Answer
$\int_C (2x-y)dx+(x+3y)dy = \frac{47}{2}$
Work Step by Step
The path C has two line segments:
$C_1:$ line segment from (0,0) to (0,-3). Here, $x=0,dx =0$
$C_2:$ line segment from (0,-3) to (2,-3). Here, $y=-3,dx =0$
$\therefore \int_C (2x-y)dx+(x+3y)dy\\
=\int_{C_1} (2x-y)dx+(x+3y)dy+\int_{C_2} (2x-y)dx+(x+3y)dy\\
= \int_{y=0}^{-3}3ydy + \int_{x=0}^2(2x+3) dx\\
=[\frac{3y^2}{2}]_0^{-3}+[x^2+3x]_0^2\\
=\frac{47}{2}$
