What are wrong top and bottom for area between curves?

Picking the wrong top yields a negative answer. The fix: test a value inside the interval and pick the larger.

What is sign error in subtraction?

F(b) - F(a) requires brackets when F has multiple terms. Sign errors here cascade.

VICMath MethodsSyllabus dot point

How is the definite integral used to compute the area under a single curve and the area between two curves?

The use of definite integrals to find the area between a curve and the $x$ -axis, and the area between two curves on a closed interval, including handling sign changes of the integrand

A focused answer to the VCE Math Methods Unit 4 key-knowledge point on areas via integration. Covers area under a curve (single function), area between two curves (top minus bottom), the sign-change handling that is the most common Paper 1 trap, and the calculator-active extensions in Paper 2.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-15

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Area under a curve (single function)
Area between two curves
Calculator-active area problems (Paper 2)
Examples in context
Try this

What this dot point is asking

VCAA wants you to compute geometric areas using definite integrals: the area between a curve and the $x$ -axis, and the area between two curves. The dot point sits at the intersection of antidifferentiation skill (Unit 4 Topic 1) and graphical interpretation (Unit 3). Sign-change handling on the interval is the most heavily marked detail.

Area under a curve (single function)

The signed area between $y = f(x)$ , the $x$ -axis, and the vertical lines $x = a$ and $x = b$ is:

\int_{a}^{b} f(x) \, dx

This integral is positive where the curve is above the axis, negative where below.

If you want the geometric area (always non-negative):

Find where $f$ changes sign on $[a, b]$ . Solve $f(x) = 0$ to get the roots; check which lie inside $(a, b)$ .
Split the interval at each sign-change root. Suppose $f$ changes sign at $r_1 < r_2 < \ldots < r_k$ in $(a, b)$ .
Compute each piece separately.

\text{Area} = \int_{a}^{r_1} f \, dx + \left\lvert \int_{r_1}^{r_2} f \, dx \right\rvert + \left\lvert \int_{r_2}^{r_3} f \, dx \right\rvert + \ldots

Each negative piece is replaced by its absolute value before summation.

Alternative compact form: $\text{Area} = \int_{a}^{b} \lvert f(x) \rvert \, dx$ . This is true mathematically but VCAA Paper 1 expects the interval-split method.

Why splitting matters

If you compute $\int_{a}^{b} f \, dx$ without splitting and the curve crosses zero, positive and negative regions cancel. The arithmetic result is the net signed area, not the geometric area.

Example. $\int_{-1}^{1} x \, dx = 0$ . The graph passes through the origin; the negative region on $(-1, 0)$ cancels the positive region on $(0, 1)$ . Net signed area is $0$ . Geometric area is $\frac{1}{2} + \frac{1}{2} = 1$ .

Area between two curves

If $f(x) \geq g(x)$ on $[a, b]$ , the area between the two curves is:

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

The rule of thumb: top curve minus bottom curve.

Procedure

Find the intersection points. Solve $f(x) = g(x)$ . These set the integration limits if the question asks for "the enclosed area".
Identify which curve is on top in the interval(s) between intersections. Pick a test value inside the interval and evaluate both functions; the larger is "top".
Set up $\int [\text{top} - \text{bottom}] \, dx$ over the interval.
Evaluate.

If the top and bottom swap on the interval

The curves may cross more than once in the interval of interest. Each sub-interval needs its own setup with the correct top and bottom.

Example. $y = \sin(x)$ and $y = \cos(x)$ on $[0, \pi]$ . They cross at $x = \pi/4$ . On $[0, \pi/4]$ , $\cos > \sin$ ; on $[\pi/4, \pi]$ , $\sin > \cos$ (well, until $\sin = 0$ at $\pi$ , where $\cos = -1$ ; need to check). For a question over the full $[0, \pi]$ , split at $\pi/4$ and apply top-minus-bottom in each piece.

Area between a curve and the $x$ -axis as a special case

If $g(x) = 0$ (the $x$ -axis), the formula reduces to $\int [f(x) - 0] \, dx = \int f(x) \, dx$ when $f \geq 0$ . This recovers the single-curve case.

Area problems: polynomial crossing the axis and two curves

We work through two representative area problems to show the full procedure: a single curve that crosses the axis (requiring an interval split), then two curves with an intersection point.

Step 1: Find the roots and split the interval (polynomial crossing)

For the area between $y = x^3 - 4x$ and the $x$ -axis on $[-2, 2]$ , begin by finding where the curve crosses zero, because the integrand changes sign there. Factorising gives $x^3 - 4x = x(x^2 - 4) = x(x-2)(x+2)$ , so the roots are at $-2, 0, 2$ . The root $x = 0$ lies inside the interval, so we must split at $x = 0$ .

Step 2: Determine the sign on each sub-interval

Knowing the sign tells us whether the definite integral over each piece is positive or negative, and therefore whether we need an absolute value. On $(-2, 0)$ , test $x = -1$ : $-1 + 4 = 3 > 0$ (curve above axis). On $(0, 2)$ , test $x = 1$ : $1 - 4 = -3 < 0$ (curve below axis).

Step 3: Antidifferentiate and evaluate each piece (polynomial)

Find an antiderivative: $F(x) = \frac{x^4}{4} - 2 x^2$ .

On $[-2, 0]$ : $F(0) - F(-2) = 0 - (4 - 8) = 4$ . Positive, so area is $4$ .

On $[0, 2]$ : $F(2) - F(0) = (4 - 8) - 0 = -4$ . Negative, so take the absolute value: area is $|-4| = 4$ .

Total area: $4 + 4 = 8$ .

Step 4: Find the intersection points for the area between two curves

For the area enclosed between $y = x^2$ and $y = x + 2$ , the limits of integration come from the intersection points. Solve $x^2 = x + 2$ , giving $x^2 - x - 2 = 0$ , which factors as $(x - 2)(x + 1) = 0$ , so $x = -1$ or $x = 2$ .

Step 5: Identify top and bottom, then integrate

On $(-1, 2)$ , test $x = 0$ : $x^2 = 0$ and $x + 2 = 2$ , so $y = x + 2$ is the top curve. The area is:

A = \int_{-1}^{2} [(x + 2) - x^2] \, dx

Antidifferentiate: $\int (x + 2 - x^2) \, dx = \frac{x^2}{2} + 2x - \frac{x^3}{3}$ .

Evaluate at $x = 2$ : $2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}$ .

Evaluate at $x = -1$ : $\frac{1}{2} - 2 + \frac{1}{3} = -\frac{7}{6}$ .

Subtract: $\frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$ .

Final answer: area enclosed between the two curves is $\dfrac{9}{2}$ .

Calculator-active area problems (Paper 2)

For non-elementary intersection points or messy antiderivatives, Paper 2 expects:

Solve the intersection equation numerically on the calculator.
Identify top and bottom by evaluating at a test point.
Use the calculator's definite integral function with the intersection points as limits.

The structural reasoning (top minus bottom, interval splitting) is the same; only the arithmetic moves to the calculator.

Common errors

Not splitting at the sign-change root: Computing $\int_{a}^{b} f \, dx$ for a function that crosses zero and reporting the result as "area" gives the wrong (smaller, possibly zero) answer.
Wrong top and bottom for area between curves: Picking the wrong top yields a negative answer. The fix: test a value inside the interval and pick the larger.
Forgetting to find intersection points: If the question asks for the "enclosed area" between two curves without giving limits, the intersection points provide the limits.
Mixing signed area and geometric area: "The value of $\int f \, dx$ " is the signed area; "the area" is geometric (non-negative). Read the question carefully.
Sign error in subtraction: $F(b) - F(a)$ requires brackets when $F$ has multiple terms. Sign errors here cascade.
Using $|F(b) - F(a)|$ instead of splitting: Taking the absolute value of the result of a single integral is wrong if the integrand changes sign on the interval. The absolute value must go on each sub-integral (or you must integrate $|f|$ piecewise).

Examples in context

Example 1. Cross-section of a garden bed. A raised garden bed's cross-section is bounded above by $y = 4 - x^2$ and below by the $x$ -axis (metres). The curve meets the axis at $x = \pm 2$ . The cross-sectional area is $\int_{-2}^{2}(4 - x^2)\,dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3} \approx 10.67 \text{ m}^2$ .

Example 2. Region between two paths. A landscape design encloses a region between the curve $y = 6x - x^2$ and the line $y = 2x$ (metres). They meet where $6x - x^2 = 2x$ , i.e. $x^2 - 4x = 0$ , so $x = 0$ and $x = 4$ . On $(0, 4)$ the curve $6x - x^2$ is on top (at $x = 2$ : $8 > 4$ ), so the area is $\int_0^4(6x - x^2 - 2x)\,dx = \int_0^4(4x - x^2)\,dx = \left[2x^2 - \frac{x^3}{3}\right]_0^4 = 32 - \frac{64}{3} = \frac{32}{3} \approx 10.67 \text{ m}^2$ .

Try this

Q1. Find the area between $y = x^2$ and the $x$ -axis on $[0, 3]$ . [2 marks]

Cue. $\int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = 9$ .

Q2. Find the total area between $y = x^2 - 1$ and the $x$ -axis on $[0, 2]$ . [4 marks]

Cue. Crosses at $x = 1$ ; $\left|\int_0^1(x^2 - 1)\,dx\right| + \int_1^2(x^2 - 1)\,dx = \frac{2}{3} + \frac{4}{3} = 2$ .

Q3. Find the area enclosed between $y = 4 - x^2$ and $y = x + 2$ . [4 marks]

Cue. Intersect at $x = -2, 1$ ; top is $4 - x^2$ ; $\int_{-2}^{1}(2 - x - x^2)\,dx = \frac{9}{2}$ .

Exam-style practice questions

Practice questions written in the style of VCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 VCAA Paper 14 marksThe graph of

y = x^2 - 2x

intersects the

x

-axis at

x = 0

and

x = 2

. Find the exact area enclosed between the curve and the

x

-axis on the interval

[0, 3]

Show worked answer →

The curve is below the $x$ -axis on $(0, 2)$ and above on $(2, 3)$ . Total area requires splitting the integral and taking the absolute value of the negative piece.

On $[0, 2]$ : the curve is below the axis, so the area is $\left\lvert \int_{0}^{2} (x^2 - 2x) \, dx \right\rvert$ .

$\int (x^2 - 2x) \, dx = \frac{x^3}{3} - x^2 = F(x)$ .

$F(2) - F(0) = \left( \frac{8}{3} - 4 \right) - 0 = -\frac{4}{3}$ .

Area on $[0, 2]$ is $\frac{4}{3}$ .

On $[2, 3]$ : the curve is above the axis.

$F(3) - F(2) = (9 - 9) - (-\frac{4}{3}) = \frac{4}{3}$ .

Total area. $\frac{4}{3} + \frac{4}{3} = \frac{8}{3}$ .

Markers reward explicit recognition that the curve changes sign at $x = 2$ , the interval split, and the absolute-value handling on the negative piece. A response that returns $\int_{0}^{3} (x^2 - 2x) \, dx = 0$ as the "area" earns no marks.

2023 VCAA Paper 24 marksFind the exact area enclosed between the curves

y = x^2

and

y = 2x

on the interval where they meet.

Show worked answer →

Find intersection points: Set $x^2 = 2x$ . So $x^2 - 2x = 0$ , $x(x - 2) = 0$ , giving $x = 0$ and $x = 2$ .
Identify top and bottom curves: On $(0, 2)$ , $2x > x^2$ (e.g. at $x = 1$ , $2 > 1$ ). So $y = 2x$ is the top, $y = x^2$ is the bottom.
Set up the area integral

$A = \int_{0}^{2} (\text{top} - \text{bottom}) \, dx = \int_{0}^{2} (2x - x^2) \, dx$ .

Evaluate.

$\int (2x - x^2) \, dx = x^2 - \frac{x^3}{3}$ .

$\left[ x^2 - \frac{x^3}{3} \right]_{0}^{2} = \left( 4 - \frac{8}{3} \right) - 0 = \frac{12 - 8}{3} = \frac{4}{3}$ .

Area is $\frac{4}{3}$ .

Markers reward correct intersection points, correct identification of top vs bottom (a quick test value), and a clean positive area at the end.