Find the Area Between curves

Question

x=y^2-4y

x=y-y^2

Point of intersection is (-15/4, 5/2)

Answer 1

There is another point of intersection at (0,0). Let's call the functions x₁ and x₂ with x₂ always being greater for the interval. Therefore the area is

Integral from 0 to 5/2 of (x₂ - x₁) dy (adding up horizontal rectangles in y)

integral from 0 to 5/2 of (y-y²-y²+4y)dy = 5y²/2 - 2y³/3 evaluated from 0 to 5/2

(5/2)³ - (2/3)(5/2)³ = (1/3)(5/2)³ = 125/24

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