Areas between two curves (TCE Mathematics Methods, Tasmania)

Syllabus dot point

“Calculate the area enclosed between two curves using definite integration.”

8 min readUpdated 2 Jun 2026

Quick answer

The area enclosed between two curves equals the integral of the upper curve minus the lower curve over the region: $\displaystyle\int_a^b \big[f(x) - g(x)\big]\,dx$ , where the limits $a$ and $b$ are the $x$ -values where the curves intersect. If the curves swap which one is on top, split the integral at the crossing point.

What this dot point is asking

You already know how to find the area between a single curve and the $x$ -axis. This dot point extends that to the region trapped between two graphs, which is one of the most common extended-response integration questions.

The core idea

If $f(x) \ge g(x)$ over an interval, the vertical gap between the curves at each $x$ is $f(x) - g(x)$ . Summing those gaps across the interval by integration gives the enclosed area.

Area between two curves

A = \int_a^b \big[f(x) - g(x)\big]\,dx

where $f$ is the upper curve, $g$ is the lower curve, and $a$ and $b$ are the $x$ -coordinates of the points of intersection.

Subtracting before integrating means the result is correct even when part of the region lies below the $x$ -axis, because the two unwanted areas cancel and only the gap between the curves remains. This is the big advantage over computing each area separately.

Finding the limits

The limits are almost never given directly; you find them by setting the curves equal.

Limits come from the intersections

Solve $f(x) = g(x)$ to find where the curves meet. These solutions are the limits $a$ and $b$ of the integral. Always solve this first, because the enclosed region is bounded precisely by the intersection points.

Deciding which curve is on top

Test a single point between the limits. Substitute an $x$ -value from inside the interval into both functions; whichever gives the larger $y$ is the upper curve $f$ . Getting this backwards produces a negative answer, which is a clear signal you have swapped them.

Area between a parabola and a line

Find the area enclosed between $y = 4 - x^2$ and the line $y = x + 2$ .

Step 1: Find the intersection points (the limits of integration)

The region is bounded by the two curves from where they meet. Set the expressions equal and solve:

4 - x^2 = x + 2 \quad\Rightarrow\quad x^2 + x - 2 = 0 \quad\Rightarrow\quad (x+2)(x-1) = 0,

so $x = -2$ and $x = 1$ .

Step 2: Identify the upper curve

Test a point inside the interval, say $x = 0$ . The parabola gives $y = 4$ and the line gives $y = 2$ , so the parabola $y = 4 - x^2$ is the upper curve throughout the interval. The integrand is therefore the upper curve minus the lower:

A = \int_{-2}^{1} \big[(4 - x^2) - (x + 2)\big]\,dx = \int_{-2}^{1} (2 - x - x^2)\,dx.

Step 3: Antidifferentiate and apply the limits

Find the antiderivative of each term, then evaluate at $x = 1$ and $x = -2$ and subtract:

\left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-2}^{1}.

At $x = 1$ : $2 - \tfrac{1}{2} - \tfrac{1}{3} = \tfrac{7}{6}$ . At $x = -2$ : $-4 - 2 + \tfrac{8}{3} = -\tfrac{10}{3}$ .

A = \tfrac{7}{6} - \left(-\tfrac{10}{3}\right) = \tfrac{7}{6} + \tfrac{20}{6} = \tfrac{27}{6} = 4.5 \text{ square units.}

Final answer: The enclosed area is $4.5$ square units.

When the curves cross inside the region

If the two curves intersect at a point between the outer limits, the upper and lower curves swap there. In that case split the integral at the crossing and write each piece as upper minus lower for that sub-interval, then add the absolute areas. Skipping the split lets the two parts partly cancel and gives too small an answer.

Curves that cross inside the interval

Find the total area enclosed between $y = x^{3}$ and $y = x$ for $-1 \le x \le 1$ .

Step 1: Find all intersection points within the interval

Setting the curves equal locates where they meet. Factoring reveals a crossing inside the given interval, which means we cannot use a single integral:

x^{3} = x \quad\Rightarrow\quad x^{3} - x = 0 \quad\Rightarrow\quad x(x - 1)(x + 1) = 0,

giving $x = -1$ , $0$ , and $1$ . The curves cross at $x = 0$ inside $[-1, 1]$ , so the upper and lower functions swap there.

Step 2: Identify which curve is on top in each sub-interval

Test one point in each sub-interval to decide the order. On $[-1, 0]$ , test $x = -0.5$ : $x^{3} = -0.125$ and $x = -0.5$ , so $x^{3} > x$ there. On $[0, 1]$ , test $x = 0.5$ : $x^{3} = 0.125$ and $x = 0.5$ , so $x > x^{3}$ there.

Step 3: Integrate upper minus lower on each piece

Write a separate integral for each sub-interval, always placing the locally upper curve first to guarantee a positive result:

A = \int_{-1}^{0}(x^{3} - x)\,dx + \int_{0}^{1}(x - x^{3})\,dx.

Step 4: Evaluate each integral and add the areas

Antidifferentiate each integrand and apply the limits:

\int_{-1}^{0}(x^{3} - x)\,dx = \left[\frac{x^{4}}{4} - \frac{x^{2}}{2}\right]_{-1}^{0} = 0 - \left(\frac{1}{4} - \frac{1}{2}\right) = \frac{1}{4}.

\int_{0}^{1}(x - x^{3})\,dx = \left[\frac{x^{2}}{2} - \frac{x^{4}}{4}\right]_{0}^{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}.

Final answer: Total area $= \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}$ square unit. By symmetry the two pieces are equal, which serves as a useful check.

Integrating with respect to y

Occasionally a region is described more naturally by horizontal strips, especially when the boundaries are functions of $y$ or the region is bounded on the left and right rather than top and bottom. The same idea applies with the roles of the axes swapped: the area is $\displaystyle\int_{c}^{d}\big[x_{\text{right}}(y) - x_{\text{left}}(y)\big]\,dy$ , where the limits $c$ and $d$ are the $y$ -coordinates of the intersections. Choosing the orientation that gives the simpler integrand often turns an awkward problem into a routine one.

Integrating without subtracting, or subtracting in the wrong order

Two errors recur. First, integrating each curve against the axis separately and then subtracting the results is far more work and easy to botch; integrate the difference $f-g$ in one go instead. Second, writing $g - f$ when $f$ is actually on top gives a negative area. If your area comes out negative, swap the order rather than just dropping the sign without thinking.

Summary

Set the curves equal to find the intersection limits, decide which curve is upper by testing a point inside, then integrate upper minus lower across the region. If the curves cross within the interval, split at that point and add the absolute areas. Subtracting the functions before integrating automatically handles regions below the axis, which is why this method is cleaner than computing separate areas.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

Calculator-assumed. Find the value of

k

, where

k > 1

, such that the area enclosed between the line

x = k

and the curves

y = \dfrac{1}{x^{3}}

and

y = \dfrac{1}{x^{2}}

is equal to

\dfrac{1}{8}

. Show all working.

Show worked answer →

For $x > 1$ , $\dfrac{1}{x^{2}}$ is the upper curve and $\dfrac{1}{x^{3}}$ is the lower curve. The region runs from $x = 1$ (where the curves meet, both equal to $1$ ) to $x = k$ .

\text{Area} = \int_{1}^{k}\left(\frac{1}{x^{2}} - \frac{1}{x^{3}}\right)dx = \left[-\frac{1}{x} + \frac{1}{2x^{2}}\right]_{1}^{k} = \frac{1}{2} - \frac{1}{k} + \frac{1}{2k^{2}}.

Set equal to $\dfrac{1}{8}$ and multiply through by $8k^{2}$ :

4k^{2} - 8k + 4 = k^{2} \quad\Rightarrow\quad 3k^{2} - 8k + 4 = 0.

Factorising:

(3k - 2)(k - 2) = 0

, giving

k = \dfrac{2}{3}

k = 2

.
Since

k > 1

, the answer is

k = 2

. Marks are for the upper-minus-lower integrand, the antiderivative, and solving the quadratic with the correct root chosen.

Practice2 marks

Calculator-free. The functions

f(x) = -x^{3} + x

(for

-1 \le x \le 1

) and

g(x) = -x^{2} + 2x

(for

0 \le x \le 2

) are sketched. Find the area enclosed between the two curves.

Show worked answer →

First find where the curves meet. Setting $-x^{3} + x = -x^{2} + 2x$ gives $-x^{3} + x^{2} - x = 0$ , i.e. $-x(x^{2} - x + 1) = 0$ . The quadratic $x^{2} - x + 1$ has no real roots (discriminant $1 - 4 < 0$ ), so the only intersection is $x = 0$ .

On $[0, 1]$ , $g(x) - f(x) = (-x^{2} + 2x) - (-x^{3} + x) = x^{3} - x^{2} + x$ , which is positive, so $g$ is the upper curve.

\text{Area} = \int_{0}^{1}(x^{3} - x^{2} + x)\,dx = \left[\frac{x^{4}}{4} - \frac{x^{3}}{3} + \frac{x^{2}}{2}\right]_{0}^{1} = \frac{1}{4} - \frac{1}{3} + \frac{1}{2} = \frac{5}{12}

square units. Marks reward identifying the upper curve and evaluating the definite integral.

Practise this

Sources & how we know this

Mathematics Methods Level 4 (MTM415117) course document (2017 onwards) — TASC