Anti-derivative Introduction for MATH 139

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Know basic antiderivatives and plus C. Reverse the derivative rules. Often the last topic before final.

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Practice Final #3

Lesson

What is an Antiderivative?

An antiderivative is the reverse of a derivative. If you know the derivative of a function, the antiderivative takes you back to the original function.

The Big Idea: Differentiation and antidifferentiation are opposite operations, just like multiplication and division.

Connection to Area

Antiderivatives are deeply connected to area under a curve. The Fundamental Theorem of Calculus (which you'll learn soon) shows that finding the area under $f(x)$ from $a$ to $b$ requires finding an antiderivative of $f(x)$. This is why antiderivatives are also called integrals — they "integrate" (add up) infinitely many tiny areas.

The Key Question

Derivatives ask: "Given $f(x)$, what is $f'(x)$?"

Antiderivatives ask: "Given $f'(x)$, what was $f(x)$?"

A Simple Example

We know that $\frac{d}{dx}[x^2] = 2x$.

So if someone asks: "What function has derivative $2x$?"

The answer is $x^2$... but also $x^2 + 1$, or $x^2 - 7$, or $x^2 + \pi$.

Any constant disappears when you differentiate!

The "+C" — Why It Matters

Since the derivative of any constant is zero, antiderivatives always include an arbitrary constant $C$.

$$\frac{d}{dx}[x^2 + C] = 2x \quad \text{for ANY value of } C$$

This is why we write:

$$\int 2x \, dx = x^2 + C$$

The $C$ represents all possible antiderivatives — there are infinitely many, forming a "family" of functions that differ only by a vertical shift.

Notation

Symbol	Meaning
$\int f(x) \, dx$	"The antiderivative of $f(x)$ with respect to $x$"
$F(x)$	An antiderivative of $f(x)$, meaning $F'(x) = f(x)$
$+ C$	The constant of integration (don't forget it!)

The $\int$ symbol is called an integral sign. When there are no bounds (limits), this is an indefinite integral.

Basic Antiderivative Rules

These are the derivative rules in reverse:

Power Rule (Reversed)

$$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$$

Why? Because $\frac{d}{dx}\left[\frac{x^{n+1}}{n+1}\right] = \frac{(n+1)x^n}{n+1} = x^n$ ✓

Special Case: $n = -1$

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

Exponential

$$\int e^x \, dx = e^x + C$$

Trigonometric

$$\int \cos(x) \, dx = \sin(x) + C$$

$$\int \sin(x) \, dx = -\cos(x) + C$$

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

Constant Multiple Rule

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

Sum/Difference Rule

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

Example 1: Basic Power Rule

Find $\displaystyle \int x^4 \, dx$.

Using the power rule with $n = 4$:

$$\int x^4 \, dx = \frac{x^{4+1}}{4+1} + C = \frac{x^5}{5} + C$$

$$\boxed{\frac{x^5}{5} + C}$$

Example 2: Polynomial

Find $\displaystyle \int (3x^2 - 5x + 2) \, dx$.

Step 1: Split into separate integrals

$$\int (3x^2 - 5x + 2) \, dx = \int 3x^2 \, dx - \int 5x \, dx + \int 2 \, dx$$

Step 2: Apply power rule to each term

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

$$= x^3 - \frac{5x^2}{2} + 2x + C$$

$$\boxed{x^3 - \frac{5x^2}{2} + 2x + C}$$

Example 3: Negative and Fractional Exponents

Find $\displaystyle \int \left( \frac{1}{x^3} + \sqrt{x} \right) dx$.

Step 1: Rewrite using exponents

$$\frac{1}{x^3} = x^{-3} \quad \text{and} \quad \sqrt{x} = x^{1/2}$$

Step 2: Apply power rule

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

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

Step 3: Combine

$$\boxed{-\frac{1}{2x^2} + \frac{2x^{3/2}}{3} + C}$$

Example 4: Trig Functions

Find $\displaystyle \int (4\cos(x) - 3\sin(x)) \, dx$.

$$\int 4\cos(x) \, dx = 4\sin(x)$$

$$\int (-3\sin(x)) \, dx = -3 \cdot (-\cos(x)) = 3\cos(x)$$

$$\boxed{4\sin(x) + 3\cos(x) + C}$$

Example 5: Initial Value Problem

Find $f(x)$ if $f'(x) = 6x^2 - 4$ and $f(1) = 5$.

Step 1: Find the general antiderivative

$$f(x) = \int (6x^2 - 4) \, dx = 6 \cdot \frac{x^3}{3} - 4x + C = 2x^3 - 4x + C$$

Step 2: Use the initial condition to find $C$

$f(1) = 5$ means when $x = 1$, $f(x) = 5$:

$$2(1)^3 - 4(1) + C = 5$$

$$2 - 4 + C = 5$$

$$C = 7$$

Step 3: Write the specific solution

$$\boxed{f(x) = 2x^3 - 4x + 7}$$

Example 6: Velocity and Position

A particle moves with velocity $v(t) = 3t^2 - 2t$ m/s. If its position at $t = 0$ is $s(0) = 4$ meters, find its position function $s(t)$.

Step 1: Position is the antiderivative of velocity

$$s(t) = \int v(t) \, dt = \int (3t^2 - 2t) \, dt = t^3 - t^2 + C$$

Step 2: Use initial condition

$s(0) = 4$:

$$0^3 - 0^2 + C = 4$$

$$C = 4$$

$$\boxed{s(t) = t^3 - t^2 + 4}$$

Example 7: Checking Your Answer

Verify that $\displaystyle \int (3x^2 - 5x + 2) \, dx = x^3 - \frac{5x^2}{2} + 2x + C$.

To check an antiderivative, differentiate the answer. If you get back the original integrand, it's correct.

$$\frac{d}{dx}\left[x^3 - \frac{5x^2}{2} + 2x + C\right] = 3x^2 - \frac{5 \cdot 2x}{2} + 2 + 0 = 3x^2 - 5x + 2 \checkmark$$

$$\boxed{\text{Verified! The antiderivative is correct.}}$$

Tips for Antiderivatives

Rewrite first — convert roots to fractional exponents, fractions to negative exponents
Don't forget $+C$ — unless you have an initial condition
Check by differentiating — the derivative of your answer should give the original
Power rule exception — when $n = -1$, use $\ln|x|$ instead
Watch the signs — especially with trig ($\int \sin x = -\cos x$, not $+\cos x$)