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Indefinite Integrals

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Reference: Stewart §4.4

Textbook Reference

Primary source OpenStax Calculus Volume 2, Section 1.4: “Integration Formulas and the Net Change Theorem”
Direct link https://openstax.org/books/calculus-volume-2/pages/1-4-integration-formulas-and-the-net-change-theorem
Textbook used in class Stewart, Calculus, Section 4.4: “Indefinite Integrals and the Net Change Theorem”

Opening Scenario

In Section 3.9, an antiderivative of $f$ was defined as any $F$ with $F' = f$. Now Stewart introduces the indefinite integral $\int f(x)\,dx$, which is just the antiderivative written with the integral sign. This is notation, not a new concept -- but the notation matters because it ties antiderivatives visually to the definite integral, reinforcing the FTC connection.


Quick Reference

Definition. $\displaystyle\int f(x)\,dx = F(x) + C$ means $F'(x) = f(x)$.

Contrast with definite integral: | | $\displaystyle\int_a^b f(x)\,dx$ | $\displaystyle\int f(x)\,dx$ | |---|---|---| | Name | Definite integral | Indefinite integral | | Result | A number | A family of functions | | Depends on | $a$, $b$, and $f$ | Only $f$ | | Constant | No $+C$ needed | Must include $+C$ |

Table of basic integration formulas:

$f(x)$ $\displaystyle\int f(x)\,dx$
$x^n$ ($n \neq -1$) $\dfrac{x^{n+1}}{n+1} + C$
$\cos x$ $\sin x + C$
$\sin x$ $-\cos x + C$
$\sec^2 x$ $\tan x + C$
$\csc^2 x$ $-\cot x + C$
$\sec x \tan x$ $\sec x + C$
$\csc x \cot x$ $-\csc x + C$
$e^x$ $e^x + C$
$\frac{1}{x}$ $\ln|x| + C$

Key Concepts

1. The Notation $\int f(x)\,dx$

The indefinite integral $\int f(x)\,dx$ is the collection of all antiderivatives of $f$. Because all antiderivatives of $f$ differ by a constant (a consequence of the MVT, proved in Section 3.2), the entire family is written as $F(x) + C$ where $F$ is any one antiderivative.

The integral sign $\int$ and the $dx$ are read together as “the antiderivative of ... with respect to $x$.” The $dx$ names the variable being integrated. This matters when expressions involve multiple variables (common in later courses).

2. Building an Integration Table

Each differentiation formula can be reversed to give an integration formula:

Every formula in the table can be verified: differentiate the right side and confirm it equals the integrand.

3. Linearity of the Indefinite Integral

$$\int [cf(x) + g(x)]\,dx = c\int f(x)\,dx + \int g(x)\,dx.$$

This lets you integrate a sum term by term and pull constants out. It is the same linearity property as for the definite integral, because the indefinite integral is defined via antiderivatives.

Common misconception

“The indefinite integral and the definite integral are the same thing.” They are related by FTC Part 2, but they are not the same. The definite integral $\int_a^b f(x)\,dx$ is a number. The indefinite integral $\int f(x)\,dx$ is a family of functions. The connection is: $\int_a^b f(x)\,dx = \left[\int f(x)\,dx\right]_a^b$, meaning evaluate any antiderivative at $b$, subtract the value at $a$.


Worked Example

Evaluate $\displaystyle\int \left(3x^4 - 2x + \frac{5}{x^2} + e^x\right)dx$.

Write $\dfrac{5}{x^2} = 5x^{-2}$ before integrating.

By linearity and the power rule:

$$= 3\cdot\frac{x^5}{5} - 2\cdot\frac{x^2}{2} + 5\cdot\frac{x^{-1}}{-1} + e^x + C$$ $$= \frac{3x^5}{5} - x^2 - \frac{5}{x} + e^x + C.$$

Verification: Differentiate the answer: $3x^4 - 2x + 5x^{-2} + e^x = 3x^4 - 2x + \dfrac{5}{x^2} + e^x$. Confirmed.


Evaluate $\displaystyle\int \frac{t^3 - 1}{t^2}\,dt$.

Rewrite: $\dfrac{t^3 - 1}{t^2} = t - t^{-2}$.

$\displaystyle\int (t - t^{-2})\,dt = \frac{t^2}{2} - \frac{t^{-1}}{-1} + C = \frac{t^2}{2} + \frac{1}{t} + C$.

Boxed answer: $\dfrac{t^2}{2} + \dfrac{1}{t} + C$.


Common Errors Summary

Error Example Correction
Omitting $+ C$ $\int x^2\,dx = \dfrac{x^3}{3}$ The complete answer is $\dfrac{x^3}{3} + C$; omitting $C$ specifies one particular antiderivative, not the family
Applying power rule to $x^{-1}$ $\int x^{-1}\,dx = \dfrac{x^0}{0} + C$ Division by zero: $\int \frac{1}{x}\,dx = \ln|x| + C$ is the special case
Forgetting to rewrite before integrating Integrating $\dfrac{3}{x^2}$ as $\dfrac{3}{x^3/3}$ Rewrite as $3x^{-2}$ first, then apply the power rule: $3\cdot\dfrac{x^{-1}}{-1} + C = -\dfrac{3}{x} + C$

Leveled Practice

Level 1 -- Power Rule

Problem 1. Evaluate $\displaystyle\int (5x^3 - 4x + 2)\,dx$.

Show answer

$\dfrac{5x^4}{4} - 2x^2 + 2x + C$.


Level 2 -- Rewriting First

Problem 2. Evaluate $\displaystyle\int \frac{x^3 + 2\sqrt{x}}{x}\,dx$.

Show answer

$\dfrac{x^3 + 2\sqrt{x}}{x} = x^2 + 2x^{-1/2}$.

$\displaystyle\int (x^2 + 2x^{-1/2})\,dx = \dfrac{x^3}{3} + 2\cdot\dfrac{x^{1/2}}{1/2} + C = \dfrac{x^3}{3} + 4\sqrt{x} + C$.


Level 3 -- Mixed Functions

Problem 3. Evaluate $\displaystyle\int (\sec^2 \theta - 3e^\theta)\,d\theta$.

Show answer

$\tan\theta - 3e^\theta + C$.


Mastery Checklist


Mental Model

The indefinite integral is the derivative run backward. Every entry in the differentiation table has a mirror entry in the integration table. The $+C$ at the end is the permanent reminder that reversing the derivative loses information about the vertical position of the curve -- the constant is the “unknown starting height.”


Connections

Looking back

Looking ahead


Back to Integration Foundations | Next: Applications of Net Change