Question

evaluate the integral of c (xydy-y^2dx) where c is the square cut from the first quadrant by the line x=1 and y=1.

Answer 1

∫(xydy-y²dx)=₀∫¹xydy-₀∫¹y²dx, because c defines the limits for x and y as the boundaries of the unit square.

₀∫¹xydy is to be integrated between y=0 and y=1. In this interval x is always 1, so xdy=dy, and the integral becomes ₀∫¹ydy=[y²/2]₀¹=½.

₀∫¹y²dx is to be integrated between x=0 and x=1. In this interval y is always 1, so y²dx=dx, and the integral becomes ₀∫¹dx=[x]₀¹=1.

∫(xydy-y²dx)=½-1=-½.