Green's theorem evaluate the line integral over c ((x^2)-(y^2))dx+(2y-x)dy

greens thm.
the bounds which are 0 to 1 and x^3 to x^2 with dydx.

∫c [(x^2 - y^2) dx + (2y - x) dy]
= ∫∫ [(∂/∂x)(2y - x) - (∂/∂y)(x^2 - y^2)] dA, via Green's Theorem
= ∫(x = 0 to 1) ∫(y = x^3 to x^2) (-1 + 2y) dy dx
= ∫(x = 0 to 1) (y^2 - y) {for y = x^3 to x^2} dx
= ∫(x = 0 to 1) (x^4 - x^2 - x^6 + x^3) dx
= 1/5 - 1/3 - 1/7 + 1/4
= -11/420.

I hope this helps!

Green's Theorem:

∫Mdx + Ndy = ∫∫ (∂N / ∂x) - (∂M / ∂y) dA
C D

∫(x² - y²)dx + (2y - x)dy


1 x²
∫ ∫ -1 + 2y dydx
0 x³


1 x²
∫ ∫ -1 + 2y dydx
0 x³


1
∫ [ -y + y² ] (from x³ to x²) dx
0

1
∫ [ -x² + x⁴ ] - [ -x³ + x⁶] dx
0

Integrate with respect to x:

[ (-x³ / 3) + (x⁵ / 5) + (x⁴ / 4) - (x⁷ / 7) ] from 0 to 1

Final Answer
-11/420