Area Between Two Curves (AI HL)
Once two curves are involved rather than one, area questions become a two-stage job: find where the curves meet, then integrate the gap between them. This page covers the top-minus-bottom formula, when it goes wrong, and how to spot the setup quickly. It's part of the broader Integration topic.
14 questions on this sub-topic.
The two formulas
Covered under IB syllabus reference AHL5.12: "Area of the region enclosed by a curve and the \(x\)- or \(y\)-axis in a given interval, including negative integrals." Treat a curve-and-axis question as the special case where one of the two "curves" is \(y=0\).
Area between two curves
\[A=\int_a^b \big(f(x)-g(x)\big)\,dx\]
\(a\) and \(b\) are the \(x\)-coordinates where the curves meet; \(f(x)\) must be the upper curve on that interval.
Not in the formula booklet - derived from the area formulaArea under a curve
\[A=\int_a^b f(x)\,dx\]
Only gives the true area directly when \(f(x)>0\) throughout \([a,b]\) - otherwise split at the roots first.
✓ In the formula bookletNumerical integration and how your GDC handles both of these formulas is covered on the parent Integration page.
Worked examples
Find the area enclosed between \(y=x^2\) and \(y=2x\).
Worked solution
Intersections: \(x^2 = 2x \Rightarrow x\) M1
\(= 0, 2.\) A1
\(\int_0^2(2x - x^2)\,dx\) M1
\(= [x^2 - \tfrac{x^3}{3}]_0^2\) A1
\(= 4 - \tfrac83 = \tfrac43 \approx 1.33.\) A1
The region between \(y=4-x^2\) and the x-axis is shown.
Find its area.
Worked solution
Curve meets axis at \(x\) M1
\(= \pm 2.\) A1
\(\int_{-2}^{2}(4 - x^2)\,dx\) M1
\(= [4x - \tfrac{x^3}{3}]_{-2}^{2}\) A1
\(= \tfrac{32}{3} \approx 10.7.\) A1
Find the area enclosed between \(y=6-x^2\) and \(y=x\).
Worked solution
Intersections: \(6 - x^2 = x \Rightarrow x^2 + x - 6 = 0 \Rightarrow x\) M1
\(= -3, 2.\) A1
\(\int_{-3}^{2}(6 - x^2 - x)\,dx\) M1
\(= [6x - \tfrac{x^3}{3} - \tfrac{x^2}{2}]_{-3}^{2}\) A1
\(= 7.333 - (-13.5) \approx 20.8.\) A1
Common mistakes
- Subtracting the curves the wrong way round. Area between two curves needs (top curve \(-\) bottom curve). Getting the order backwards gives a negative answer that should have been positive.
- Treating a definite integral as area without checking the sign. If the curve dips below the \(x\)-axis anywhere in the interval, the raw integral gives a signed (net) value, not the true area - split at the roots and take magnitudes first.
- Assuming one curve stays on top the whole way. If the two curves cross somewhere between the given limits, "top" and "bottom" swap there - split the integral at the crossing point and subtract in the opposite order on each piece.
Ready to practise properly?
14 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, then integrate (top curve minus bottom curve) between those limits: \(A=\int_a^b(f(x)-g(x))\,dx.\)
What if the curves swap which one is on top?
Split the integral at the point where they cross, and subtract in the opposite order on each piece, so both pieces come out positive.