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A Math · O-Level · Integration · Exam Technique

How to Find the Area Between Two Curves Using Integration

1 August 2026 · Dojo Education · 5 min read

How to Find the Area Between Two Curves Using Integration

Area between two curves is one of the most predictable question types in Paper 2 of O-Level Additional Mathematics (4049). It looks intimidating because the diagram is often shaded and messy, but the method is almost always the same four steps. Once you drill it, these questions become some of the fastest marks on the paper.

The Core Idea

If one graph sits above another between two x-values, the area of the region trapped between them is:

Area = ∫ from a to b of (top curve minus bottom curve) dx

That is it. The only real skills being tested are:

  1. Finding the limits a and b, usually the points of intersection.
  2. Deciding which function is on top.
  3. Integrating correctly.
  4. Substituting limits without sign errors.

Step by Step With a Worked Example

Find the area of the region enclosed by the curve y = x² and the line y = x + 2.

Step 1: Find the Points of Intersection

Set them equal:

x² = x + 2 x² − x − 2 = 0 (x − 2)(x + 1) = 0 x = −1 or x = 2

These are your limits. Never guess limits from the diagram. Examiners award a method mark for correctly solving the simultaneous equations.

Step 2: Decide Which Is on Top

Pick any x value between −1 and 2, say x = 0. The line gives y = 2, the curve gives y = 0. The line is above the curve, so the integrand is (line − curve).

Step 3: Set Up and Integrate

Area = ∫ from −1 to 2 of (x + 2 − x²) dx = [x²/2 + 2x − x³/3] from −1 to 2

Step 4: Substitute Carefully

At x = 2: 2 + 4 − 8/3 = 10/3 At x = −1: 1/2 − 2 + 1/3 = −7/6

Area = 10/3 − (−7/6) = 27/6 = 4.5 units²

Notice how the second bracket is subtracted as a whole. Losing the bracket is the single most common careless error we see in marked scripts.

When the Curves Cross in the Middle

If the two graphs swap positions inside your interval, you cannot use one integral. Consider y = x³ and y = x. They meet at x = −1, 0 and 1. Between −1 and 0 the cubic is above the line, and between 0 and 1 the line is above the cubic.

Split it:

Area = ∫ from −1 to 0 of (x³ − x) dx + ∫ from 0 to 1 of (x − x³) dx = 1/4 + 1/4 = 0.5 units²

If you had integrated straight through from −1 to 1, the two halves would cancel and you would write 0, which earns almost nothing. Always sketch or test a value in each sub-interval.

Integrating With Respect to y

Some regions are far easier with horizontal strips. Take the curve y² = x and the line y = x − 2.

Rewrite both in terms of y: x = y² and x = y + 2. Solve y² = y + 2 to get y = −1 and y = 2.

Area = ∫ from −1 to 2 of (y + 2 − y²) dy = 4.5 units²

Doing this one in terms of x would force you to split the region into two parts because of the two branches of the square root. Look for this whenever the boundary curve is written as y² = something, or when the region is bounded on the left and right rather than top and bottom.

Shortcuts Examiners Accept

When one boundary is a straight line, you can often replace an integral with the area of a triangle or trapezium.

This is fully acceptable in the mark scheme and is usually faster. Just make sure you state clearly what you are subtracting from what.

Common Mistakes to Avoid

A Quick Checklist for the Exam

  1. Sketch or label the diagram, mark the intersections.
  2. Solve simultaneously for the limits.
  3. Test one x value to confirm which graph is on top.
  4. Write the full integral with limits before integrating. This secures the method mark even if the arithmetic slips.
  5. Substitute with brackets, simplify to an exact fraction, and state units².

Practise five of these back to back from past papers and you will start recognising the setup within seconds of reading the question.

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