The area of a region between two curves
In Year 12 Mathematical Methods (Queensland, QCAA), the area of a region between two curves is \(\displaystyle A=\int_{a}^{b}\big(f(x)-g(x)\big)\,dx\), the integral of the upper curve minus the lower curve. You first solve \(f(x)=g(x)\) to find the limits \(a\) and \(b\), and you split the interval wherever the curves cross so each piece is positive. The area between a curve and the \(x\)-axis is the same rule with \(g(x)=0\).
The region between two curves \(y=f(x)\) and \(y=g(x)\) over \(a\le x\le b\) is the area trapped between them. If \(f(x)\ge g(x)\) on that interval (\(f\) is the upper curve, \(g\) the lower), then
\(\displaystyle A=\int_{a}^{b}\big(f(x)-g(x)\big)\,dx.\)
Each thin strip has height (top \(-\) bottom) \(=f(x)-g(x)\) and width \(dx\); adding the strips gives the integral. Because you always subtract the lower value from the upper value, the answer is positive.
The limits \(a\) and \(b\) are the points of intersection: solve \(f(x)=g(x)\) first. If the curves cross inside the interval, the top and bottom swap over, so the interval must be split at each crossing and (top \(-\) bottom) taken on each piece. The area between a curve and the \(x\)-axis is the same idea with the axis as the line \(y=0\).
Area between two curves, with \(f(x)\ge g(x)\) on \([a,b]\):
If the curves cross at \(x=c\) inside \([a,b]\), split the interval:
Area between a curve and the \(x\)-axis (take \(g(x)=0\); split at each \(x\)-intercept):
How to find the area between two curves
- Find the limits. Solve \(f(x)=g(x)\) to get the \(x\)-values \(a\) and \(b\) where the curves meet.
- Decide which curve is on top. Test a value between the limits (or sketch): the higher curve is \(f\), the lower is \(g\).
- Check for a crossing inside the interval. If the curves cross between \(a\) and \(b\), split there so each piece keeps a fixed top and bottom.
- Integrate (upper \(-\) lower) on each piece. Evaluate \(\int(\text{top}-\text{bottom})\,dx\); every piece comes out positive.
- Add the pieces. The total is the area, in square units.
Solve \(x^{2}=x+2\) for the limits; the line is on top.
| \(x^{2}-x-2\) | \(=\) | \(0\Rightarrow x=-1,\ 2\) |
| \(A\) | \(=\) | \(\displaystyle\int_{-1}^{2}\big((x+2)-x^{2}\big)\,dx\) |
| \(=\) | \(\left[\dfrac{x^{2}}{2}+2x-\dfrac{x^{3}}{3}\right]_{-1}^{2}=\dfrac{9}{2}\) |
Solve for the limits; \(y=4-x^{2}\) is the upper curve.
| \(x^{2}\) | \(=\) | \(4-x^{2}\Rightarrow x=\pm\sqrt{2}\) |
| \(A\) | \(=\) | \(\displaystyle\int_{-\sqrt{2}}^{\sqrt{2}}\big(4-2x^{2}\big)\,dx\) |
| \(=\) | \(\left[4x-\dfrac{2x^{3}}{3}\right]_{-\sqrt{2}}^{\sqrt{2}}=\dfrac{16\sqrt{2}}{3}\) |
The curves cross at \(x=0\), so split; add positive pieces.
| \(A\) | \(=\) | \(\displaystyle\int_{-1}^{0}\!(x^{3}-x)\,dx+\int_{0}^{1}\!(x-x^{3})\,dx\) |
| \(=\) | \(\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\) |
A single \(\int_{-1}^{1}\) would cancel to \(0\).
The \(x\)-intercepts give the limits; the curve is above the axis.
| \(6x-x^{2}\) | \(=\) | \(0\Rightarrow x=0,\ 6\) |
| \(A\) | \(=\) | \(\displaystyle\int_{0}^{6}\big(6x-x^{2}\big)\,dx\) |
| \(=\) | \(\left[3x^{2}-\dfrac{x^{3}}{3}\right]_{0}^{6}=36\) |
Common pitfalls
Frequently asked questions
How do you find the area between two curves?
Integrate the upper curve minus the lower curve between the limits: \(A=\displaystyle\int_{a}^{b}\big(f(x)-g(x)\big)\,dx\), where the limits come from solving \(f(x)=g(x)\).
Why find the points of intersection first?
Because they are the limits of the integral. Solving \(f(x)=g(x)\) locates where the region starts and ends, so it is always the first step.
When do you split the interval?
Whenever the curves cross inside it, because the top and bottom swap. Integrate (top \(-\) bottom) on each piece so each part is positive, then add.
How do you find the area between a curve and the x-axis?
Use the axis as \(y=0\): find the \(x\)-intercepts, split where the curve changes sign, take each piece as positive, and add.
Does it work for exponential and trig curves?
Yes. The upper-minus-lower rule works for a parabola and a line, two exponentials such as \(e^{2x}\) and \(e^{x}\), or a cosine curve and a horizontal line — just integrate the difference.
What is the difference between the signed integral and the area?
The signed integral can be negative or zero because area below the axis counts negative. The true area is always positive, found by splitting at each crossing and adding the absolute value of each piece.