Antidifferentiation: Indefinite Integrals Revision Notes for VCE SSCE Mathematical Methods

Antidifferentiation: Indefinite Integrals

What is antidifferentiation?

Antidifferentiation is the process of finding a function from its derivative. This is essentially "undoing" differentiation.

For example, we know that the derivative of $x^2$ with respect to $x$ is $2x$ . If we work backwards from $2x$ , we can determine that the original function was $x^2$ . This reverse process is called antidifferentiation.

Multiple antiderivatives

An important feature of antidifferentiation is that there isn't just one answer. Consider these two functions:

$f(x) = x^2 + 1$
$g(x) = x^2 - 7$

Both have the same derivative: $f'(x) = 2x$ and $g'(x) = 2x$ .

This means that both $x^2 + 1$ and $x^2 - 7$ are antiderivatives of $2x$ .

Why can different functions have the same derivative?

When two functions have the same derivative, they differ only by a constant. Graphically, this means one graph is a vertical shift of the other.

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The diagram shows several antiderivatives of $2x$ . Each parabola is a vertical translation of $y = x^2$ . They all have the same shape but are shifted up or down the $y$ -axis. This visual representation demonstrates why we need the constant $c$ in our general antiderivative.

Notation for indefinite integrals

The general antiderivative of $2x$ is written as $x^2 + c$ , where $c$ represents any constant value. We use special notation to express this:

\int 2x \, dx = x^2 + c

This is read as:

"the general antiderivative of $2x$ with respect to $x$ is equal to $x^2 + c$ ", or
"the indefinite integral of $2x$ with respect to $x$ is $x^2 + c$ "

Understanding the notation

The symbol $\int$ is called the integral sign
The expression $2x$ is the function we're integrating
$dx$ indicates we're integrating with respect to $x$
$c$ is the constant of integration (an arbitrary real number)

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General Formula for Indefinite Integrals

If $F'(x) = f(x)$ , then:

\int f(x) \, dx = F(x) + c

where $c$ is an arbitrary real number.

This formula tells us that if $F(x)$ is an antiderivative of $f(x)$ , then the indefinite integral of $f(x)$ gives us $F(x)$ plus a constant.

The power rule for integration

When integrating powers of $x$ , we can use a general formula, provided the power is not equal to $-1$ .

Building understanding through examples

Let's look at some differentiation results and reverse them:

Since $f(x) = x^3$ implies $f'(x) = 3x^2$ , we can reverse this: $\int 3x^2 \, dx = x^3 + c$
Since $f(x) = x^8$ implies $f'(x) = 8x^7$ , we can reverse this: $\int 8x^7 \, dx = x^8 + c$
Since $f(x) = x^{\frac{3}{2}}$ implies $f'(x) = \frac{3}{2}x^{\frac{1}{2}}$ , we can reverse this: $\int \frac{3}{2}x^{\frac{1}{2}} \, dx = x^{\frac{3}{2}} + c$
Since $f(x) = x^{-4}$ implies $f'(x) = -4x^{-5}$ , we can reverse this: $\int -4x^{-5} \, dx = x^{-4} + c$

We can also write:

\int x^2 \, dx = \frac{1}{3}x^3 + c

\int x^{\frac{1}{2}} \, dx = \frac{2}{3}x^{\frac{3}{2}} + c

\int x^{-5} \, dx = -\frac{1}{4}x^{-4} + c

The general power rule

From these examples, we can see a pattern:

\int x^r \, dx = \frac{x^{r+1}}{r+1} + c, \quad r \in \mathbb{Q} \setminus \{-1\}

In words: to integrate $x^r$ , add one to the power, then divide by the new power, and don't forget to add the constant $c$ .

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Critical Restriction

This rule cannot be used when $r = -1$ (as division by zero would occur). The case of $r = -1$ requires a different approach that will be covered later.

Domain considerations

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The power rule can only be applied for suitable values of $x$ depending on the value of $r$ :

If $r = \frac{1}{2}$ , then $x$ must be positive ( $x \in \mathbb{R}^+$ )
If $r = -2$ , then $x$ can be any real number except zero ( $x \in \mathbb{R} \setminus \{0\}$ )
If $r = 3$ , then $x$ can be any real number ( $x \in \mathbb{R}$ )

Always consider the domain of the original function when applying integration rules.

Rules for integration

These rules follow directly from the corresponding differentiation rules:

Sum rule:

\int [f(x) + g(x)] \, dx = \int f(x) \, dx + \int g(x) \, dx

Difference rule:

\int [f(x) - g(x)] \, dx = \int f(x) \, dx - \int g(x) \, dx

Constant multiple rule:

\int k f(x) \, dx = k \int f(x) \, dx

where $k$ is a real number.

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These rules allow us to integrate more complex expressions by breaking them into simpler parts. You can integrate term by term and factor out constants before integrating.

Worked examples

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Worked Example 1: Finding the general antiderivative of $3x^5$

Find the general antiderivative (indefinite integral) of $3x^5$ .

Step 1: Apply the constant multiple rule

\int 3x^5 \, dx = 3 \int x^5 \, dx

Step 2: Apply the power rule

= 3 \times \frac{x^6}{6} + c

Step 3: Simplify

= \frac{x^6}{2} + c

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Worked Example 2: Finding the general antiderivative of $3x^2 + 4x^{-2} + 3$

Find the general antiderivative (indefinite integral) of $3x^2 + 4x^{-2} + 3$ .

Step 1: Apply the sum rule to split the integral

\int (3x^2 + 4x^{-2} + 3) \, dx = 3 \int x^2 \, dx + 4 \int x^{-2} \, dx + 3 \int 1 \, dx

Step 2: Apply the power rule to each term

= \frac{3x^3}{3} + \frac{4x^{-1}}{-1} + \frac{3x}{1} + c

Step 3: Simplify

= x^3 - \frac{4}{x} + 3x + c

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Worked Example 3: Finding $y$ in terms of $x$ if $\frac{dy}{dx} = \frac{1}{x^2}$

Step 1: Rewrite in power form

\int \frac{1}{x^2} \, dx = \int x^{-2} \, dx

Step 2: Apply the power rule

= \frac{x^{-1}}{-1} + c

Step 3: Simplify and write the solution

\therefore y = -\frac{1}{x} + c

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Worked Example 4: Finding $y$ in terms of $x$ if $\frac{dy}{dx} = 3\sqrt{x}$

Step 1: Rewrite in power form

\int 3\sqrt{x} \, dx = 3 \int x^{\frac{1}{2}} \, dx

Step 2: Apply the power rule

= 3 \times \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + c

Step 3: Simplify

\therefore y = 2x^{\frac{3}{2}} + c

Finding a specific antiderivative

When we're given additional information about a function, we can find the exact value of the constant $c$ and determine a unique antiderivative.

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Worked Example 5: Finding a specific function with a given condition

It is known that $f'(x) = x^3 + 4x^2$ and $f(0) = 0$ . Find $f(x)$ .

Step 1: Find the general antiderivative

\int (x^3 + 4x^2) \, dx = \frac{x^4}{4} + \frac{4x^3}{3} + c

\therefore f(x) = \frac{x^4}{4} + \frac{4x^3}{3} + c

Step 2: Use the condition $f(0) = 0$ to find $c$

As $f(0) = 0$ , we have $c = 0$ .

Step 3: Write the final answer

f(x) = \frac{x^4}{4} + \frac{4x^3}{3}

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Worked Example 6: Finding the equation of a curve

If the gradient of the tangent at a point $(x, y)$ on a curve is given by $2x$ and the curve passes through the point $(-1, 4)$ , find the equation of the curve.

Step 1: Set up the problem

Let the curve have equation $y = f(x)$ . Then $f'(x) = 2x$ .

Step 2: Find the general antiderivative

\int 2x \, dx = \frac{2x^2}{2} + c = x^2 + c

\therefore f(x) = x^2 + c

Step 3: Use the condition that the curve passes through $(-1, 4)$

Since $f(-1) = 4$ , we have:

4 = (-1)^2 + c

Hence $c = 3$ .

Step 4: Write the final answer

f(x) = x^2 + 3

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Key Points to Remember:

Antidifferentiation is the reverse process of differentiation - finding a function from its derivative
The indefinite integral $\int f(x) \, dx$ represents all possible antiderivatives of $f(x)$
Always include the constant of integration $+c$ when finding indefinite integrals
The power rule for integration states: $\int x^r \, dx = \frac{x^{r+1}}{r+1} + c$ (where $r \neq -1$ )
You can find a specific antiderivative when given an additional condition by solving for the constant $c$

Antidifferentiation: Indefinite Integrals (VCE SSCE Mathematical Methods): Revision Notes