Problem 6 - Entrance Test

The curves intersect where x^2 = x + 1. Rearranging gives x^2 - x - 1 = 0, which has roots x = (1 ± √5)/2. The area of the region bounded by the curves is ∫[(1 + √5)/2, (1 - √5)/2] (x + 1 - x^2) dx. Evaluating the integral gives ((x^2)/2 + x - (x^3)/3) from (1 + √5)/2 to (1 - √5)/2 = ((1 - √5)^2)/4 + (1 - √5)/2 - ((1 - √5)^3)/6 - (((1 + √5)^2)/4 + (1 + √5)/2 - ((1 + √5)^3)/6) = ((1 - 2√5 + 5)/4 + (1 - √5)/2 - (1 - 3√5 + 3*5)/6) - ((1 + 2√5 + 5)/4 + (1 + √5)/2 - (1 + 3√5 + 3*5)/6) = ((6 - 12√5 + 30)/12 + (6 - 6√5)/12 - (1 - 3√5 + 15)/6) - ((6 + 12√5 + 30)/12 + (6 + 6√5)/12 - (1 + 3√5 + 15)/6) = 5/3.