Lesson 30 (11/09/26)

Lesson 30 (11/09/26)

Antiderivatives (4.9)

Warmup: \( f(x) = 2x \), find \( F(x) \) such that \( F'(x) = f(x) \)

The progression is shown as:

? &xrightarrow{derivative} \( 2x \) &xrightarrow{derivative} 2

The following expressions are grouped with a bracket labeled "Antiderivatives of \( 2x \)" :

• \( x^2 \)
• \( x^2 + 5 \)
• \( x^2 + 79865 \)

The progression is shown as:

? &xrightarrow{derivative} \( \sin x \) &xrightarrow{derivative} \( \cos x \)

The following expressions are grouped with a bracket labeled "Antiderivatives of \( \sin x \)" :

• \( -\cos x \)
• \( -\cos x - 17589 \)
• \( -\cos x + 573 \)

Antiderivative

Definition:

If \(F(x)\) is an antiderivative of \(f(x)\) then \(F(x) + C\) is also an antiderivative for any \(C\).

Indefinite integral notation for all antiderivatives

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

Annotation: The integral symbol \(\int\) is an elongated S which means "Sum".

Annotation: The function \(f(x)\) is the integrand, defined as "the function whose antiderivative we are finding".

Annotation: The variable \(x\) is the variable of integration.

Annotation: The term \(F(x)\) is one antiderivative of \(f(x)\).

Annotation: The term \(C\) is the constant of integration.

Examples:

1. \(\int 2x \, dx = x^2 + C\)

Margin Note: Verify: \(\frac{d}{dx}[x^2 + C] = 2x\)

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

Margin Note: \(\frac{d}{dx}[-\cos x + C] = \sin x + C\)

3. \(\int 2x + \sin x \, dx = x^2 - \cos x + C\)

Note: Underneath the integral, there are two arrows pointing to the terms in the result. The first arrow points from the \(2x\) term in the integrand to \(x^2\), labeled: \(x^2\) is Antiderivative. The second arrow points from the \(\sin x\) term in the integrand to \(-\cos x\), labeled: \(-\cos x\) is Antiderivative.

4. Find \(\int 5x^3 + 7x^{-2} + \cos x \, dx\)

Note: The problem uses arrows to indicate finding the antiderivative (finding the function whose derivative is the given term).

For the term \(x^3\):

Find function whose derivative is \(x^3\)

\[ \frac{d}{dx} x^4 = 4x^3 \]

\[ \frac{d}{dx} \left[ \frac{x^4}{4} \right] = x^3 \]

Therefore, the derivative of \(\frac{5x^4}{4}\) is \(5x^3\).

For the term \(7x^{-2}\):

\[ \frac{d}{dx} x^{-1} = -1 \cdot x^{-2} \]

\[ \frac{d}{dx} (-7x^{-1}) = 7x^{-2} \]

For the term \(\cos x\):

\[ \frac{d}{dx} [\sin x] = \cos x \]

Combining these, the calculation is as follows:

\[ \int 5x^3 + 7x^{-2} + \cos x \, dx = 5 \int x^3 \, dx + 7 \int x^{-2} \, dx + \int \cos x \, dx \]

\[ = 5 \frac{x^4}{4} - 7x^{-1} + \sin x + C \]

\[ \int x^n \, dx \]

Case \( n = -1 \):

\[ \int x^{-1} \, dx = \int \frac{1}{x} \, dx = \ln|x| + C \]

Case \( n \neq -1 \):

derivative Reduces power by 1

\[ \frac{d}{dx} [x^{n+1}] = (n+1) x^{(n+1)-1} = (n+1) \cdot x^n \quad (n \neq -1) \]

\[ \Rightarrow \frac{d}{dx} \left[ \frac{x^{n+1}}{n+1} \right] = x^n \]

\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1 \]

Indefinite Integral of Some Common Functions

\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1 \]

\[ \int x^{-1} \, dx = \int \frac{1}{x} \, dx = \ln|x| + C \]

\[ \int e^x \, dx = e^x + C \]

\[ \int \sin x \, dx = -\cos x + C \]

\[ \int \cos x \, dx = \sin x + C \]

\[ \int \sec^2 x \, dx = \tan x + C \]

\[ \int \sec x \tan x \, dx = \sec x + C \]

\[ \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) + C \]

\( x > 0 \): \[ \frac{d}{dx} \ln x = \frac{1}{x} \]

\( x < 0 \): \( -x > 0 \)

\[ \frac{d}{dx} (\ln(-x)) = \frac{1}{-x} \cdot \frac{d}{dx}(-x) = \frac{1}{x} \]

\[ \ln|x| = \begin{cases} \ln x & x > 0 \\ \ln(-x) & x < 0 \end{cases} \]

\( x \neq 0 \): \[ \frac{d}{dx} \ln|x| = \frac{1}{x} \]

Initial Value Problem

Given a condition to find ONE antiderivative of the function, satisfying the condition.

Example: Find \(F(x)\) such that \(F'(x) = 2x\), \(F(0) = 5\)

\[F(x) = x^2 + C\]

\[F(0) = 5\]

\[5 = 0 + C \implies C = 5\]

\[F(x) = x^2 + 5\]

Example: Find \( f(x) \) such that \( f'(x) = \sec^2 x \), \( f(\pi) = 3 \)

\[ \int \sec^2 x \, dx = \tan x + c \]

\[ f(x) = \tan x + c \]

Note: want \( f(\pi) = 3 \)

\[ 3 = f(\pi) = \tan \pi + c = 0 + c \]

\[ \implies C = 3 \]

\[ f(x) = \tan x + 3 \]

Higher Order Derivative

\[ F''(x) = 2x \]

Note: There is an arrow pointing from \( F''(x) = 2x \) to \( F'(x) = x^2 + C \) labeled "derivative" and another arrow pointing from \( F'(x) = x^2 + C \) to \( F(x) = \frac{x^3}{3} + Cx + D \) also labeled "derivative".

Note: There is a downward arrow from \( F''(x) = 2x \) to \( F'(x) \) labeled "Antiderivative".

\[ F'(x) = \int 2x \, dx = x^2 + C \]

Note: There is a downward arrow from \( F'(x) = x^2 + C \) to \( F(x) = \dots \) labeled "Antiderivative".

\[ F(x) = \int x^2 + C \, dx = \frac{x^3}{3} + Cx + D \]

Remark: Because you started with \( F'' \) and found antiderivative twice, we have two constants of integration.

Remark: Need 2 conditions to find \( C \) and \( D \).

Example: Find \( F(x) \) such that \( F''(x) = 2x \), \( F(0) = 3 \), \( F'(1) = 4 \)

To find the general form of \( F(x) \), we integrate \( F''(x) = 2x \) twice:

First integration gives:

\[ F'(x) = x^2 + C \]

Second integration gives:

\[ F(x) = \frac{x^3}{3} + Cx + D \]

Now we use the given conditions to solve for constants \( C \) and \( D \).

Using the condition \( F(0) = 3 \):

\[ 3 = F(0) = \frac{0^3}{3} + C(0) + D \implies D = 3 \]

Using the condition \( F'(1) = 4 \):

\[ 4 = F'(1) = 1^2 + C \] \[ 4 = 1 + C \implies C = 3 \]

Substituting the values of \( C \) and \( D \) back into the expression for \( F(x) \):

\[ F(x) = \frac{x^3}{3} + 3x + 3 \]

Example: Find \( f(x) \), such that \( f''(x) = \sin x \), \( f(0) = -11 \), \( f'(0) = 12 \)

\( f''(x) = \sin x \)

Annotation: Anti derivative

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

Note: want \( f'(0) = 12 \implies 12 = -\cos(0) + C \implies C = 13 \)

\( f'(x) = -\cos x + 13 \)

Annotation: Anti derivative

\( f(x) = \int -\cos x + 13 \, dx = -\sin x + 13x + D \)

Note: want \( f(0) = -11 \implies -11 = -\sin(0) + 13(0) + D \implies D = -11 \)

\( f(x) = -\sin x + 13x - 11 \)