Areas, volumes and motion · Integration

A definite integral gives an area under a curve (see calculus), but only when the curve stays above the $x$ -axis. This lesson deals with the parts below it, then with volumes and with motion.

Area below the x-axis

Where the curve is below the axis, $y$ is negative, so the integral there is negative. An area can’t be negative: take its size.

Above and below the axisThe integral counts the part below the axis as negative

When the curve crosses the axis inside the region, find where it crosses, integrate each part separately, and add the sizes. Integrating straight across would let the parts cancel:

Net value or total area?Pick a curve, slide the upper limit

0∫ from 0 to 32.667total area

upper limit = 3

The curve crosses the axis at 1. The parts give 1.333, −1.333: the integral adds them (0), but the area adds their sizes (2.667).

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q12

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

area of the finite region bounded by the curve and the $x$ -axis;

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

Where it crosses the axis
- ${x^2 - 4 = 0}$ , so ${x = -2}$ or ${x = 2}$ . These are the limits.
Think first. Solve x² − 4 = 0.
Integrate
- At ${x = 2}$ : ${\frac83 - 8 = -\frac{16}{3}}$ .
- At ${x = -2}$ : ${-\frac83 + 8 = \frac{16}{3}}$ .
- So ${\int_{-2}^2 (x^2 - 4)\,dx = -\frac{16}{3} - \frac{16}{3} = -\frac{32}{3}}$ .
Think first. Work out [x³/3 − 4x] at 2 and at −2.
The area
- The region is below the axis, so the integral is negative.
- The area is ${\frac{32}{3} \approx 10.67}$ square units.
Think first. The integral is negative. Why?
The volume
- ${V = \pi\int_{-2}^2 (x^2 - 4)^2\,dx}$ .
- Expand the square: ${(x^2 - 4)^2 = x^4 - 8x^2 + 16}$ .
- Integrate: ${V = \pi\left[\frac{x^5}{5} - \frac{8x^3}{3} + 16x\right]_{-2}^2}$ .
- ${= 2\pi\left(\frac{32}{5} - \frac{64}{3} + 32\right) = \frac{512}{15}\pi}$ .
- With ${\pi = \frac{22}{7}}$ : ${V \approx 107.28}$ cubic units.
Think first. V = π∫y² dx. Square x² − 4 first.

Areas between curves and volumes

For the area between two curves, find where they meet, then integrate top minus bottom. For a volume of revolution about the $x$ -axis, each thin slice is a disc of radius $y$ :

Area between two graphsIntegrate (top − bottom) between the meeting points

Volume of revolutionV = π∫ y² dx: each slice is a disc of radius y

More: areas and definite integrals

Volume about the y-axis

When the region is spun about the $y$ -axis instead, each thin slice is a horizontal disc of radius $x$ and thickness $\delta y$ . So write $x^2$ in terms of $y$ and integrate between two $y$ -values:

V = \pi\int_a^b x^2\,dy

About the y-axisV = π∫ x² dy: discs of radius x, limits a and b are y-values

Worked example

The region bounded by

y = x^2

, the

y

-axis and the lines

y = 1

and

y = 4

is rotated through

360^\circ

about the

y

-axis. Find the volume, in terms of

\pi

x² in terms of y
- ${y = x^2}$ , so ${x^2 = y}$ .
Think first. Make x² the subject of the curve's equation.
The limits
- The slices are stacked up the $y$ -axis, so the limits are $y$ -values.
- They are ${y = 1}$ and ${y = 4}$ .
Think first. Are the limits x-values or y-values?
Integrate
- ${V = \pi\int_1^4 y\,dy = \pi\left[\frac{y^2}{2}\right]_1^4}$ .
- ${= \pi\left(8 - \frac12\right)}$ .
- ${= \frac{15}{2}\pi}$ cubic units.
Think first. Work out π∫ y dy from 1 to 4.

More: volume about the y-axis

WAEC 2009 · Paper 2 · Q12Two linear transformations are defined by $P : (x, y) \to (-2x + 3y,\ 5x - 4y)$ and $T : (x, y) \to (-3x - 4y,\ 6x - 5y)$ . …

Motion: integrate to go back

Differentiating goes from displacement to velocity to acceleration. Integrating goes back:

v = \int a\,dt \qquad s = \int v\,dt

Each integral brings a constant. “Starts from rest” means $v = 0$ when $t = 0$ , and “from the origin” means $s = 0$ when $t = 0$ .

distance sd/dt →velocity vd/dt →acceleration a

differentiate to go right; integrate to come back

Displacement, velocity, accelerationDifferentiate down the chain; integrate back up it

Worked example · WAEC 2023

WAEC 2023 · Paper 2 · Q8

The acceleration, $a$ , of a particle starting from rest and moving at any time $t$ seconds is given by $a = (20t - 3t^2)\text{ m s}^{-2}$ . Find the:

time taken for the particle to come to rest again;

distance covered by the particle after 5 seconds.

The velocity
- ${v = \int (20t - 3t^2)\,dt = 10t^2 - t^3 + c}$ .
- It starts from rest, so ${v = 0}$ at ${t = 0}$ , and ${c = 0}$ .
Think first. Integrate a, then use 'starting from rest'.
At rest again
- ${10t^2 - t^3 = 0}$ , so ${t^2(10 - t) = 0}$ .
- So it is at rest again at ${t = 10}$ s.
Think first. Solve v = 0 for t > 0.
Distance in 5 seconds
- ${s = \int_0^5 (10t^2 - t^3)\,dt = \left[\frac{10t^3}{3} - \frac{t^4}{4}\right]_0^5}$ .
- ${= \frac{1250}{3} - \frac{625}{4} = \frac{3125}{12} \approx 260.42}$ m.
- $v$ is positive from $t = 0$ to $10$ , so this is the distance travelled.
Think first. Integrate v from 0 to 5.

Your turn

WAEC 2022 · Paper 2 · Q14 (a)

(a)
A particle initially at rest moves in a straight line with an acceleration of $(10t - 4t^2)\text{ m s}^{-2}$ . Find the: (i) velocity of the particle after $t$ seconds; (ii) average acceleration of the particle during the 4th second.

Worked solution (try it first)

(a)(i)

$v = \displaystyle\int (10t - 4t^2)\,dt$
$= 5t^2 - \frac43t^3 + k$ .
It starts at rest, so $k = 0$ and $v = 5t^2 - \frac43t^3$ .

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