Antidifferentiation is the reverse of differentiation!

The Basic Question

Differentiation asks: Given $F(x)$, what is $F'(x)$?

Antidifferentiation asks: Given $f(x)$, what is $F(x)$ such that $F'(x) = f(x)$?

Example

Differentiation: If $F(x) = x^2$, then $F'(x) = 2x$

Antidifferentiation: If $f(x) = 2x$, then $F(x) = x^2$ is an antiderivative

Notation

If $F'(x) = f(x)$, we write:

• "$F(x)$ is an antiderivative of $f(x)$"
• Or: "$F(x)$ is the integral of $f(x)$"

We'll learn the integral notation $\int$ soon!

Simple Examples

Example 1: Power Function

Question: Find an antiderivative of $f(x) = x^3$

Think: What function has derivative $x^3$?

Try $F(x) = x^4$: $F'(x) = 4x^3 \neq x^3$ ✗

Try $F(x) = \frac{1}{4}x^4$: $F'(x) = \frac{1}{4}(4x^3) = x^3$ ✓

Answer: $F(x) = \frac{1}{4}x^4$ is an antiderivative of $f(x) = x^3$

Example 2: Trigonometric Function

Question: Find an antiderivative of $f(x) = \cos x$

Think: What function has derivative $\cos x$?

We know: $\frac{d}{dx}[\sin x] = \cos x$ ✓

Answer: $F(x) = \sin x$ is an antiderivative of $f(x) = \cos x$

Example 3: Exponential Function

Question: Find an antiderivative of $f(x) = e^x$

Think: What function has derivative $e^x$?

We know: $\frac{d}{dx}[e^x] = e^x$ ✓

Answer: $F(x) = e^x$ is an antiderivative of $f(x) = e^x$

The "+C" Problem

Here's something important: antiderivatives are not unique!

Why Not?

If $F(x) = x^2$, then $F'(x) = 2x$

But also:

• If $G(x) = x^2 + 5$, then $G'(x) = 2x$
• If $H(x) = x^2 - 7$, then $H'(x) = 2x$
• If $K(x) = x^2 + 100$, then $K'(x) = 2x$

All of these are antiderivatives of $f(x) = 2x$!

The General Solution

The most general antiderivative of $f(x) = 2x$ is: $F(x) = x^2 + C$

where $C$ is any constant (called the constant of integration).

Why Does +C Work?

Theorem: If $F(x)$ is an antiderivative of $f(x)$, then the most general antiderivative is: $F(x) + C$

where $C$ is any constant.

Reason: The derivative of any constant is zero! $\frac{d}{dx}[F(x) + C] = F'(x) + 0 = f(x)$

Different Functions, Same Derivative

Two functions with the same derivative differ by a constant.

Theorem: If $F'(x) = G'(x)$ for all $x$, then $F(x) = G(x) + C$ for some constant $C$.

This is why we need the $+C$!

Finding Antiderivatives: The Reverse Rules

To find antiderivatives, we reverse our derivative rules!

Power Rule (Reversed)

Derivative: $\frac{d}{dx}[x^n] = nx^{n-1}$

Antiderivative: If $f(x) = x^n$ (where $n \neq -1$), then: $F(x) = \frac{x^{n+1}}{n+1} + C$

Remember:

• Add 1 to the exponent
• Divide by the new exponent
• Don't forget +C!

Examples Using Power Rule

Example 1: Find an antiderivative of $f(x) = x^5$

$F(x) = \frac{x^{5+1}}{5+1} + C = \frac{x^6}{6} + C$

Check: $F'(x) = \frac{6x^5}{6} = x^5$ ✓

Example 2: Find an antiderivative of $f(x) = x$

$F(x) = \frac{x^{1+1}}{1+1} + C = \frac{x^2}{2} + C$

Check: $F'(x) = \frac{2x}{2} = x$ ✓

Example 3: Find an antiderivative of $f(x) = 1 = x^0$

$F(x) = \frac{x^{0+1}}{0+1} + C = \frac{x^1}{1} + C = x + C$

Check: $F'(x) = 1$ ✓

Example 4: Find an antiderivative of $f(x) = \frac{1}{x^2} = x^{-2}$

$F(x) = \frac{x^{-2+1}}{-2+1} + C = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C$

Check: $F'(x) = -(-1)x^{-2} = \frac{1}{x^2}$ ✓

Common Antiderivatives to Know

Basic Functions

Trigonometric Functions

Note: These will be covered in more detail in later lessons!

Linearity Properties

Antiderivatives follow the same linearity rules as derivatives:

Constant Multiple Rule

If $F(x)$ is an antiderivative of $f(x)$, then: $k \cdot F(x) \text{ is an antiderivative of } k \cdot f(x)$

Example: Antiderivative of $5x^2$: $F(x) = 5 \cdot \frac{x^3}{3} + C = \frac{5x^3}{3} + C$

Sum/Difference Rule

If $F(x)$ is an antiderivative of $f(x)$ and $G(x)$ is an antiderivative of $g(x)$, then: $F(x) \pm G(x) \text{ is an antiderivative of } f(x) \pm g(x)$

Example: Antiderivative of $x^2 + x$: $F(x) = \frac{x^3}{3} + \frac{x^2}{2} + C$

Complete Example: Polynomial

Find the most general antiderivative of $f(x) = 3x^4 - 2x^3 + 5x - 7$.

Step 1: Break it down

Use sum rule and constant multiple rule:

• Antiderivative of $3x^4$ is $3 \cdot \frac{x^5}{5} = \frac{3x^5}{5}$
• Antiderivative of $-2x^3$ is $-2 \cdot \frac{x^4}{4} = -\frac{x^4}{2}$
• Antiderivative of $5x$ is $5 \cdot \frac{x^2}{2} = \frac{5x^2}{2}$
• Antiderivative of $-7$ is $-7x$

Step 2: Combine

$F(x) = \frac{3x^5}{5} - \frac{x^4}{2} + \frac{5x^2}{2} - 7x + C$

Step 3: Check (optional but recommended!)

$F'(x) = \frac{3 \cdot 5x^4}{5} - \frac{4x^3}{2} + \frac{5 \cdot 2x}{2} - 7$

$= 3x^4 - 2x^3 + 5x - 7$ ✓

The Special Case: $\frac{1}{x}$

Why can't we use the power rule?

For $f(x) = \frac{1}{x} = x^{-1}$, the power rule would give: $F(x) = \frac{x^{-1+1}}{-1+1} = \frac{x^0}{0} = \frac{1}{0}$

Division by zero! ❌

The answer: $\int \frac{1}{x}\,dx = \ln|x| + C$

This is a special case you must memorize!

⚠️ Common Mistakes

Mistake 1: Forgetting +C

WRONG: The antiderivative of $x^2$ is $\frac{x^3}{3}$

RIGHT: The antiderivative of $x^2$ is $\frac{x^3}{3} + C$

Always include the constant of integration!

Mistake 2: Wrong Power Rule

WRONG: Antiderivative of $x^3$ is $\frac{x^3}{3}$ (forgot to add 1 to exponent!)

RIGHT: Antiderivative of $x^3$ is $\frac{x^4}{4} + C$

Mistake 3: Treating Product Like Sum

WRONG: Antiderivative of $x \cdot x^2 = x^3$ is... (treating separately)

RIGHT: Simplify first! $x \cdot x^2 = x^3$, so antiderivative is $\frac{x^4}{4} + C$

Mistake 4: Sine and Cosine Signs

Remember:

• Antiderivative of $\sin x$ is $-\cos x + C$ (negative!)
• Antiderivative of $\cos x$ is $\sin x + C$ (positive!)

Checking Your Work

Always check by differentiating your answer!

If you found $F(x)$ as an antiderivative of $f(x)$, verify: $F'(x) = f(x)$

This is a foolproof way to catch errors!

Initial Value Problems

Sometimes we can find the specific value of $C$ using an initial condition.

Example

Find $F(x)$ such that $F'(x) = 2x$ and $F(1) = 5$.

Step 1: Find general antiderivative $F(x) = x^2 + C$

Step 2: Use initial condition $F(1) = 5$ $F(1) = 1^2 + C = 5$ $1 + C = 5$ $C = 4$

Step 3: Write specific solution $F(x) = x^2 + 4$

Physical Interpretations

Position, Velocity, Acceleration

If $a(t)$ is acceleration:

• $v(t)$ is velocity (antiderivative of $a$)
• $s(t)$ is position (antiderivative of $v$)

Example: If $a(t) = -32$ ft/s² (gravity), then: $v(t) = -32t + C_1$ $s(t) = -16t^2 + C_1 t + C_2$

The constants are determined by initial conditions!

📝 Practice Strategy

1. Memorize basic antiderivatives (powers, trig, exponential)
2. Always include +C in your answer
3. Use linearity to break complex problems into simpler parts
4. Check your work by differentiating
5. For power rule: Add 1 to exponent, divide by new exponent
6. Watch for special cases like $\frac{1}{x}$
7. For initial value problems: Find general solution first, then solve for $C$

For $-6x$: $-6 \cdot \frac{x^2}{2} = -3x^2$

General solution: $f(x) = x^4 - 3x^2 + C$

Step 2: Use initial condition $f(2) = 10$

$f(2) = (2)^4 - 3(2)^2 + C = 10$

$16 - 3(4) + C = 10$

$16 - 12 + C = 10$

$4 + C = 10$

$C = 6$

Step 3: Write the specific solution

$f(x) = x^4 - 3x^2 + 6$

Check:

• $f'(x) = 4x^3 - 6x$ ✓
• $f(2) = 16 - 12 + 6 = 10$ ✓

Answer: $f(x) = x^4 - 3x^2 + 6$

$v(t) = \frac{12t^2}{2} - 6t + C_1 = 6t^2 - 6t + C_1$

Step 2: Use initial velocity $v(0) = 2$

$v(0) = 6(0)^2 - 6(0) + C_1 = 2$

$C_1 = 2$

So: $v(t) = 6t^2 - 6t + 2$

Step 3: Find position from velocity

$s(t) = \int v(t)\,dt = \int (6t^2 - 6t + 2)\,dt$

$s(t) = \frac{6t^3}{3} - \frac{6t^2}{2} + 2t + C_2$

$s(t) = 2t^3 - 3t^2 + 2t + C_2$

Step 4: Use initial position $s(0) = 5$

$s(0) = 2(0)^3 - 3(0)^2 + 2(0) + C_2 = 5$

$C_2 = 5$

Answer: $s(t) = 2t^3 - 3t^2 + 2t + 5$ meters

Summary:

• Acceleration: $a(t) = 12t - 6$
• Velocity: $v(t) = 6t^2 - 6t + 2$
• Position: $s(t) = 2t^3 - 3t^2 + 2t + 5$