§-Syllabus dot point

SAMath MethodsSyllabus dot point

How do we find the exact area between a curve and the x-axis?

The area between a curve and the x-axis equals the definite integral, with regions below the axis requiring a sign adjustment.

How to compute the area between a curve and the x-axis using definite integrals, why regions below the axis need absolute values, and how to handle curves that cross the axis.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
When the curve is above the axis
When the curve is below the axis
When the curve crosses the axis
Common errors
Estimating area before integrating
Area to a numerical answer
Why it matters

What this dot point is asking

The definite integral $\int_a^b f(x)\,dx$ measures signed area: regions above the x-axis count positively and regions below count negatively. Geometric area is always positive, so you must handle the sign carefully.

When the curve is above the axis

If $f(x)\ge 0$ on $[a,b]$ , the area between the curve and the x-axis is simply the integral:

Area under a parabola

Find the area between $y = 4 - x^2$ and the x-axis for $-2 \le x \le 2$ .

Step 1: Confirm the curve is above the axis on the interval

Before integrating, check whether any sign adjustments will be needed. The parabola equals zero at $x = -2$ and $x = 2$ (its x-intercepts) and is positive between them, so the curve lies entirely above the x-axis on the interval. This means the definite integral directly gives the geometric area, with no need to split or take absolute values.

Step 2: Set up and antidifferentiate

A = \int_{-2}^{2}(4 - x^2)\,dx = \Big[4x - \tfrac{x^3}{3}\Big]_{-2}^{2}.

Step 3: Evaluate at each limit

Substitute the upper limit $x = 2$ and the lower limit $x = -2$ into the antiderivative.

At $x = 2$ : $4(2) - \tfrac{(2)^3}{3} = 8 - \tfrac{8}{3} = \tfrac{24}{3} - \tfrac{8}{3} = \tfrac{16}{3}$ .

At $x = -2$ : $4(-2) - \tfrac{(-2)^3}{3} = -8 + \tfrac{8}{3} = -\tfrac{24}{3} + \tfrac{8}{3} = -\tfrac{16}{3}$ .

Step 4: Subtract lower from upper

A = \tfrac{16}{3} - \left(-\tfrac{16}{3}\right) = \frac{32}{3} \approx 10.67 \text{ square units}.

Final answer: The area is $\dfrac{32}{3} \approx 10.67$ square units.

When the curve is below the axis

If $f(x)\le 0$ on the interval, $\int_a^b f(x)\,dx$ is negative. The area is its absolute value:

A=\left|\int_a^b f(x)\,dx\right|.

When the curve crosses the axis

This is the trap. If the curve is above the axis on part of the interval and below on another, a single integral gives the net signed area, which is not the geometric area. You must:

Find the x-intercepts (where $f(x)=0$ ) inside the interval.
Integrate over each sub-interval separately.
Take the absolute value of any negative piece, then add.

A cubic that crosses the axis

Find the total area between $y = x^3$ and the x-axis from $x = -1$ to $x = 2$ .

Step 1: Find where the curve crosses the x-axis inside the interval

Setting $x^3 = 0$ gives $x = 0$ , which lies inside $[-1, 2]$ . This means the curve is not entirely on one side of the axis, so we cannot use a single integral: below-axis portions would contribute negatively and cancel part of the above-axis area, understating the total.

Step 2: Determine the sign of the curve on each sub-interval

At $x = -0.5$ (a test point in $[-1, 0]$ ): $(-0.5)^3 = -0.125 < 0$ , so the curve is below the axis there.

At $x = 1$ (a test point in $[0, 2]$ ): $1^3 = 1 > 0$ , so the curve is above the axis there.

Split at $x = 0$ and integrate each piece separately.

Step 3: Integrate the left piece (below axis) and take the absolute value

\int_{-1}^{0} x^3\,dx = \Big[\tfrac{x^4}{4}\Big]_{-1}^{0} = 0 - \tfrac{1}{4} = -\tfrac{1}{4}.

The result is negative because the curve lies below the axis here. The geometric area of this piece is $\left|-\tfrac{1}{4}\right| = \tfrac{1}{4}$ .

Step 4: Integrate the right piece (above axis)

\int_{0}^{2} x^3\,dx = \Big[\tfrac{x^4}{4}\Big]_{0}^{2} = \tfrac{16}{4} - 0 = 4.

Step 5: Add the areas of both pieces

A = \tfrac{1}{4} + 4 = \frac{17}{4} = 4.25 \text{ square units}.

Note: the single integral $\int_{-1}^{2} x^3\,dx = \tfrac{15}{4}$ gives only the net (signed) area and is not the correct answer for total geometric area.

Final answer: The total area is $\dfrac{17}{4} = 4.25$ square units.

Common errors

Estimating area before integrating

Before the Fundamental Theorem, area was approximated by rectangles - the idea behind the integral itself. Splitting $[a,b]$ into strips and summing rectangle areas gives a numerical estimate. Using the left endpoint of each strip as the height underestimates an increasing function and overestimates a decreasing one; right endpoints do the reverse. The estimate improves as the strips get narrower, and in the limit of infinitely many infinitely thin strips it becomes the exact definite integral. SACE questions sometimes ask for a two- or three-rectangle estimate and then for the exact integral, so you can see how close the approximation is.

Area to a numerical answer

When the antiderivative is awkward, SACE allows the calculator to evaluate a definite integral directly. State the integral first - for instance "Area $=\displaystyle\int_1^7 (\ln(2x-1))^2\,dx$ " - then give the calculator value to the required accuracy. Marks are awarded for the correct integral set-up even when the evaluation is done by technology, so never skip writing the integral with its limits.

Why it matters

Area under a curve is the most common applied use of the definite integral and a frequent exam question. The sign-handling skill carries directly into the next dot point - areas between two curves - and into interpreting accumulated change in modelling tasks.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20231 marksCalculator-assumed. Let

A

be the area bounded by the graph of

y = (\ln(2x - 1))^2

, the

x

-axis, and the vertical lines at

x = 1

and

x = 7

. Write an integral expression for the exact value of

A

Show worked answer →

The exact area between a curve $y = f(x)$ and the $x$ -axis from $x = a$ to $x = b$ , where the curve lies above the axis, is the definite integral of $f(x)$ over $[a, b]$ .

Here $f(x) = (\ln(2x - 1))^2$ is positive on the interval, with $a = 1$ and $b = 7$ , so

A = \int_1^7 (\ln(2x - 1))^2\,dx.

The single mark is for the correct integrand and limits. No evaluation is required.

SACE 20222 marksCalculator-assumed. An overestimate of the area bounded by

f(x) = x e^x

, the

x

-axis, and the lines

x = 1

and

x = 5

is found using two rectangles each 2 units wide. Calculate this overestimate

S

, correct to three decimal places.

Show worked answer →

Because $f(x) = x e^x$ is increasing on $[1, 5]$ , the right-hand endpoint of each strip gives the tallest rectangle, so right endpoints overestimate.

The strips are $[1, 3]$ and $[3, 5]$ , each width $2$ , with right-endpoint heights $f(3) = 3e^3$ and $f(5) = 5e^5$ :

S = 2f(3) + 2f(5) = 6e^3 + 10e^5.

Numerically $6e^3 \approx 120.513$ and $10e^5 \approx 1484.132$ , so $S \approx 1604.645$ (to three decimal places).

Marks: one for right-endpoint heights with width $2$ , one for the value.

SACE 20213 marksCalculator-free. Find the total geometric area between

y = x^2 - 4

and the

x

-axis from

x = 0

x = 3

Show worked answer →

The curve crosses the axis at $x = 2$ (where $x^2 - 4 = 0$ ): it is below the axis on $[0, 2]$ and above on $[2, 3]$ , so split.

Below piece: $\int_0^2 (x^2 - 4)\,dx = \Big[\tfrac{x^3}{3} - 4x\Big]_0^2 = \tfrac{8}{3} - 8 = -\tfrac{16}{3}$ , area $\tfrac{16}{3}$ .

Above piece: $\int_2^3 (x^2 - 4)\,dx = \Big[\tfrac{x^3}{3} - 4x\Big]_2^3 = (9 - 12) - (\tfrac{8}{3} - 8) = -3 + \tfrac{16}{3} = \tfrac{7}{3}$ .

Total area $= \tfrac{16}{3} + \tfrac{7}{3} = \tfrac{23}{3}$ square units.

Marks: one for finding the intercept and splitting, one for each piece evaluated with the below-axis part made positive.