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Year 12 Maths Advanced (2027) Integral calculus

Areas: Between a Curve and the Axes

20 practice questions 0 video lessons Theory + worked examples
NSW · Year 12 Mathematics Advanced · Integral calculus

Areas between a curve and the axes are found with a definite integral, taking the size of any part below the axis so the area is never negative.

Part of the NSW Year 12 Mathematics Advanced course, in the Calculus area of study (Integral calculus focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.

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Theory

The area between a curve and the \(x\)-axis is a definite integral. This Year 12 Mathematics Advanced topic (MAV-12-05) finds areas, taking the size of any part below the axis.

For \(f(x)\ge0\) on \([a,b]\), Area \(=\displaystyle\int_a^b f(x)\,dx\). Where the region is below the axis the integral is negative, so take its absolute value.

If the curve crosses the axis, split at the intercept and add the sizes of the pieces.

Area under a parabolaThe shaded area under y = x squared from 0 to 3 is 9. xy y=x^2 3
Area under \(y=x^{2}\) from \(0\) to \(3\) is \(9\).
\[\text{Area}=\int_a^b f(x)\,dx\quad(f\ge0)\]
area equals the integral from a to b of f of x dx when f is non negative

Method

  1. Integrate \(f\) between the limits.
  2. Take the size of any negative piece (below the axis).
  3. Add the pieces if the curve crosses the axis.
Example 1 — Under a parabola
Find the area under \(y=x^{2}\) from \(0\) to \(3\).
Solution
\(\ \)\(=\)\(\big[\tfrac{x^{3}}{3}\big]_0^3=9\)
Example 2 — Under a line
Find the area under \(y=2x\) from \(0\) to \(4\).
Solution
\(\ \)\(=\)\(\big[x^{2}\big]_0^4=16\)
Example 3 — A strip
Find the area under \(y=x^{2}\) from \(1\) to \(2\).
Solution
\(\ \)\(=\)\(\big[\tfrac{x^{3}}{3}\big]_1^2=\tfrac73\)
Example 4 — Below the axis
Find the area between \(y=-x\) and the axis from \(0\) to \(2\).
Solution
\(\int_0^2(-x)dx\)\(=\)\(-2\)
\(\text{Area}\)\(=\)\(2\)

Common pitfalls

Below the axis is negative — take its size.
Split at intercepts and add the pieces.
Area is never negative.

Frequently asked questions

How do you find the area under a curve?

Integrate the function between the two x-values. For a curve above the axis, the definite integral is the area.

What if the curve is below the x-axis?

The integral will be negative, so take its absolute value to get the area.

What if the curve crosses the x-axis?

Split the interval at the intercept, find each piece's area separately, and add the sizes.

What is the area under y = x squared from 0 to 3?

It is the integral, x cubed over 3 evaluated from 0 to 3, which is 9.