JAMB 2001 · UME · Q39

Find the area bounded by the curve $y = 4 - x^2$ and the $x$ -axis.

Worked solution (try it first)

The curve meets the $x$ -axis where $4 - x^2 = 0$ , at $x = -2$ and $x = 2$ .
These are the limits.
$\int_{-2}^{2} (4 - x^2)\,dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2}$ .
At $x = 2$ this is $8 - \frac83 = \frac{16}{3}$ , and at $x = -2$ it is $-\frac{16}{3}$ .
Subtract: $\frac{16}{3} + \frac{16}{3} = \frac{32}{3}$
$= 10\frac23$ square units, option B.