Answer
$\int_C (2x-y)dx+(x+3y)dy = -\frac{11}{6}$
Work Step by Step
The path is given by,
$C: \text{arc on } y = 1-x^{2}$ from (0,1) to (1,0)
$\therefore dy = -2x dx$
$\int_C (2x-y)dx + (x+3y)dy \\
= \int_{x=0}^{1}[2x-(1-x^2)]dx + [x+3(1-x^2)].(-2x) dx\\
= \int_{0}^{1} (6x^3-x^2-4x-1)dx\\
=[\frac{6x^4}{4}-\frac{x^3}{3}-\frac{4x^2}{2}-x]_0^{1}\\
=-\frac{11}{6}$
