JAMB Mathematics 2023 Paper

Concept:Area between two curves is found by integrating the difference between the upper and lower functions over their intersection points.Explanation:Find points of intersection by equating $y = 2x^2 + 10$ and $y = 4x + 16$.$2x^2 + 10 = 4x + 16$$2x^2 - 4x - 6 = 0$$2(x + 1)(x - 3) = 0$So $x = -1$ and $x = 3$.Between $x = -1$ and $x = 3$, the line $4x + 16$ lies above the curve $2x^2 + 10$.Therefore, area $= \int_{-1}^{3} [(4x + 16) - (2x^2 + 10)] \, dx$$= \int_{-1}^{3} (-2x^2 + 4x + 6) \, dx$$= \left[-\frac{2x^3}{3} + 2x^2 + 6x\right]_{-1}^{3}$Evaluate at $x = 3$: $-\frac{2(3)^3}{3} + 2(3)^2 + 6(3) = 18$Evaluate at $x = -1$: $-\frac{2(-1)^3}{3} + 2(-1)^2 + 6(-1) = -\frac{10}{3}$Area $= 18 - \left(-\frac{10}{3}\right) = 18 + \frac{10}{3} = \frac{64}{3}$Answer:The area of the region is $\frac{64}{3}$ square units.Correct option: D. $\frac{64}{3}$