Skip to main content
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 4.78 min answer

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

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. When the curve is above the axis
  3. When the curve is below the axis
  4. When the curve crosses the axis
  5. Common errors
  6. Why it matters

What this dot point is asking

The definite integral ∫abf(x) dx\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)≥0f(x)\ge 0 on [a,b][a,b], the area between the curve and the x-axis is simply the integral:

When the curve is below the axis

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

A=∣∫abf(x) dx∣.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:

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

Common errors

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.

2023 SACE Stage 21 marksLet 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, the left limit is a = 1 and the right limit is b = 7, so:

A = integral from 1 to 7 of (ln(2x - 1))^2 dx.

The single mark is for the correct integral with the correct integrand and the limits 1 and 7. No evaluation is required - this part only tests expressing an area as a definite integral.

2017 SACE Stage 22 marksAn 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 produce an overestimate.

The two strips are [1, 3] and [3, 5], each width 2. Their right-endpoint heights are f(3) = 3e^3 and f(5) = 5e^5.

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

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

Marks: one for using right-endpoint heights with the correct width of 2, and one for the correct value. The rectangles overestimate because each one rises to the curve's highest point on its strip.