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Module 3 · L12 of 15 ~40 min +95 XP available

Areas

Integration lets us measure the space under a curve. Like calculating the total rainfall from a rate graph, the definite integral accumulates area between a function and the axis, revealing totals that simple geometry cannot. The fundamental theorem of calculus is the key that unlocks the calculation.

Today's hook, A rainfall gauge records the rate of rain falling (in mm per hour) throughout the day. The total rainfall is the area under the rate graph. How do you calculate the total when the rate changes continuously? Integration is the answer.

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Orient to definite area

Work through the visible teaching and complete each embedded check.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall, your gut answer first

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Before we start, what do you already know about measuring the space under a curve? How would you estimate the area under $y = x^2$ between $x = 0$ and $x = 2$?

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The two moves

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There are only two moves in this entire lesson. Lock them into muscle memory and the rest is just calculation.

Move 1, Integrate. Find the antiderivative $F(x)$ of $f(x)$. This is the reverse of differentiation: increase the power by one and divide.

Move 2, Evaluate at limits and subtract. Compute $F(b) - F(a)$. This gives the signed area between the curve and the $x$-axis from $x = a$ to $x = b$.

The definite integral is the shaded area under y = f(x) from x = a to x = b.

$$\int_a^b f(x)\,dx = F(b) - F(a) \quad \text{where } F'(x) = f(x)$$

Sketch the curve first

Always sketch to see where the curve crosses the $x$-axis. Regions below the axis give negative signed area, you must handle them separately for total area.

Split at $x$-intercepts

For total area when the curve crosses the axis, integrate each piece separately and take absolute values before adding.

Area between curves

Always integrate (top curve $-$ bottom curve). Find intersection points first to set the correct limits.

What you will master

Know

Key facts

The fundamental theorem: $\int_a^b f(x)\,dx = F(b) - F(a)$
Area above the axis is positive; area below is negative (signed area)
Area between curves: $\int_a^b (\text{top} - \text{bottom})\,dx$

Understand

Concepts

Why the definite integral gives signed area
How to handle regions that cross the $x$-axis
Why we need to identify the “top” curve when finding area between two curves

Can do

Skills

Evaluate definite integrals using the fundamental theorem
Find the area between a curve and the $x$-axis
Find the area enclosed between two curves

Build the area language

Work through the visible teaching and complete each embedded check.

Key terms

Definite integralAn integral with upper and lower limits: $\int_a^b f(x)\,dx$, giving a numerical value.

Fundamental theorem of calculus$\int_a^b f(x)\,dx = F(b) - F(a)$ where $F$ is any antiderivative of $f$.

Signed areaArea above the $x$-axis contributes positively; area below contributes negatively. The definite integral measures signed area.

Total areaThe sum of absolute values of signed areas in each sub-interval. Always non-negative.

AntiderivativeA function $F(x)$ such that $F'(x) = f(x)$. Used in the fundamental theorem to evaluate definite integrals.

Area between curves$\int_a^b (f(x) - g(x))\,dx$ where $f(x) \ge g(x)$ on $[a, b]$. Find intersection points first.

Connect integrals and area

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Areas and the definite integral

core concept

The definite integral $\int_a^b f(x)\,dx$ measures the signed area between the curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$. The fundamental theorem of calculus converts this into a calculation with antiderivatives:

$$\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a)$$

Areas above the $x$-axis contribute positively. Areas below the $x$-axis contribute negatively. To find total area, split at $x$-intercepts and take absolute values.

Area above the axis counts as positive and area below as negative in the signed integral.

Green (above axis): positive signed area. Red (below axis): negative signed area. For total area, take the absolute value of each piece.

Area between two curves

When two curves intersect, the enclosed area is found by integrating the difference between the top and bottom curves over the interval between their intersection points:

$$\text{Area} = \int_a^b \bigl(f(x) - g(x)\bigr)\,dx \quad \text{where } f(x) \ge g(x) \text{ on } [a,b]$$

Why the fundamental theorem is remarkable. Before calculus, finding the area under a curve required approximating it with hundreds of thin rectangles. The fundamental theorem, discovered independently by Newton and Leibniz, turns this infinite process into a simple subtraction: $F(b) - F(a)$. It connects the two big ideas of calculus, differentiation and integration.

Fundamental theorem: $\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$ where $F'(x) = f(x)$; Signed area: above axis $= +$; below axis $= -$

Pause, copy the Fundamental Theorem $\int_a^b f(x)\,dx = F(b) - F(a)$ and the signed-area rule (above $x$-axis = positive, below = negative) into your book.

Quick check: True or false, $\displaystyle\int_0^2 (x^2 - 1)\,dx$ gives the total (unsigned) area between $y = x^2 - 1$ and the $x$-axis on $[0, 2]$.

Evaluate area examples

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Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · EVALUATE A DEFINITE INTEGRAL

Evaluate $\displaystyle\int_1^3 (2x + 1)\,dx$.

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$\displaystyle\int (2x + 1)\,dx = x^2 + x$

Find the antiderivative. (No $+C$ needed for definite integrals.)

PROBLEM 2 · AREA WITH CURVE BELOW AXIS

Find the total area bounded by $y = x^2 - 4$, the $x$-axis, $x = 0$ and $x = 3$.

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$y = x^2 - 4 = 0$ when $x = 2$ (in the interval $[0,3]$).

Find where the curve crosses the $x$-axis within the bounds. This divides $[0,3]$ into $[0,2]$ (below axis) and $[2,3]$ (above axis).

PROBLEM 3 · AREA BETWEEN TWO CURVES

Find the area enclosed by $y = x^2$ and $y = x$.

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$x^2 = x \Rightarrow x^2 - x = 0 \Rightarrow x(x-1) = 0 \Rightarrow x = 0$ or $x = 1$

Find the intersection points. These give the limits of integration.

Quick check: What is $\displaystyle\int_0^2 3x^2\,dx$?

Avoid signed-area traps

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Common errors · the 3 traps that cost marks

Trap 01

Forgetting that area below the axis is negative

The definite integral gives signed area. If part of the region is below the axis, you must split the integral at the $x$-intercepts and take absolute values to find total area. Failing to do this gives an answer that is too small (or even negative).

Trap 02

Integrating the wrong way around for area between curves

Always subtract the lower curve from the upper curve: $\int (\text{top} - \text{bottom})\,dx$. Subtracting the wrong way gives a negative area. Test a point to confirm which curve is higher.

Trap 03

Not finding all intersection points

When finding the area between two curves, you must find all intersection points to set the correct bounds of integration. Missing an intersection means missing part of the area or using the wrong limits entirely.

Odd one out: Three of the following describe the definite integral correctly. Which one is WRONG?

Practise and reflect

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Quick-fire practice · 5 problems

Evaluate $\displaystyle\int_0^2 (3x^2 + 2x)\,dx$.

Find the area bounded by $y = x^2$, the $x$-axis, $x = 1$ and $x = 3$.

Find the total area between $y = x^2 - 1$ and the $x$-axis from $x = 0$ to $x = 2$.

Find the area enclosed by $y = x^2$ and $y = 2x$.

Find the area between $y = x^3$ and $y = x$ for $x \ge 0$.

Fill the blanks: drag each token into the matching blank.

antiderivative subtract absolute intersection

The fundamental theorem says: find the ___ $F(x)$, then ___ $F(a)$ from $F(b)$. For total area when the curve dips below the axis, take the ___ value of each piece. For area between two curves, first find their ___ points.

Match each integral to its correct value.

$\displaystyle\int_0^1 x^2\,dx$
$\displaystyle\int_1^3 2x\,dx$
$\displaystyle\int_0^2 (x+1)\,dx$
$\displaystyle\int_0^3 1\,dx$

$3$
$4$
$8$
$\tfrac{1}{3}$

Revisit your thinking

Earlier you were asked about measuring space under curves. The definite integral does exactly this. The fundamental theorem lets us calculate it using antiderivatives, avoiding the need for approximation methods. The key rule to remember: the integral gives signed area, so you must handle below-axis regions with care.

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Complete the mixed practice

Complete the question-bank round and short-answer practice, then compare your responses with the comprehensive answers.

Multiple choice

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Short answer

ApplyBand 4

Q1. Evaluate $\displaystyle\int_1^4 (2x - 3)\,dx$. Show all working. 2 MARKS

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Q2. Find the total area bounded by $y = x^2 - 4x + 3$, the $x$-axis, and the lines $x = 0$ and $x = 4$. Show all working. 4 MARKS

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ApplyBand 4

Q3. Find the area enclosed by the curves $y = x^2$ and $y = 2 - x^2$. Show all working. 4 MARKS

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📖 Comprehensive answers (click to reveal)

Drill 1: $\left[x^3 + x^2\right]_0^2 = 12$.

Drill 2: $\left[\frac{x^3}{3}\right]_1^3 = 9 - \frac{1}{3} = \frac{26}{3}$ sq units.

Drill 3: Intercept at $x = 1$. $\left|\int_0^1(x^2-1)\,dx\right| + \int_1^2(x^2-1)\,dx = \frac{2}{3} + \frac{4}{3} = 2$ sq units.

Drill 4: Intersect at $x = 0, 2$. $\int_0^2(2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$ sq units.

Drill 5: $x^3 = x$ at $x = 0, 1$. On $[0,1]$, $x \ge x^3$. Area $= \int_0^1(x-x^3)\,dx = \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ sq units.

Q1 (2 marks): $\left[x^2 - 3x\right]_1^4 = (16-12) - (1-3) = 4 - (-2) = 6$ [2].

Q2 (4 marks): $y = (x-1)(x-3)$, intercepts at $x = 1, 3$ [0.5]. $\int_0^1 = \frac{4}{3}$; $\left|\int_1^3\right| = \frac{4}{3}$; $\int_3^4 = \frac{4}{3}$ [2.5]. Total area $= \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$ sq units [1].

Q3 (4 marks): $x^2 = 2 - x^2 \Rightarrow x = \pm 1$ [1]. Top curve: $y = 2 - x^2$ [0.5]. Area $= \int_{-1}^{1}(2 - 2x^2)\,dx = \left[2x - \frac{2x^3}{3}\right]_{-1}^{1} = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$ sq units [2.5].

Review and complete

Retrieve the central ideas, then mark the lesson complete or continue to the module quiz.

Retrieve the lesson before you leave

State the Fundamental Theorem rule used to evaluate a definite integral.
Explain the difference between signed area and total area.
Describe how to set up an area-between-curves integral.

Answer from memory first, then return to the matching Learn checkpoint to check and correct your response.

Take the full module quiz

quiz

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Mark lesson as complete

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Module overview · Maths Advanced · Checkpoint 1