avoid' doing some integrals:

$$I=\int_{\gamma} (x+y) dx -(x-y) dy = \int_{\gamma} x dx + y dy + \int_{\gamma} ydx -x dy = \int_{\gamma} d(xy) -\int_{\gamma}x^2 d( \frac{y}{x})= (xy)|_{\gamma} - \frac{y}{x} x^2|_{\gamma} +\int_{\gamma} 2ydx $$

Ultimately:

$$ I = 2 \int_{\gamma} y dx$$

Which is just area under curve, also area of the triangle which is $1$ , hence the final answer is $I=2$