Area between curves and applications

Covers AS91579

Area between two curves

To find the area enclosed between two curves, integrate the difference of the functions over the interval where they overlap:

$\text{area} = \int_a^b \big(\text{top curve} - \text{bottom curve}\big)\,dx$

Method:
- Find the intersection points by setting the two functions equal — these give the limits $a$ and $b$ .
- Decide which curve is on top between those limits (test a value in between).
- Integrate (top − bottom) and evaluate.

Integration as "undoing a rate"

Because integration reverses differentiation, integrating a rate of change gives the total change.
For example, integrating a velocity function over a time interval gives the distance travelled (or displacement).

Worked ExampleArea between a line and a parabola

Find the area enclosed between $y = 2x$ and $y = x^2$ .

Step 1 — Find where the curves meet

Set them equal: $x^2 = 2x \Rightarrow x^2 - 2x = 0 \Rightarrow x(x - 2) = 0$ , so $x = 0$ and $x = 2$ . These are the limits.

Step 2 — Decide which is on top

Test $x = 1$ : the line gives $2$ , the parabola gives $1$ , so the line $y = 2x$ is on top.

Step 3 — Integrate (top − bottom)

$\text{area} = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = \left(4 - \frac{8}{3}\right) - 0 = \frac{4}{3}$

The enclosed area is $\tfrac{4}{3}$ square units.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

The curves $y = x^2$ and $y = 4$ meet at $x = -2$ and $x = 2$ . Write down the integral (with limits) for the area between them, taking $y = 4$ as the top curve.

Merit

Find the area between the line $y = x + 2$ and the curve $y = x^2$ .

Excellence

Water flows into a tank at a rate of $r = 6t - t^2$ litres per minute, for $0 \le t \le 6$ . Find the total volume of water added during the first 6 minutes.