Areas: Between a Curve and the Axes
Learn areas between a curve and the axes for NSW Year 12 Mathematics Advanced (NESA outcome MAV-12-05). A definite integral counts area above the \(x\)-axis as positive and area below it as negative, so finding a true area needs care wherever a curve crosses the axis.
You will learn to find the \(x\)-intercepts, split the interval at sign changes, take the absolute value of any part below the axis, find areas against the \(y\)-axis too, and use odd and even symmetry to save work — the method behind every area question.
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.
For a region between a curve and the \(y\)-axis, write \(x\) in terms of \(y\) and use Area \(=\displaystyle\int_c^d x\,dy\) between the lines \(y=c\) and \(y=d\), again taking the size of any part to the left of the axis.
Note: areas enclosed between two curves are studied in Mathematics Extension 1, not in this course.
Method
- Integrate \(f\) between the limits.
- Take the size of any negative piece (below the axis).
- Add the pieces if the curve crosses the axis.
| \(\ \) | \(=\) | \(\big[\tfrac{x^{3}}{3}\big]_0^3=9\) |
| \(\ \) | \(=\) | \(\big[x^{2}\big]_0^4=16\) |
| \(\ \) | \(=\) | \(\big[\tfrac{x^{3}}{3}\big]_1^2=\tfrac73\) |
| \(\int_0^2(-x)dx\) | \(=\) | \(-2\) |
| \(\text{Area}\) | \(=\) | \(2\) |
Common pitfalls
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.