Answer
$\int_C (2x-y)dx +(x+3y)dy = \frac{592}{5}$
Work Step by Step
The path is given by,
$C: \text{arc on } y = x^{3/2}$ from (0,0) to (4,8)
$\therefore dy = \frac{3}{2}x^{1/2} dx$
$\int_C (2x-y)dx + (x+3y)dy \\
= \int_{x=0}^{4}[2x-x^{3/2}]dx + [x+3x^{3/2}].(\frac{3}{2}x^{1/2}) dx\\
= \int_{0}^{4} (2x+\frac{1}{2}x^{3/2}+\frac{9}{2}x^2)dt\\
=[x^2+\frac{x^{5/2}}{5}+\frac{3x^3}{2}]_0^{4}\\
=\frac{592}{5}$
