Answer
$\int_C (2x-y)dx+(x+3y)dy = \frac{63}{2}$
Work Step by Step
The path C has two line segments:
$C_1:$ line segment from (0,0) to (3,0). Here, $y=0,dy =0$
$C_2:$ line segment from (3,0) to (3,3). Here, $x=3,dx =0$
$\therefore \int_C (2x-y)dx+(x+3y)dy\\
=\int_{C_1} (2x-y)dx+(x+3y)dy+\int_{C_2} (2x-y)dx+(x+3y)dy\\
= \int_{x=0}^{3}2xdx + \int_{y=0}^3(3+3y) dy\\
=[\frac{2x^2}{2}]_0^{3}+[3y+\frac{3y^2}{2}]_0^3\\
=\frac{63}{2}$
