WAEC 2011 · Paper 2 · Q12

Integration

Given the curve $y = x^2 - 4$ , calculate, correct to two decimal places, the:

(a)
area of the finite region bounded by the curve and the $x$ -axis;
(b)
volume generated by rotating the region in (a) through $360^\circ$ about the $x$ -axis. $\left[\text{Take } \pi = \frac{22}{7}\right]$

Try it on a graph

The shaded region is below the axis — so its integral is negative. Change the curve to explore.

Worked solution (try it first)

(a)

The curve meets the $x$ -axis where $x^2 - 4 = 0$ : $x = -2$ and $x = 2$ .
$\displaystyle\int_{-2}^2 (x^2 - 4)\,dx = \left[\frac{x^3}{3} - 4x\right]_{-2}^2$
$= \left(\frac83 - 8\right) - \left(-\frac83 + 8\right)$
$= -\frac{32}{3}$ .
The region is below the axis, so the integral is negative.
The area is $\dfrac{32}{3} \approx 10.67$ square units.

(b)

$V = \pi\displaystyle\int_{-2}^2 (x^2 - 4)^2\,dx$
$= \pi\int_{-2}^2 (x^4 - 8x^2 + 16)\,dx$ .
$= \pi\left[\dfrac{x^5}{5} - \dfrac{8x^3}{3} + 16x\right]_{-2}^2$
$= 2\pi\left(\dfrac{32}{5} - \dfrac{64}{3} + 32\right)$
$= \dfrac{512}{15}\pi$ .
With $\pi = \frac{22}{7}$ : $V = \dfrac{512}{15} \times \dfrac{22}{7}$
$\approx 107.28$ cubic units.

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