Area of a region between two curves
Find the area between two curves in Year 12 Specialist Mathematics for Queensland (QCAA). The area of a region enclosed by two graphs is the integral of the upper curve minus the lower curve, taken between the points where the curves intersect.
You will learn to find intersection points to set the limits, decide which curve is on top, and split the integral wherever the curves cross — a key application of integral calculus that also underpins volumes of revolution and later modelling.
Theory
The area between two curves is a core application of integral calculus in Year 12 Specialist Mathematics (QCAA, Queensland). You find where the curves intersect to fix the limits, decide which graph is the upper curve, then integrate the upper minus the lower curve. Where the curves cross inside the interval, you split the integral and add the pieces.
The area between two curves \(y=f(x)\) and \(y=g(x)\) is the area of the region they enclose. On an interval where \(f\) lies above \(g\), that area is \(\displaystyle\int_a^b\big(f(x)-g(x)\big)\,dx\) — the integral of the upper curve minus the lower curve. Because you subtract lower from upper, the integrand is never negative, so the answer is a genuine area.
The limits \(a\) and \(b\) are usually the \(x\)-coordinates of the intersection points, found by solving \(f(x)=g(x)\). These are the two ends of the enclosed region, so you rarely need extra information to set them.
Deciding which curve is upper matters: test an \(x\)-value between the intersections (or read it off a sketch). The curve with the larger \(y\) there is the upper one. Getting this backwards makes the integral negative.
If the curves cross inside the interval, the upper and lower curves swap over. Then you split the integral at each interior intersection, integrate each piece with its own upper curve, and add the absolute areas. When a region is described by \(x\) as a function of \(y\), integrate with respect to \(y\) instead: \(\displaystyle\int_c^d\big(\text{right}-\text{left}\big)\,dy\).
For an interval \([a,b]\) on which \(f(x)\ge g(x)\), the area enclosed between the curves is the integral of the upper minus the lower:
The limits are the intersection abscissae, found by solving:
If the curves cross at interior points \(c_1 For a region bounded by curves written as \(x\) in terms of \(y\), integrate with respect to \(y\) between the \(y\)-limits, right curve minus left:
Finding the area between two curves
- Find the intersections: solve \(f(x)=g(x)\) to get the \(x\)-values where the curves meet — these are the limits \(a\) and \(b\).
- Decide which is upper: test a point between the intersections; the curve with the larger \(y\) is the upper one.
- Set up upper minus lower: write \(A=\displaystyle\int_a^b\big(\text{upper}-\text{lower}\big)\,dx\) and simplify the integrand.
- Integrate and evaluate: anti-differentiate, substitute the limits, and do the arithmetic. If the curves cross inside \([a,b]\), split at each interior intersection and add the absolute pieces.
Set the curves equal to find the limits:
| \(x^2+1\) | \(=\) | \(x+3\) |
| \(x^2-x-2\) | \(=\) | \(0\) |
| \((x-2)(x+1)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-1,\ 2\) |
Test \(x=0\) to see which curve is upper:
| \(\text{line: } y\) | \(=\) | \(0+3=3\) |
| \(\text{parabola: } y\) | \(=\) | \(0+1=1\) |
| \(\Rightarrow\ \text{upper}\) | \(\text{ is }\) | \(y=x+3\) |
Integrate upper minus lower over \([-1,2]\):
| \(A\) | \(=\) | \(\int_{-1}^{2}\big((x+3)-(x^2+1)\big)\,dx\) |
| \(=\) | \(\int_{-1}^{2}(x+2-x^2)\,dx\) | |
| \(=\) | \(\left[\dfrac{x^2}{2}+2x-\dfrac{x^3}{3}\right]_{-1}^{2}\) | |
| \(=\) | \(\left(2+4-\dfrac{8}{3}\right)-\left(\dfrac{1}{2}-2+\dfrac{1}{3}\right)\) | |
| \(=\) | \(\dfrac{10}{3}-\left(-\dfrac{7}{6}\right)\) | |
| \(=\) | \(\dfrac{9}{2}\) |
The area is \(\dfrac{9}{2}\) square units.
Set the curves equal to find the limits:
| \(8-x^2\) | \(=\) | \(x^2\) |
| \(2x^2\) | \(=\) | \(8\) |
| \(x^2\) | \(=\) | \(4\) |
| \(x\) | \(=\) | \(\pm 2\) |
Test \(x=0\) for the upper curve:
| \(8-x^2\) | \(=\) | \(8\) |
| \(x^2\) | \(=\) | \(0\) |
| \(\Rightarrow\ \text{upper}\) | \(\text{ is }\) | \(y=8-x^2\) |
Integrate upper minus lower over \([-2,2]\):
| \(A\) | \(=\) | \(\int_{-2}^{2}\big((8-x^2)-x^2\big)\,dx\) |
| \(=\) | \(\int_{-2}^{2}(8-2x^2)\,dx\) | |
| \(=\) | \(\left[8x-\dfrac{2x^3}{3}\right]_{-2}^{2}\) | |
| \(=\) | \(\left(16-\dfrac{16}{3}\right)-\left(-16+\dfrac{16}{3}\right)\) | |
| \(=\) | \(32-\dfrac{32}{3}\) | |
| \(=\) | \(\dfrac{64}{3}\) |
The area is \(\dfrac{64}{3}\) square units.
Find every intersection:
| \(x^3\) | \(=\) | \(9x\) |
| \(x^3-9x\) | \(=\) | \(0\) |
| \(x(x-3)(x+3)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-3,\ 0,\ 3\) |
On \((0,3)\) test \(x=1\): \(9(1)=9>1=1^3\), so \(y=9x\) is upper there; the two lobes are equal by symmetry, so integrate one and double:
| \(A\) | \(=\) | \(2\int_{0}^{3}(9x-x^3)\,dx\) |
| \(=\) | \(2\left[\dfrac{9x^2}{2}-\dfrac{x^4}{4}\right]_{0}^{3}\) | |
| \(=\) | \(2\left(\dfrac{81}{2}-\dfrac{81}{4}\right)\) | |
| \(=\) | \(2\times\dfrac{81}{4}\) | |
| \(=\) | \(\dfrac{81}{2}\) |
The total area is \(\dfrac{81}{2}\) square units.
Set the curves equal in \(y\) to find the \(y\)-limits:
| \(y^2\) | \(=\) | \(y+2\) |
| \(y^2-y-2\) | \(=\) | \(0\) |
| \((y-2)(y+1)\) | \(=\) | \(0\) |
| \(y\) | \(=\) | \(-1,\ 2\) |
Test \(y=0\) for the right-hand curve:
| \(x=y+2\) | \(=\) | \(2\) |
| \(x=y^2\) | \(=\) | \(0\) |
| \(\Rightarrow\ \text{right}\) | \(\text{ is }\) | \(x=y+2\) |
Integrate right minus left over \([-1,2]\):
| \(A\) | \(=\) | \(\int_{-1}^{2}\big((y+2)-y^2\big)\,dy\) |
| \(=\) | \(\left[\dfrac{y^2}{2}+2y-\dfrac{y^3}{3}\right]_{-1}^{2}\) | |
| \(=\) | \(\left(2+4-\dfrac{8}{3}\right)-\left(\dfrac{1}{2}-2+\dfrac{1}{3}\right)\) | |
| \(=\) | \(\dfrac{10}{3}-\left(-\dfrac{7}{6}\right)\) | |
| \(=\) | \(\dfrac{9}{2}\) |
The area is \(\dfrac{9}{2}\) square units.
Common pitfalls
Frequently asked questions
How do you find the area between two curves?
Solve \(f(x)=g(x)\) to find the intersection points, which give the limits. Decide which curve is upper, then evaluate \(\displaystyle\int_a^b(\text{upper}-\text{lower})\,dx\).
How do you know which curve is the upper one?
Test an \(x\)-value between the intersections (or read it from a sketch). The curve with the larger \(y\) there is the upper curve, which you subtract the lower one from.
What if the two curves cross between the limits?
The upper and lower curves swap where they cross. Split the integral at each interior intersection, integrate each piece with its own upper curve, and add the absolute values of the pieces.
What are the limits of integration when finding area between curves?
They are the \(x\)-coordinates of the points where the curves meet, found by solving \(f(x)=g(x)\). These are the ends of the enclosed region.
When should you integrate with respect to y?
When the curves are naturally written as \(x\) in terms of \(y\) (for example \(x=y^2\)), integrate right curve minus left curve with respect to \(y\), between the \(y\)-values where they meet.
Why subtract the lower curve instead of finding two separate areas?
Subtracting lower from upper gives the height of the strip between the curves directly, so \(\int_a^b(f-g)\,dx\) is the enclosed area in one integral — even where part of the region is below the \(x\)-axis.