Area between x²=4y and x=4y−2

Curves: parabola x² = 4y and line x = 4y − 2, i.e. y = (x + 2)/4.

Intersection. Substitute 4y = x² into x = 4y − 2: x = x² − 2, so x² − x − 2 = 0, (x − 2)(x + 1) = 0. Hence x = 2 or x = −1, giving the points (−1, 1/4) and (2, 1).

Between x = −1 and x = 2 the line lies above the parabola (e.g. at x = 0, line y = 1/2, parabola y = 0).

A = ∫_(−1)^2 [ (x + 2)/4 − x²/4 ] dx
  = (1/4) [ x²/2 + 2x − x³/3 ]_(−1)^2
  = (1/4) [ (2 + 4 − 8/3) − (1/2 − 2 + 1/3) ]
  = (1/4) [ 10/3 + 7/6 ]
  = (1/4)(27/6)

Area = 9/8 square units.