The Definite Integral and Its Properties Revision Notes for HSC SSCE Mathematics Advanced

The Definite Integral and Its Properties

Introduction

This note extends the theory of definite integrals to functions that take negative values. We'll explore key properties of definite integrals, including how to handle regions below the x-axis, and several important rules that simplify integration calculations.

Signed areas under the curve

When calculating a definite integral, we need to consider whether the function sits above or below the $x$ -axis. This gives us the concept of signed areas.

When a function has negative values, its graph lies below the $x$ -axis. The very thin rectangles in our area approximation have negative heights, which means areas below the $x$ -axis contribute negative values to the integral.

In the diagram above, where $a < h < k < b$ :

Region $A$ is above the $x$ -axis, so it contributes a positive value
Region $B$ is below the $x$ -axis (where $f(x)$ is negative), so it contributes a negative value
Region $C$ is above the $x$ -axis, so it contributes a positive value

Therefore:

$\int_{a}^{b} f(x)\, dx = \text{area } A - \text{area } B + \text{area } C$

infoNote

Key principle: The definite integral as the sum of signed areas

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

For regions where the curve is above the $x$ -axis, attach a positive sign to the area
For regions where the curve is below the $x$ -axis, attach a negative sign to the area
These areas with signs attached are called signed areas under the curve
The definite integral $\int_{a}^{b} f(x)\, dx$ equals the sum of these signed areas in the interval $[a, b]$

lightbulbExample

Worked Example: Evaluating integrals with signed areas

Consider the function $y = x - 4$ .

a) Evaluate $\int_{0}^{4} (x - 4)\, dx$

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

$= (8 - 16) - (0 - 0)$

$= -8$

b) Evaluate $\int_{4}^{6} (x - 4)\, dx$

$\int_{4}^{6} (x - 4)\, dx = \left[\frac{1}{2}x^2 - 4x\right]_{4}^{6}$

$= (18 - 24) - (8 - 16)$

$= -6 - (-8)$

$= 2$

c) Evaluate $\int_{0}^{6} (x - 4)\, dx$

$\int_{0}^{6} (x - 4)\, dx = \left[\frac{1}{2}x^2 - 4x\right]_{0}^{6}$

$= (18 - 24) - (0 - 0)$

$= -6$

Interpretation:

Triangle $OAB$ has area $8$ square units and lies below the $x$ -axis, so the integral in part a is $-8$
Triangle $BMC$ has area $2$ square units and lies above the $x$ -axis, so the integral in part b is $2$
The integral in part c represents area of $\triangle BMC$ minus area of $\triangle OAB$ , giving $2 - 8 = -6$

Properties of odd and even functions

The symmetry of odd and even functions creates useful shortcuts when integrating over symmetric intervals.

Odd functions

An odd function has point symmetry about the origin. For example, $y = x^3 - 4x$ is odd.

When we integrate an odd function from $-a$ to $a$ , the areas above and below the $x$ -axis have equal magnitude but opposite signs. They cancel each other out completely.

infoNote

Property: Integration of odd functions

If $f(x)$ is odd, then:

$\int_{-a}^{a} f(x)\, dx = 0$

This is because the positive and negative areas cancel exactly.

Even functions

An even function has line symmetry in the y-axis. For example, $y = x^2 + 1$ is even.

When we integrate an even function from $-a$ to $a$ , the areas on both sides of the $y$ -axis are identical. Therefore, we can calculate the integral from $0$ to $a$ and double it.

infoNote

Property: Integration of even functions

If $f(x)$ is even, then:

$\int_{-a}^{a} f(x)\, dx = 2\int_{0}^{a} f(x)\, dx$

This allows you to calculate half the region and double it.

lightbulbExample

Worked Example: Using symmetry

a) Evaluate $\int_{-2}^{2} (x^3 - 4x)\, dx$

Since $f(x) = x^3 - 4x$ is an odd function:

$\int_{-2}^{2} (x^3 - 4x)\, dx = 0$

We can verify this by direct calculation:

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

$= (4 - 8) - (4 - 8)$

$= 0$

b) Evaluate $\int_{-2}^{2} (x^2 + 1)\, dx$

Since $f(x) = x^2 + 1$ is an even function:

$\int_{-2}^{2} (x^2 + 1)\, dx = 2\int_{0}^{2} (x^2 + 1)\, dx$

$= 2\left[\frac{1}{3}x^3 + x\right]_{0}^{2}$

$= 2\left(\frac{8}{3} + 2 - 0\right)$

$= 2 \times \frac{14}{3}$

$= 9\frac{1}{3}$

chatImportant

Exam tip: Always check for odd or even symmetry before calculating—it can save significant time!

Dissection of the interval

When a region is divided into parts, the total area equals the sum of the individual areas. We can apply this principle to definite integrals by splitting the interval of integration.

If $f(x)$ is being integrated over the closed interval $[a, b]$ and the number $c$ lies within this interval, then we can break up the interval at any intermediate point.

infoNote

Property: Dissection of intervals

$\int_{a}^{b} f(x)\, dx = \int_{a}^{c} f(x)\, dx + \int_{c}^{b} f(x)\, dx$

This property allows us to break complex integrals into simpler parts, particularly useful when a function changes character at certain points (for example, where it crosses the $x$ -axis).

Intervals of zero width

If we integrate a function over an interval from $a$ to $a$ (width zero), and the function is defined at $x = a$ , the region also has zero width.

infoNote

Property: Zero-width intervals

$\int_{a}^{a} f(x)\, dx = 0$

This makes intuitive sense: there's no area to calculate when the interval has no width.

Reversing the interval

When we reverse the limits of integration (swapping the upper and lower bounds), the sign of the integral changes.

infoNote

Property: Reversing integration limits

$\int_{a}^{b} f(x)\, dx = -\int_{b}^{a} f(x)\, dx$

This agrees with calculations using the fundamental theorem, because:

$F(a) - F(b) = -(F(b) - F(a))$

lightbulbExample

Worked Example: Reversed limits

a) Evaluate $\int_{2}^{4} (x - 1)\, dx$

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

$= (8 - 4) - (2 - 2)$

$= 4$

This is positive because the region lies above the $x$ -axis.

b) Evaluate $\int_{4}^{2} (x - 1)\, dx$

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

$= (2 - 2) - (8 - 4)$

$= -4$

This is the opposite of part a, because the integral runs backwards from right to left, from $x = 4$ to $x = 2$ .

Sums of functions

When two functions are added together, their graphs stack vertically. The integral of the sum equals the sum of the individual integrals.

infoNote

Property: Integration of sums

$\int_{a}^{b} (f(x) + g(x))\, dx = \int_{a}^{b} f(x)\, dx + \int_{a}^{b} g(x)\, dx$

Integration distributes over addition.

lightbulbExample

Worked Example: Sum property

Show that the following two expressions are equal:

a) $\int_{0}^{1} (x^2 + x + 1)\, dx$

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

$= \left(\frac{1}{3} + \frac{1}{2} + 1\right) - (0 + 0 + 0)$

$= 1\frac{5}{6}$

b) $\int_{0}^{1} x^2\, dx + \int_{0}^{1} x\, dx + \int_{0}^{1} 1\, dx$

$\int_{0}^{1} x^2\, dx + \int_{0}^{1} x\, dx + \int_{0}^{1} 1\, dx = \left[\frac{x^3}{3}\right]_{0}^{1} + \left[\frac{x^2}{2}\right]_{0}^{1} + [x]_{0}^{1}$

$= \left(\frac{1}{3} - 0\right) + \left(\frac{1}{2} - 0\right) + (1 - 0)$

$= 1\frac{5}{6}$

Both expressions give the same result, confirming the sum property.

Multiples of functions

When a function is multiplied by a constant, the region is stretched vertically by that constant factor. The constant can be factored outside the integral.

infoNote

Property: Constant multiples

$\int_{a}^{b} kf(x)\, dx = k\int_{a}^{b} f(x)\, dx$

This property allows you to pull constant factors outside the integral, simplifying calculations.

lightbulbExample

Worked Example: Constant multiple

Show that these two expressions are equal:

a) $\int_{1}^{3} 10x^3\, dx$

$\int_{1}^{3} 10x^3\, dx = \left[\frac{10x^4}{4}\right]_{1}^{3}$

$= \frac{810}{4} - \frac{10}{4}$

$= \frac{800}{4}$

$= 200$

b) $10\int_{1}^{3} x^3\, dx$

$10\int_{1}^{3} x^3\, dx = 10 \times \left[\frac{x^4}{4}\right]_{1}^{3}$

$= 10 \times \left(\frac{81}{4} - \frac{1}{4}\right)$

$= 10 \times \frac{80}{4}$

$= 200$

Both methods produce the same answer, as expected.

Inequalities with definite integrals

If one curve always lies beneath another curve over an interval, then the area under the lower curve is less than the area under the higher curve.

infoNote

Property: Inequalities with integrals

If $f(x) \leq g(x)$ throughout the closed interval $[a, b]$ , then:

$\int_{a}^{b} f(x)\, dx \leq \int_{a}^{b} g(x)\, dx$

This property is essential for bounding integrals and proving inequalities.

lightbulbExample

Worked Example: Inequality application

a) Sketch the graph of $f(x) = 4 - x^2$ for $-2 \leq x \leq 2$

b) Explain why $0 \leq \int_{-2}^{2} (4 - x^2)\, dx \leq 16$

Solution:

The parabola $y = 4 - x^2$ and the horizontal line $y = 4$ are shown in the diagram.

Over the interval $-2 \leq x \leq 2$ , we can see that: $0 \leq 4 - x^2 \leq 4$

Therefore, the region associated with the integral fits inside a square with side length $4$ .

The area of this square is $4 \times 4 = 16$ square units, and the area cannot be negative, so: $0 \leq \int_{-2}^{2} (4 - x^2)\, dx \leq 16$

bookmarkSummary

Key Points to Remember:

Signed areas: Regions above the x-axis contribute positive values to the integral, while regions below the x-axis contribute negative values.
Odd and even functions:
- For odd functions integrated over symmetric intervals, $\int_{-a}^{a} f(x)\, dx = 0$ (areas cancel)
- For even functions, $\int_{-a}^{a} f(x)\, dx = 2\int_{0}^{a} f(x)\, dx$ (areas double)
Essential properties to memorise:
- Dissection: $\int_{a}^{b} f(x)\, dx = \int_{a}^{c} f(x)\, dx + \int_{c}^{b} f(x)\, dx$
- Reversing: $\int_{a}^{b} f(x)\, dx = -\int_{b}^{a} f(x)\, dx$
- Sum: $\int_{a}^{b} (f(x) + g(x))\, dx = \int_{a}^{b} f(x)\, dx + \int_{a}^{b} g(x)\, dx$
- Multiple: $\int_{a}^{b} kf(x)\, dx = k\int_{a}^{b} f(x)\, dx$
Exam strategy: Always sketch the graph when dealing with signed areas or inequalities—visualisation helps prevent sign errors and provides geometric insight into the problem.

The Definite Integral and Its Properties (HSC SSCE Mathematics Advanced): Revision Notes