Evaluating a line integeral on the sides of a triangle

Question

Let $C$ be the triangle with vertices $(0,0);(1,0);(0,1)$ , traversed in a counterclockwise direction ;

then what is the value of $\int_C(x+y)ds$ ? I'm having trouble with the range of parameter for each side of the triangle , Please help .

Answer 1

EDITED

In this version the counterclockwise nature of the traversal will be paid attention to.

The triangle has $3$ sides. The parametric equations of the sides are

1. $x_1(t)=t,\,y_1(t)=0;\,\,0\le t \le 1;\,\,r_1(t)=\vec i t;\,\,|r'_1(t)|=1,$
2. $x_2(t)=0,\ y_2(t)=1-t;\,\,0\le t \le 1;\,\,r_2(t)=\vec j (1-t);\,\,|r'_2(t)|=1,$
3. $x_3(t)=1-t,\ y_3(t)=t;\,\,0\le t \le 1;\,\,r_3(t)=\vec i (1-t)+\vec jt;\,\,|r'_3(t)|=\left |-\vec i+\vec j\right |=\sqrt2.$

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The line integral has three parts then

$$\int_{\triangle}x+y\ ds=$$ $$\int_0^1 (x_1(t)+y_1(t))|r'_1(t)|\ dt+\int_0^1 (x_2(t)+y_2(t))|r'_2(t)|\ dt+\sqrt2\int_0^1 (x_3(t)+y_3(t))|r'_3(t)| \ dt=$$ $$=\int_0^1 t\ dt+\int_0^1 1-t\ dt+\sqrt2\int_0^1 1 \ dt=1+\sqrt2.$$