Areas and the Definite Integral Revision Notes for HSC SSCE Mathematics Advanced

Areas and the Definite Integral

Introduction to area calculation

Every method we use to calculate area builds on two fundamental principles:

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The Two Principles of Area Calculation

Area of a rectangle = length × breadth
When a region is dissected, the area remains unchanged

For regions bounded by straight lines (such as triangles or trapeziums), we can calculate the area by cutting the shape into rectangles using a few strategic cuts. However, when dealing with curved boundaries, we need to imagine dividing the region into infinitely many rectangles. This requires a limiting process, similar to what we use in differentiation.

Developing the definite integral

Building from thin strips

Consider the region between a continuous curve $y = f(x)$ and the $x$ -axis, from $x = a$ to $x = b$ (where $a \leq b$ ). For now, assume the function is never negative in this interval.

The diagram above shows three stages in developing the integral:

Left panel: The complete region we want to find the area of, shaded in blue
Middle panel: The region divided into many thin vertical strips. Each strip approximates a rectangle, though the curved top boundary makes this approximation imperfect
Right panel: A close-up of a single strip at position $x$ on the $x$ -axis

Calculating the area of one strip

For a single strip at position $x$ :

The height at the left edge is $f(x)$
The width is $\delta x$ (a very small amount)
If the strip is thin enough, the height remains approximately $f(x)$ across its entire width

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The approximation becomes more accurate as we make the strip thinner. The key insight is that for an infinitesimally thin strip, the approximation becomes exact.

Therefore:

$\text{Area of strip} \approx \text{height} \times \text{width}$

$\text{Area of strip} \approx f(x) \delta x$

Summing all the strips

To find the total area, we add up all the strips using sigma notation:

$\text{Area of region} \approx \sum_{x=a}^{b} f(x) \delta x$

where the symbol $\sum$ means "sum of all".

Taking the limit

Now imagine infinitely many strips, each infinitesimally thin. The inaccuracy in our approximation disappears as $\delta x$ approaches zero. We might expect to write:

$\text{Area} = \lim_{\delta x \to 0} \sum_{x=a}^{b} f(x) \delta x$

However, instead of this notation, we use the elegant system introduced by Leibnitz:

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Leibnitz Notation

Replace $\delta x$ with $dx$ (suggesting infinitesimal width)
Replace the sigma symbol $\sum$ with $\int$ (an elongated S, suggesting a smooth infinite sum)

Think of it this way: $\int$ is a stretched S for "Sum", and $dx$ is $\delta x$ shrunk to zero!

This gives us the definite integral:

$\int_a^b f(x) \, dx = \text{area of shaded region}$

The definite integral - formal definition

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Definition: The Definite Integral

Let $f(x)$ be a function that is continuous in a closed interval $\[a, b]$ , where $a \leq b$ .

For now, suppose that $f(x)$ is never negative in the interval.

The definite integral $\int_a^b f(x) \, dx$ is defined as the area of the region between the curve and the $x$ -axis, from $x = a$ to $x = b$
The function $f(x)$ is called the integrand
The values $x = a$ and $x = b$ are called the lower and upper limits (or bounds) of the integral

Key insight: The term 'integration' reflects the process of combining many small pieces to form a complete whole. The notation captures the idea of building up a region from infinitely many infinitesimally thin strips.

Evaluating definite integrals using area formulae

When the function is linear or involves circles, we can calculate the definite integral by sketching the graph and using standard geometric area formulas.

Standard area formulas

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Essential Area Formulas

Triangle:

$\text{Area} = \frac{1}{2}bh = \frac{1}{2} \times \text{base} \times \text{height}$

Trapezium:

$\text{Area} = \frac{1}{2}(a + b)h = \text{average of parallel sides} \times \text{width}$

Note: For a trapezium, $h$ is the perpendicular distance between the parallel sides. Depending on orientation, we might call this the "height" or "width".

Circle:

$\text{Area} = \pi r^2 = \pi \times \text{(square of the radius)}$

Worked example: linear functions

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Worked Example 1a: Triangle Region

Evaluate $\int_1^4 (x - 1) \, dx$

Step 1: Sketch the function

The function $y = x - 1$ is a straight line with gradient $1$ and $y$ -intercept $-1$ .

Step 2: Identify the shape

The shaded region is a triangle with:

Base = $4 - 1 = 3$
Height = $3$

Step 3: Apply the area formula

$\int_1^4 (x - 1) \, dx = \frac{1}{2} \times \text{base} \times \text{height}$

$= \frac{1}{2} \times 3 \times 3$

$= \frac{9}{2}$

$= 4.5$

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Worked Example 1b: Trapezoidal Region

Evaluate $\int_2^4 (x - 1) \, dx$

Step 1: Use the same function

Using the same function $y = x - 1$ :

Step 2: Identify the shape

The shaded region is a trapezium with:

Width = $4 - 2 = 2$
Parallel sides of length $1$ and $3$

Step 3: Apply the area formula

$\int_2^4 (x - 1) \, dx = \text{average of parallel sides} \times \text{width}$

$= \frac{1 + 3}{2} \times 2$

$= 2 \times 2$

$= 4$

Worked example: absolute value and circular functions

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Worked Example 2a: Absolute Value Function

Evaluate $\int_{-2}^2 |x| \, dx$

Step 1: Sketch the function

The function $y = |x|$ forms a V-shape with vertex at the origin.

Step 2: Identify the shape

The shaded region consists of two identical triangles, each with base $2$ and height $2$ .

Step 3: Calculate the total area

$\int_{-2}^2 |x| \, dx = 2 \times \left(\frac{1}{2} \times \text{base} \times \text{height}\right)$

$= 2 \times \left(\frac{1}{2} \times 2 \times 2\right)$

$= 2 \times 2$

$= 4$

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Worked Example 2b: Semicircle Region

Evaluate $\int_{-5}^5 \sqrt{25 - x^2} \, dx$

Step 1: Recognize the function

The function $y = \sqrt{25 - x^2}$ can be rewritten as $x^2 + y^2 = 25$ , which represents a circle with centre at the origin and radius 5.

Since we're taking the positive square root, we get the upper semicircle.

Step 2: Identify the shape

The shaded region is a semicircle with radius $5$ .

Step 3: Apply the area formula

$\int_{-5}^5 \sqrt{25 - x^2} \, dx = \frac{1}{2} \times \pi r^2$

$= \frac{1}{2} \times \pi \times 5^2$

$= \frac{1}{2} \times 25\pi$

$= \frac{25\pi}{2}$

Using upper and lower rectangles to approximate integrals

We can estimate the value of a definite integral by trapping it between two bounds using rectangles.

Consider the integral $\int_0^4 (25 - x^2) \, dx$ .

We can construct:

Lower (inner) rectangles: These sit entirely inside the region
Upper (outer) rectangles: These contain the entire region

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The true value of the integral always lies between the lower rectangle sum and the upper rectangle sum. This method gives us a guaranteed range for the answer.

Worked example: successive approximations

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Worked Example 3: Progressive Rectangle Approximations

We'll evaluate $\int_0^4 (25 - x^2) \, dx$ using increasingly refined subdivisions.

Part (a): Single rectangle approximation

Using one rectangle on the interval $[0, 4]$ :

Lower rectangle area: $9 \times 4 = 36$

Upper rectangle area: $25 \times 4 = 100$

Therefore: $36 < \int_0^4 (25 - x^2) \, dx < 100$

Part (b): Two rectangle approximation

Subdividing the interval as $[0, 2] \cup [2, 4]$ :

Lower rectangles area: $21 \times 2 + 9 \times 2 = 42 + 18 = 60$

Upper rectangles area: $25 \times 2 + 21 \times 2 = 50 + 42 = 92$

Therefore: $60 < \int_0^4 (25 - x^2) \, dx < 92$

Part (c): Four rectangle approximation

Subdividing into four subintervals of width $1$ :

Lower rectangles area: $24 + 21 + 16 + 9 = 70$

Upper rectangles area: $25 + 24 + 21 + 16 = 86$

Therefore: $70 < \int_0^4 (25 - x^2) \, dx < 86$

Observation: As we increase the number of rectangles, the bounds get progressively tighter. The exact value is $78\frac{2}{3}$ .

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The Rectangle Approximation Method

More rectangles always give better approximations
The upper and lower bounds converge to the true value as the number of subdivisions increases
This method is particularly useful when you cannot use standard geometric formulas
In exams, you may be asked to find bounds using a specified number of rectangles

Special case: the area of a circle

The formula $A = \pi r^2$ for the area of a circle is itself derived using integration principles. Here's the classic geometric proof:

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Geometric Proof of Circle Area

The proof involves:

Dividing the circle into many thin sectors (pie-shaped wedges)
Rearranging these sectors alternately (yellow and blue) to approximate a rectangle
As the sectors become infinitesimally thin, the arrangement becomes a perfect rectangle

The resulting rectangle has:

Height = $r$ (the radius)
Length = $\pi r$ (half the circumference $2\pi r$ )

Therefore:

$\text{Area of circle} = \text{length} \times \text{height} = \pi r \times r = \pi r^2$

This demonstrates how dissecting a curved region and taking a limit leads to an exact area formula, just like the definite integral does for regions under curves.

Remember!

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Key Points to Remember:

Two fundamental principles underpin all area calculations: area of a rectangle = length × breadth, and dissection doesn't change total area
The definite integral $\int_a^b f(x) \, dx$ represents the area under the curve $y = f(x)$ from $x = a$ to $x = b$ . The symbol $\int$ is an elongated S representing an infinite sum
For linear and circular functions, evaluate definite integrals by sketching the graph and using standard area formulas (triangle, trapezium, circle)
Upper and lower rectangles can trap an integral between two bounds. More subdivisions give tighter bounds and better approximations
Integration is a limiting process involving infinitely many infinitesimally thin strips, similar to how differentiation involves limits

Areas and the Definite Integral (HSC SSCE Mathematics Advanced): Revision Notes