Answer
$\int_C (2x-y)dx +(x+3y)dy = \frac{316}{3}$
Work Step by Step
The path is given by,
$x = t , y = 2t^2$ from (0,0) to (2,8)
$\therefore dx = dt, dy = 4t dt, 0\le t \le 2$
$\int_C (2x-y)dx + (x+3y)dy \\
= \int_{t=0}^{2}[2t-2t^2]dt - [t+6t^2].4t dt\\
= \int_{0}^{2} (24t^3+2t^2+2t)dt\\
=[24 \frac{t^4}{4}+2\frac{t^3}{3}+2\frac{t^2}{2}]_0^{2}\\
=\frac{316}{3}$
