Area under and between curves: VCE Math Methods Unit 4

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:

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

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

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

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 Exam 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:

The rule of thumb: top curve minus bottom curve.

Procedure

1. Find the intersection points. Solve $f(x) = g(x)$. These set the integration limits if the question asks for "the enclosed area".
2. 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".
3. Set up $\int [\text{top} - \text{bottom}] \, dx$ over the interval.
4. 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.

Calculator-active area problems (Paper 2)

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

1. Solve the intersection equation numerically on the calculator.
2. Identify top and bottom by evaluating at a test point.
3. 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.

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