To find the point of intersection, equate the two equations ⇒ 2x\(^2\) + 10 = 4x + 16 ⇒ 2x\(^2\) + 10 - 4x - 16 = 0 ⇒ 2x\(^2\) - 4x - 6 = 0 Factorize ⇒ 2(x + 1)(x - 3) = 0 ∴ x = -1 or 3 The curves will intersect at -1 and 3 Area = \(\int^b_a\) [upper function] - [lower function] dx = \(\int^3_1 4x + 16 - (2x^2 + 10) dx\) = \(\int^3_1 -2x^2 + 4x + 6dx\) = \((\frac {-2x^3}{3} + 2x^2 + 6x)^3_1\) = \((\frac {-2(3)^3}{3} + 2(3)^2 + 6(3)) - (\frac {-2(-1)^3}{3} + 2(-1)^2 + 6(-1))\) = 18 - \((\frac {10}{3})\) = 18 + \(\frac {10}{3}\) = \(\frac {64}{3}\)