Here's the integral:
$$\int\limits_C(x^2+y^2)\,\mathrm{d}x+(x^2-y^2)\,\mathrm{d}y$$
Where:
$$\begin{cases}y=1-|1-x| \\ 0\le x \le 2 \end{cases}$$
The integral itself seems easy, but I am more confused about how to deal with module inside the function.
On
Break it up as the integral over two curves $C_1$ in which $0\le x\le 1$ and $C_2$ in which $1\le x\le 2$. Then $\int_C=\int_{C_1}+\int_{C_2}$.
On
If you graph $y = 1-|1-x|$, you'll see that it's two line segments of different slopes attached at $x=1$. One way to attack this is to write $C = C_1 \cup C_2$, where $C_1$ and $C_2$ are the two line segments. Then we can take advantage of the fact that $\displaystyle \int_C \text{ stuff } = \int_{C_1} \text{ stuff } + \int_{C_2} \text{ stuff }$.
Wee can parametrize $C_1$ as $\alpha_1(t) = (t, \ t)$ for $0 \leq t \leq 1$, so $\displaystyle \int_{C_1} (x^2 + y^2)\ dx + (x^2-y^2)\ dy = \int_0^1 2t^2 \ dt$.
Do something similar for $C_2$, compute the integrals, and sum.
$$ y=\begin{cases} x&\text{ for }0\le x<1\\ 2-x&\text{ for }1\le x\le2\end{cases}$$
$$ dy=\begin{cases} \phantom{-}dx&\text{ for }0\le x<1\\ -dx&\text{ for }1\le x\le2\end{cases}$$
\begin{eqnarray} \int\limits_C(x^2+y^2)\,\mathrm{d}x+(x^2-y^2)\,\mathrm{d}y &=&\int\limits_{C_1}2x^2-(x^2-x^2)\,\mathrm{d}x+\int\limits_{C_2}(x^2+(2-x)^2)-(x^2-(2-x)^2)\,\mathrm{d}x\\ &=&2\int\limits_{C_1}x^2\,\mathrm{d}x+2\int\limits_{C_2}(2-x)^2\,\mathrm{d}x \end{eqnarray}
Where $C_1$ is the portion of the path where $0\le x<1$ and $C_2$ the portion where $1\le x\le2$.