JAMB 2017 · UTME · Q33

Find the area bounded by the curves $y = 4 - x^2$ and $y = 2x + 1$ .

Worked solution (try it first)

The curves meet where $4 - x^2 = 2x + 1$ , that is $x^2 + 2x - 3 = 0$ , so $(x + 3)(x - 1) = 0$ and $x = -3$ or $x = 1$ .
Between them the parabola is on top, so integrate top minus bottom: $(4 - x^2) - (2x + 1) = 3 - 2x - x^2$ .
$\left[3x - x^2 - \frac{x^3}{3}\right]_{-3}^{1}$ : at $x = 1$ it is $\frac53$ , and at $x = -3$ it is $-9$ .
Subtract: $\frac53 + 9 = \frac{32}{3} = 10\frac23$ square units, option B.