4.10 Antiderivatives

Antiderivatives and Indefinite Integrals

Antiderivatives reverse the process of differentiation. Instead of asking "what's the derivative of this function?", you're asking "what function has this as its derivative?" This concept bridges everything you've learned about derivatives with the integration techniques you'll use throughout the rest of calculus.

General Antiderivatives of Functions

A function $F(x)$ is an antiderivative of $f(x)$ if $F'(x) = f(x)$. You're working backwards from a derivative to recover the original function.

For example, if $f(x) = 3x^2$, then $F(x) = x^3$ is an antiderivative because the derivative of $x^3$ is $3x^2$.

But here's the catch: $x^3 + 5$ also has a derivative of $3x^2$, and so does $x^3 - 17$. Since the derivative of any constant is zero, there are infinitely many antiderivatives for a given function. They all differ by a constant vertical shift.

The general antiderivative accounts for this by including an arbitrary constant $C$:

$F(x) = x^3 + C$

The value of $C$ can't be determined from the function alone. You need an additional piece of information, like an initial condition, to pin it down.

Indefinite Integrals and Notation

The indefinite integral is the notation used to represent the general antiderivative. It's written as:

$\int f(x) \, dx = F(x) + C$

Here's what each piece means:

• $\int$ is the integral sign, indicating you're finding an antiderivative
• $f(x)$ is the integrand, the function you're integrating
• $dx$ tells you the variable of integration (you're integrating with respect to $x$)
• $C$ is the constant of integration, representing the entire family of antiderivatives

For example: $\int 3x^2 \, dx = x^3 + C$

An indefinite integral gives you a family of functions, not a single number. Don't forget the $+ C$. Leaving it off is one of the most common mistakes on exams.

Power Rule for Integration

The power rule for integration is the reverse of the power rule for derivatives. For $f(x) = x^n$ where $n \neq -1$:

$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$

The steps are straightforward:

1. Add 1 to the exponent
2. Divide by the new exponent
3. Add $C$

Example: $\int x^4 \, dx = \frac{x^5}{5} + C$

You can verify this by differentiating: the derivative of $\frac{x^5}{5}$ is $\frac{5x^4}{5} = x^4$. Checking your answer by differentiating is a reliable way to catch errors.

For sums of terms, integrate each term separately:

$\int (3x^2 + 2x) \, dx = x^3 + x^2 + C$

Notice you only need one $C$ at the end, not one per term. The individual constants just combine into a single arbitrary constant.

The restriction $n \neq -1$ matters because plugging in $n = -1$ would give division by zero. The antiderivative of $x^{-1} = \frac{1}{x}$ is $\ln|x| + C$, which you'll use more in Calculus II.

Initial-Value Problems via Antidifferentiation

An initial-value problem gives you a derivative and a specific point the original function must pass through. This lets you solve for the exact value of $C$.

Steps:

1. Find the general antiderivative (with $C$)
2. Plug in the given condition to solve for $C$
3. Write the specific antiderivative with that value of $C$

Example: Find $F(x)$ given that $F'(x) = 4x^3$ and $F(1) = 2$.

1. General antiderivative: $F(x) = \int 4x^3 \, dx = x^4 + C$
2. Apply the condition: $F(1) = 2 \implies 1^4 + C = 2 \implies C = 1$
3. Specific solution: $F(x) = x^4 + 1$

You can verify: $F'(x) = 4x^3$ ✓ and $F(1) = 1 + 1 = 2$ ✓

Relationship Between Derivatives and Antiderivatives

Differentiation and integration are inverse operations. If you differentiate a function and then integrate the result, you get back to the original function (up to a constant). If you integrate and then differentiate, you get back exactly where you started:

$\frac{d}{dx}\left[\int f(x)\,dx\right] = f(x)$

This inverse relationship is formalized by the Fundamental Theorem of Calculus, which you'll encounter soon. It connects the indefinite integrals you're learning now with definite integrals that compute accumulated quantities like area.