Determine the area of the region bounded by y 2... - JAMB Mathematics 2023 Question

To find the area of the region bounded by the curves \(y = 2x^2 + 10\) and \(y = 4x + 16\), we need to set them equal to each other and find the points of intersection.

\[2x^2 + 10 = 4x + 16\]

Rearrange it to a quadratic equation:

\[2x^2 - 4x - 6 = 0\]

Now, factor the quadratic equation:

\[2(x^2 - 2x - 3) = 0\]

\[(x - 3)(x + 1) = 0\]

So, the points of intersection are \(x = 3\) and \(x = -1\).

Now, integrate to find the area:

\[A = \int_{-1}^{3} (4x + 16 - (2x^2 + 10)) \,dx\]

Simplify the integrand:

\[A = \int_{-1}^{3} (-2x^2 + 4x + 6) \,dx\]

Now integrate with respect to \(x\):

\[A = \left[-\frac{2}{3}x^3 + 2x^2 + 6x\right]_{-1}^{3}\]

Evaluate the expression at the upper and lower limits:

\[A = \left[-\frac{2}{3}(3)^3 + 2(3)^2 + 6(3)\right] - \left[-\frac{2}{3}(-1)^3 + 2(-1)^2 + 6(-1)\right]\]

\[A = \left[-18 + 18 + 18\right] - \left[\frac{2}{3} + 2 - 6\right]\]

\[A = 18 - \left[-\frac{4}{3}\right]\]

\[A = 18 + \frac{4}{3}\]

\[A = \frac{54}{3} + \frac{4}{3} = \frac{58}{3}\]

Therefore, the correct answer is \(\frac{64}{3}\).