# Lecture 18: Antiderivatives

## Calculus I — Differential Calculus

---

## Learning Objectives

By the end of this lecture, students will be able to:

1. Define an antiderivative and the indefinite integral
2. Apply basic antidifferentiation rules (power rule, trig, exponential, logarithmic)
3. Use the "+C" convention and explain why it is necessary
4. Solve initial value problems using antiderivatives
5. Relate antiderivatives to motion (position from velocity, velocity from acceleration)

---

## Lecture Content

### I. Definition of Antiderivative

A function F is an **antiderivative** of f on an interval I if F'(x) = f(x) for all x in I. For example, F(x) = x^3 is an antiderivative of f(x) = 3x^2, because d/dx[x^3] = 3x^2. Antiderivatives are not unique: if F is an antiderivative of f, then so is F(x) + C for any constant C.

A theorem guarantees that if F is any one antiderivative of f on an interval, then every antiderivative of f has the form F(x) + C. This follows from the Mean Value Theorem: if F'(x) = G'(x) for all x, then F(x) - G(x) must be a constant.

### II. Indefinite Integral Notation

The **indefinite integral** of f(x) is written as integral of f(x) dx = F(x) + C, where F'(x) = f(x) and C is the **constant of integration**. The integral sign is an elongated S (hinting at the connection to summation that will appear with definite integrals). The function f(x) is called the **integrand**, and dx indicates the variable of integration. It is essential to **always include +C** for indefinite integrals -- omitting it is one of the most common errors in calculus.

### III. Basic Antidifferentiation Rules

The **Power Rule** in reverse states that the integral of x^n dx = x^{n+1}/(n+1) + C for n not equal to -1. For instance, the integral of x^4 dx = x^5/5 + C, the integral of x^{-3} dx = -1/(2x^2) + C, and the integral of sqrt(x) dx = (2/3)x^{3/2} + C.

The **special case n = -1** requires separate treatment: the integral of 1/x dx = ln|x| + C. The **constant multiple rule** says the integral of c * f(x) dx = c * integral of f(x) dx, and the **sum/difference rule** allows term-by-term integration.

### IV. Antiderivatives of Common Functions

The trigonometric antiderivatives are: integral of cos x dx = sin x + C, integral of sin x dx = -cos x + C, integral of sec^2 x dx = tan x + C, integral of csc^2 x dx = -cot x + C, integral of sec x tan x dx = sec x + C, and integral of csc x cot x dx = -csc x + C.

The exponential antiderivatives are: integral of e^x dx = e^x + C, and integral of a^x dx = a^x / ln a + C (for a > 0, a not equal to 1).

Two inverse-trig-related antiderivatives are particularly important: integral of 1/sqrt(1 - x^2) dx = arcsin x + C, and integral of 1/(1 + x^2) dx = arctan x + C.

<image>A two-column reference table of basic antiderivatives. Left column shows the integrand f(x); right column shows the antiderivative F(x) + C. Includes: power rule (x^n for general n and n = -1), trig functions (sin, cos, sec^2, csc^2, sec*tan, csc*cot), exponentials (e^x, a^x), and inverse trig (1/sqrt(1-x^2), 1/(1+x^2)). Title: "Table of Basic Antiderivatives."</image>

### V. Antidifferentiation by Rewriting

Many integrands must be rewritten before a basic rule applies. **Expanding products** is one strategy: the integral of (x + 1)^2 dx = integral of (x^2 + 2x + 1) dx = x^3/3 + x^2 + x + C. **Dividing each term** is another: the integral of (x^3 + 2x)/x dx = integral of (x^2 + 2) dx = x^3/3 + 2x + C. **Rewriting radicals and fractions as powers** makes the power rule applicable: the integral of 1/x^4 dx = integral of x^{-4} dx = x^{-3}/(-3) + C. **Using trig identities** can also simplify integrands: the integral of tan^2 x dx = integral of (sec^2 x - 1) dx = tan x - x + C.

### VI. Initial Value Problems

An **initial value problem (IVP)** provides a differential equation together with an initial condition, which determines the constant C. The general process is: first, find the general antiderivative (with +C); second, substitute the initial condition to solve for C; third, write the particular solution.

For example, given f'(x) = 6x - 2 and f(1) = 5, the general antiderivative is f(x) = 3x^2 - 2x + C. The initial condition gives f(1) = 3 - 2 + C = 5, so C = 4, and the particular solution is f(x) = 3x^2 - 2x + 4.

### VII. Motion Applications

Antiderivatives connect acceleration to velocity and velocity to position. Given acceleration a(t), the velocity is v(t) = integral of a(t) dt + C_1, where the initial velocity v(0) determines C_1. The position is then s(t) = integral of v(t) dt + C_2, where the initial position s(0) determines C_2.

For **free fall**, a(t) = -g = -9.8 m/s^2, and integrating twice with initial conditions v(0) = v_0 and s(0) = s_0 gives v(t) = -9.8t + v_0 and s(t) = -4.9t^2 + v_0 t + s_0. As another example, if a particle has acceleration a(t) = 12t - 6, initial velocity v(0) = 4, and initial position s(0) = 1, then v(t) = 6t^2 - 6t + 4 and s(t) = 2t^3 - 3t^2 + 4t + 1.

<image>A three-panel vertical stack showing the relationships between acceleration, velocity, and position. Panel A (top): a(t) = constant (horizontal line), with an arrow labeled "integrate" pointing down to Panel B. Panel B (middle): v(t) = linear function, with an arrow labeled "integrate" pointing down to Panel C. Panel C (bottom): s(t) = quadratic function (parabola). Reverse arrows labeled "differentiate" point upward. The initial conditions v(0) and s(0) are shown determining the integration constants. Title: "From acceleration to position via antiderivatives."</image>