Definite integrals and area under a curve

Covers AS91579

What a definite integral is

A definite integral has limits $a$ (bottom) and $b$ (top) and gives a number, not a function:

$\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a)$

Method: integrate as usual (no $+c$ needed), then substitute the top limit and subtract the value at the bottom limit.
The $+c$ cancels in the subtraction, which is why definite integrals do not need it.

Area under a curve

For a curve that lies above the $x$ -axis between $x = a$ and $x = b$ , the area between the curve and the axis is exactly the definite integral:

$\text{area} = \int_a^b f(x)\,dx$

If part of the curve is below the axis, the integral there is negative — split the calculation at the $x$ -intercepts and take the size (absolute value) of each piece.

Worked ExampleEvaluating a definite integral

Evaluate $\displaystyle\int_1^2 3x^2\,dx$ .

Step 1 — Integrate (no constant needed)

$\int 3x^2\,dx = x^3$

Step 2 — Substitute the limits, top minus bottom

$\Big[x^3\Big]_1^2 = (2)^3 - (1)^3 = 8 - 1 = 7$

Worked ExampleArea under a curve

Find the area between $y = x^2$ , the $x$ -axis, and the lines $x = 0$ and $x = 3$ .

Step 1 — The curve is above the axis, so integrate

$\text{area} = \int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3$

Step 2 — Substitute the limits

$= \frac{3^3}{3} - \frac{0^3}{3} = \frac{27}{3} - 0 = 9$

The area is $9$ 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

Evaluate $\displaystyle\int_0^2 2x\,dx$ .

Merit

Evaluate $\displaystyle\int_1^3 (3x^2 + 2)\,dx$ .

Excellence

Find the area enclosed between the curve $y = 4 - x^2$ and the $x$ -axis.