Area Between Two Curves (AA SL)
When two curves cross, the region trapped between them is found by integrating the top curve minus the bottom curve across the interval where they overlap. The intersection points supply the limits, so most of the work happens before you even reach the integral. It's part of the broader Integration topic.
10 questions on this sub-topic.
Finding the area
Covered under IB syllabus reference SL5.11: definite integrals and areas between curves, where the enclosed function can be positive or negative. Neither idea below sits in the formula booklet - both are applications of the definite integral you're expected to set up yourself.
Top minus bottom
Once you know which curve lies above the other on the interval \([a,b]\), the enclosed area is \(\displaystyle\int_a^b \big(f(x)-g(x)\big)\,dx\), with \(f(x)\ge g(x)\) throughout.
Finding the limits
Set the two expressions equal, \(f(x)=g(x)\), and solve. The resulting \(x\)-values are almost always your limits of integration - sketch the curves if it isn't obvious which one is on top.
Need the full syllabus wording, GDC steps or the definite-integral notation itself? See Integration and its GDC guidance.
Worked examples
Find the area enclosed between \(y = x^2\) and \(y = 2x\).
Worked solution
\(x^2=2x\Rightarrow x^2-2x=0\Rightarrow x\) M1 \(=0,2.\) A1
On \([0,2]\) the line \(2x\) is above \(x^2\): \(\int_0^2(2x-x^2)\,dx=\Big[x^2-\tfrac{x^3}{3}\Big]_0^2=4-\tfrac83=\tfrac43\) M1 \(\approx 1.33.\) A1
Find the area enclosed between \(y = 6x - x^2\) and \(y = x\).
Worked solution
\(6x-x^2=x\Rightarrow 5x-x^2=0\Rightarrow x(5-x)=0\Rightarrow x\) M1 \(=0,5.\) A1
On \([0,5]\) the parabola is above \(y=x\): \(\int_0^5(5x-x^2)\,dx=\Big[\tfrac{5x^2}{2}-\tfrac{x^3}{3}\Big]_0^5=\tfrac{125}{2}-\tfrac{125}{3}=\tfrac{125}{6}\) M1 \(\approx 20.8.\) A1
Find the area enclosed between the curve \(y = 4 - x^2\) and the \(x\)-axis.
Worked solution
The curve meets the \(x\)-axis where \(y=0\): \(4-x^2=0\Rightarrow x\) M1 \(=\pm 2.\) A1
\(\int_{-2}^{2}(4-x^2)\,dx=\Big[4x-\tfrac{x^3}{3}\Big]_{-2}^{2}=\Big(8-\tfrac83\Big)-\Big(-8+\tfrac83\Big)=\tfrac{32}{3}\) M1 \(\approx 10.7.\) A1
Common mistakes
- Integrating in the wrong order. If you subtract the top curve's expression from the bottom one by mistake, the integral comes out negative - always check which curve has the larger \(y\)-value on the interval first, by sketching or testing a point.
- Skipping the intersection step. The limits of integration for area-between-curves questions come from solving \(f(x)=g(x)\), not from limits given elsewhere in the question - forgetting to find them is the most common way to lose the method mark.
- Treating a negative result as the final answer. Even with the correct order of subtraction, an arithmetic slip can leave a negative value - area itself is always positive, so a negative answer is a signal to recheck the setup, not something to just make positive.
Ready to practise properly?
11 area-between-curves questions, marked instantly like the real exam.
Quick answers
How do you find the area between two curves?
Find where the curves intersect to get the limits of integration, then integrate \(f(x)-g(x)\) between those limits, with \(f(x)\) being the curve on top.
Do I need to know which curve is on top before integrating?
Yes. Sketch or reason out which curve has the larger \(y\)-value on the interval between the intersection points, then subtract the lower curve from the upper one so the integral comes out positive.