A definite integral measures signed area between the curve and the x-axis. Where the curve is above the axis the contribution is positive; where it is below the axis the contribution is negative. So a definite integral can be positive, negative or zero.

When does the definite integral equal the area?

Only when the curve stays on or above the x-axis across the whole interval, that is f(x) is greater than or equal to zero. Then the signed area and the geometric area are the same positive number.

What is the difference between net displacement and total distance?

Net displacement is the signed integral of velocity, so backward motion cancels forward motion. Total distance splits where the velocity changes sign and adds the absolute values, so it is always positive.

A definite integral measures signed area between the curve and the x-axis. Where the curve is above the axis the contribution is positive; where it is below the axis the contribution is negative. So a definite integral can be positive, negative or zero.

When does the definite integral equal the area?

Only when the curve stays on or above the x-axis across the whole interval, that is f(x) is greater than or equal to zero. Then the signed area and the geometric area are the same positive number.

What is the difference between net displacement and total distance?

Net displacement is the signed integral of velocity, so backward motion cancels forward motion. Total distance splits where the velocity changes sign and adds the absolute values, so it is always positive.

Year 12 Methods (Unit 3 & 4) Integration

Signed area

Q: How do you find the total area between a curve and the x-axis?

Find the x-intercepts in the interval, integrate over each piece separately, then add the absolute values of the piece integrals. This stops the below-axis and above-axis pieces from cancelling.

Q: Why can a definite integral be negative?

Because it is signed area, not geometric area. When the curve lies below the x-axis, f(x) is negative, so its integral is negative. That is correct, not an error.

Q: How does the fundamental theorem of calculus evaluate a definite integral?

If F is an antiderivative of f, then the integral from a to b of f equals F(b) minus F(a). Antidifferentiate, substitute the upper and lower limits, and subtract.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), the definite integral \(\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)\) measures signed area: above the \(x\)-axis it counts positively, below the axis it counts negatively. To evaluate an integral, use the fundamental theorem of calculus. To find the total area between a curve and the \(x\)-axis, split at every \(x\)-intercept and add the absolute values of the pieces.

A definite integral \(\displaystyle\int_a^b f(x)\,dx\) is a single number, evaluated by the fundamental theorem of calculus: if \(F'(x)=f(x)\) then the integral equals \(F(b)-F(a)\). Geometrically it is the signed area between the curve \(y=f(x)\) and the \(x\)-axis from \(x=a\) to \(x=b\).

Where the curve is above the axis the region counts as positive; where the curve is below the axis (\(f(x)<0\)) it counts as negative. So a definite integral can come out positive, negative, or exactly zero when the pieces cancel. It equals the ordinary (geometric) area only when \(f(x)\ge 0\) across the whole interval.

The total area enclosed between a curve and the \(x\)-axis is always positive. If the curve crosses the axis inside \([a,b]\), split the interval at every \(x\)-intercept, integrate each piece, and add the absolute values. This is the “find the area” question, as opposed to “evaluate the definite integral”.

Key idea. The definite integral is signed area (below the axis is negative). Total area splits at the intercepts and sums absolute values: \(\text{Area}=\displaystyle\sum\left|\int_{\text{piece}} f\right|\).

\(y=x^2-1\): the below-axis piece \((0,1)\) is negative and partly cancels the above-axis piece \((1,2)\) — signed area vs total area

\(y=x^2-4x\): the whole region lies below the axis, so \(\int_0^4 f<0\) and the area is \(\left|\int_0^4 f\right|\)

The fundamental theorem of calculus (evaluate a definite integral):

\[\int_a^b f(x)\,dx=F(b)-F(a),\qquad F'(x)=f(x)\]

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

Area under a curve that stays above the axis (\(f(x)\ge 0\)):

\[\text{Area}=\int_a^b f(x)\,dx\]

Total area when the curve crosses the axis (split at the intercepts):

\[\text{Area}=\sum_{\text{pieces}}\left|\int_{x_i}^{x_{i+1}} f(x)\,dx\right|\]

A = \sum | \int_{x_{i}}^{x_{i + 1}} f |

Motion. For a velocity \(v(t)\), net displacement is \(\displaystyle\int_a^b v\,dt\) (signed); total distance is \(\displaystyle\int_a^b |v|\,dt\), found by splitting where \(v=0\).

How to handle a signed-area / total-area question

Read the wording. “Evaluate the definite integral” wants a signed value (may be negative or zero). “Find the area” wants a positive number.
Find where the curve meets the axis. Solve \(f(x)=0\) for the \(x\)-intercepts inside \([a,b]\).
Antidifferentiate. Find \(F(x)\) with \(F'(x)=f(x)\); for a plain integral, compute \(F(b)-F(a)\).
For total area, split and take sizes. Integrate over each piece between consecutive intercepts, then add the absolute values.
Check the sign. A piece below the axis should give a negative integral; if the whole region is above the axis, the signed value already is the area.

Symmetry shortcut. For an odd function over a symmetric interval \([-a,a]\), the signed integral is \(0\) (the below-axis part cancels the above-axis part) — but the total area is not zero.

Example 1 — evaluate a definite integral

Evaluate \(\displaystyle\int_0^3 \left(x^2-4x\right)dx\).

Solution

Antidifferentiate, then compute \(F(3)-F(0)\).

\(\displaystyle\int_0^3\!\left(x^2-4x\right)dx\)	\(=\)	\(\left[\dfrac{x^3}{3}-2x^2\right]_0^3\)
	\(=\)	\((9-18)-0=-9\)

Negative, because the region lies below the \(x\)-axis.

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Example 2 — total area below the axis

Find the area between \(y=x^2-4x\) and the \(x\)-axis from \(x=0\) to \(x=4\).

Solution

The whole region is below the axis, so take the absolute value.

\(\displaystyle\int_0^4\!\left(x^2-4x\right)dx\)	\(=\)	\(\dfrac{64}{3}-32=-\dfrac{32}{3}\)
\(\text{Area}\)	\(=\)	\(\left\|-\dfrac{32}{3}\right\|=\dfrac{32}{3}\ \text{u}^2\)

\frac{32}{3}

Example 3 — curve crossing the axis

For \(y=x^2-1\) on \([0,2]\) (crossing at \(x=1\)), compare the definite integral with the total area.

Solution

Split at the intercept \(x=1\).

\(\displaystyle\int_0^2\!\left(x^2-1\right)dx\)	\(=\)	\(\dfrac{8}{3}-2=\dfrac{2}{3}\ (\text{signed})\)
\(\displaystyle\int_0^1,\ \int_1^2\)	\(=\)	\(-\dfrac{2}{3},\ \dfrac{4}{3}\)
\(\text{Area}\)	\(=\)	\(\dfrac{2}{3}+\dfrac{4}{3}=2\ \text{u}^2\)

2

Example 4 — trigonometric case

For \(y=\sin x\) on \([0,2\pi]\), find the definite integral and the total area.

Solution

Use \(\displaystyle\int\sin x\,dx=-\cos x\).

\(\displaystyle\int_0^{\pi}\sin x\,dx,\ \int_{\pi}^{2\pi}\sin x\,dx\)	\(=\)	\(2,\ -2\)
\(\displaystyle\int_0^{2\pi}\sin x\,dx\)	\(=\)	\(2+(-2)=0\)
\(\text{Area}\)	\(=\)	\(\|2\|+\|-2\|=4\ \text{u}^2\)

4

Common pitfalls

A negative integral is not a mistake. When the curve is below the \(x\)-axis, \(\int_a^b f\,dx\) is negative — that is the correct signed value. Only take an absolute value if the question asks for an area.

Do not integrate straight across for total area. If the curve crosses the axis, split at every \(x\)-intercept first; otherwise the below-axis and above-axis pieces partly cancel and you undercount the area.

Symmetry can give zero. For an odd function over \([-a,a]\), \(\int_{-a}^{a} f\,dx=0\); the area is still positive. Do not report \(0\) as the area.

Frequently asked questions

What is signed area?

A definite integral counts area above the \(x\)-axis as positive and area below it as negative, so it can be positive, negative or zero.

How do you find the total area between a curve and the x-axis?

Find the \(x\)-intercepts, integrate over each piece, then add the absolute values so the pieces do not cancel.

Why can a definite integral be negative?

Because it is signed area. Where \(f(x)<0\) the curve is below the axis, so that part of the integral is negative.

When does the integral equal the area?

Only when \(f(x)\ge 0\) across the whole interval; then the signed area and the geometric area are the same.

Net displacement vs total distance?

Net displacement is \(\int v\,dt\) (signed); total distance is \(\int |v|\,dt\), split where \(v=0\).

How does the fundamental theorem evaluate an integral?

If \(F'(x)=f(x)\), then \(\int_a^b f\,dx=F(b)-F(a)\): antidifferentiate, substitute the limits, subtract.

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