Area Between Two Curves (AA HL)
When two curves cross, the region trapped between them has an area you can find by integrating the gap between them - top curve minus bottom curve - from one intersection point to the next. This page covers the method, worked examples, and the mistakes that cost marks. It's part of the broader Integration topic.
10 questions on this sub-topic.
The method
Covered under IB syllabus reference SL5.11: definite integrals, including the analytical approach \(\int_a^b g'(x)\,dx = g(b)-g(a)\), and areas between curves. There's no separate formula-booklet entry for this - it's a two-step technique built on the definite integral.
Areas between curves
Integrate the difference of the two functions, top minus bottom, between their points of intersection.
Not in the formula booklet - techniqueArea below the axis
Where a curve dips below the \(x\)-axis, the definite integral over that interval comes out negative. Take the modulus of that piece to get a genuine area.
Not in the formula booklet - reasoning stepNeed the full syllabus wording and formula-booklet reference table? See Integration.
Worked examples
A parabola \(y=x^2-4x\) and a line \(y=2x\) intersect at two points. Determine the area of the finite region enclosed between the parabola and the line.
Worked solution
\(x^2-4x = 2x \Rightarrow x^2-6x=0 \Rightarrow x(x-6)\) M1 \(=0 \Rightarrow x=0,6.\) A1
on \((0,6),\) so \(\displaystyle\int_0^6\big[2x-(x^2-4x)\big]\,dx = \int_0^6(6x-x^2)\,dx = \left[3x^2-\tfrac{x^3}{3}\right]_0^6\) M1
\(= 108-72\) A1 \(= 36.\)
The curves \(y=\sqrt{x}\) and \(y=x^2\) bound a finite region in the first quadrant. Work out the area of this region.
Worked solution
\(\sqrt{x}=x^2 \Rightarrow x=x^4 \Rightarrow x(x^3-1)\) M1 \(=0 \Rightarrow x=0,1.\) A1
is \(y=\sqrt{x}\) on \((0,1).\) M1
\(\displaystyle\int_0^1(x^{1/2}-x^2)\,dx = \left[\tfrac{2}{3}x^{3/2}-\tfrac{x^3}{3}\right]_0^1\) A1
\(= \tfrac23-\tfrac13\) A1 \(= \tfrac13.\)
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
\(= 4 - \tfrac83\) A1
\(= \tfrac43 \approx 1.33.\) A1
Common mistakes
- Skipping straight to integrating without finding the intersections first. The intersection points are your limits of integration - without solving for them, you don't know where the enclosed region begins and ends.
- Integrating bottom minus top instead of top minus bottom. Subtracting in the wrong order gives a negative area - always check which curve lies above the other on the interval before setting up the integral.
- Not taking the modulus when a curve dips below the axis. A definite integral over a region where \(f(x)<0\) comes out negative - if the question asks for an area, take the modulus of that piece before adding it to the rest.
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, then integrate the top curve minus the bottom curve between those intersection points.
Why do I need to find intersection points first?
The intersection points give the limits of integration - without them you don't know where the enclosed region starts and ends, or which curve is on top.