where C is the curve:
x = t, y = t^2, z = 0
y = x^2 + 2x - 3
A = ∫[0,2] (x^2 + 2x - 3) dx = [(1/3)x^3 + x^2 - 3x] from 0 to 2 = (1/3)(2)^3 + (2)^2 - 3(2) - 0 = 8/3 + 4 - 6 = 2/3 where C is the curve: x = t,
Solution:
from t = 0 to t = 1.
∫(2x^2 + 3x - 1) dx = (2/3)x^3 + (3/2)x^2 - x + C y = t^2