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

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 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:

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.

Worked examples

Example 1. Polynomial curve crossing the axis

Area between $y = x^3 - 4x$ and the $x$-axis on $[-2, 2]$.

Find roots. $x^3 - 4x = x(x^2 - 4) = x(x-2)(x+2)$. Roots at $-2, 0, 2$.

Sign analysis on $[-2, 2]$. The curve crosses zero at $0$ inside the interval. On $(-2, 0)$, test $x = -1$: $-1 + 4 = 3 > 0$. On $(0, 2)$, test $x = 1$: $1 - 4 = -3 < 0$.

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 area is $|-4| = 4$.

Total area: $4 + 4 = 8$.

Example 2. Area between two curves with intersection

Area enclosed between $y = x^2$ and $y = x + 2$.

Intersection. $x^2 = x + 2$, so $x^2 - x - 2 = 0$, $(x - 2)(x + 1) = 0$, $x = -1$ or $x = 2$.

Top vs bottom on $(-1, 2)$. Test $x = 0$: $x^2 = 0$, $x + 2 = 2$. So $y = x + 2$ is the top.

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

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

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

Area is $\frac{9}{2}$.

Example 3. Curves intersecting more than twice

Area enclosed between $y = \sin(x)$ and $y = \cos(x)$ on $[0, 2\pi]$ requires finding all intersections and splitting accordingly. The intersections on $[0, 2\pi]$ are at $x = \pi/4$ and $x = 5\pi/4$.

The strategy is the same: identify top in each sub-interval, integrate top minus bottom, sum up.

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.

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).

In one sentence

Area between a single curve and the $x$-axis is computed by splitting the interval at sign-change roots and summing the absolute values of the definite integral over each sub-interval; area between two curves is the integral of top minus bottom between intersection points, with further sub-intervals required if the top and bottom swap.