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Area Under a Curve

Q: Why can a definite integral be negative if area is always positive?

The integral measures SIGNED area — positive above the x-axis, negative below. The physical area is the magnitude (modulus) of that signed value.

Imagine slicing the region under a curve into thousands of impossibly thin rectangles and adding up all their areas. That is exactly what a definite integral computes — and it is why $\int_{a}^{b} f (x) d x$ gives the EXACT area under $y = f (x)$ between $x = a$ and $x = b$ .

32 min Video by TLMaths Integration

The big picture

This lesson is where integration earns its geometric meaning: the definite integral IS the area under a curve, found exactly rather than by an approximation of rectangles. But this only works cleanly above the $x$ -axis — below it, the "area" the integral returns is negative, and between two curves you must subtract one integral from another. Mastering all three cases (single curve, negative region, area between curves) is essential for exam questions.

A-Level Maths H3-02: Finding the Exact Area under a CurveWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Use a definite integral to find the exact area between a curve and the x-axis
Recognise that area below the x-axis gives a negative integral value
Find the area between two curves by subtracting integrals
Handle regions that cross the x-axis by splitting the integral

Area under a curve (above the x-axis)

When $f (x) \geq 0$ throughout $[a, b]$ , the definite integral $\int_{a}^{b} f (x) d x$ gives EXACTLY the area enclosed between the curve, the $x$ -axis, and the vertical lines $x = a$ and $x = b$ .

1Set up the integral:

\int_{0}^{3} x^{2} d x

2Antiderivative:

\frac{x ^{3}}{3}

3Evaluate:

\frac{27}{3} - 0 = 9

AnswerArea

= 9

square units

Tip — Always sketch the curve first (even roughly) — it instantly reveals whether the region is entirely above the axis or dips below, which changes your method.

Regions below the x-axis

When the curve dips below the $x$ -axis over an interval, the definite integral there comes out NEGATIVE — but a physical area is always positive. To find the true (positive) area, take the modulus of the negative integral value.

1Set up the integral:

\int_{- 2}^{2} (x^{2} - 4) d x

2Antiderivative:

\frac{x ^{3}}{3} - 4 x

3Evaluate:

(\frac{8}{3} - 8) - (- \frac{8}{3} + 8) = (- \frac{16}{3}) - (\frac{16}{3}) = - \frac{32}{3}

4The integral is negative because the region is below the axis; take the modulus for the actual area.

AnswerArea

= \frac{32}{3}

square units

Why this works

A negative integral value is not a mistake — it correctly signals that the region lies below the axis. The SIGN carries information; the AREA is always its magnitude.

Area between two curves

To find the area enclosed between two curves, integrate the DIFFERENCE between the upper and lower functions over the interval where they enclose a region — the parts where each curve individually dips below the axis take care of themselves automatically.

A = \int_{a}^{b} [f (x) - g (x)] d x, f (x) \geq g (x) on [a, b]

Upper curve minus lower curve.

1Set up:

\int_{- 1}^{2} [(x + 2) - x^{2}] d x = \int_{- 1}^{2} (x + 2 - x^{2}) d x

2Antiderivative:

\frac{x ^{2}}{2} + 2 x - \frac{x ^{3}}{3}

3At

x = 2

2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}

4At

x = - 1

\frac{1}{2} - 2 + \frac{1}{3} = - \frac{7}{6}

5Subtract:

\frac{10}{3} - (- \frac{7}{6}) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6}

Answer

\frac{27}{6} = \frac{9}{2}

square units

Think like an examiner

Always sketch the region (or at least identify where the curve is above/below the axis) before setting up the integral — this determines whether you need a modulus.
For "area between two curves", find the intersection points first (by setting the two functions equal) to establish the correct limits.
State your final area as a positive value, taking the modulus of any negative integral result.

Common misconceptions

Reporting a negative integral value as the final "area".

Area is always positive — take the modulus of a negative integral result.

Subtracting the curves the wrong way round (lower minus upper).

Always subtract the LOWER function from the UPPER function:

\int (f_{upper} - f_{lower}) d x

Using the wrong limits for "area between two curves" without finding the intersection points first.

Set the two functions equal to each other and solve to find where the enclosed region actually starts and ends.

Area formulas

A = \int_{a}^{b} f (x) d x

Area above the x-axis.

A = \int_{a}^{b} [f (x) - g (x)] d x

Area between two curves, f ≥ g.

Stretch yourself

Find the total area enclosed between the curve $y = x^{3} - 4 x$ and the x-axis, between $x = - 2$ and $x = 2$ .

Hint — The curve crosses the x-axis at $x = - 2, 0, 2$ within this range, changing sign — split the integral at these crossing points and take the modulus of each part before adding.

Questions students ask

Why can a definite integral be negative if area is always positive?

The integral measures SIGNED area — positive above the

x

-axis, negative below. The physical area is the magnitude (modulus) of that signed value.

How do I find the area between two curves?

Integrate the difference (upper function minus lower function) between their points of intersection, which you find by setting the two functions equal to each other.

What if a region crosses the x-axis part-way through?

Split the integral at every point where the curve crosses the axis, evaluate each part separately, take the modulus of each result, then add them together — do not just integrate across the whole interval in one go.

Key takeaways

A definite integral gives the SIGNED area under a curve: positive above the axis, negative below.
For the true (positive) area, take the modulus of a negative result.
Area between two curves: integrate (upper − lower) between their intersection points.
If a region crosses the axis, split the integral at each crossing point first.

How this fits the course

Build on

Definite Integrals

Integration by Substitution

Leads to

Integration by Substitution

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