
Session 06-01 - Antiderivatives & Indefinite Integrals
Section 06: Integral Calculus
Entry Quiz - 10 Minutes
Quick Review from Section 05
Test your understanding of Differential Calculus
Find \(f'(x)\) for \(f(x) = 3x^4 - 2x^2 + 5x - 7\)
A profit function is \(P(x) = -x^2 + 80x - 400\). Find the production level that maximizes profit.
For \(g(x) = x^3 - 6x^2 + 9x + 1\), find all critical points and classify them.
If \(C'(x) = 6x + 10\) represents marginal cost, what does \(C'(5) = 40\) mean in business terms?
Homework Discussion - 15 Minutes
Your questions from Section 05
Focus on differential calculus applications
- Challenges with optimization problems
- Curve sketching and second derivative test
- Function determination (Funktionsscharen)
- Any remaining questions before we move forward
. . .
Today we begin Integral Calculus - the reverse of differentiation!
Learning Objectives
What You’ll Master Today
- Understand the antiderivative concept as reversing differentiation
- Apply the power rule for integration: \(\int x^n \, dx\)
- Use sum and constant rules to integrate polynomials
- Interpret the constant of integration (+C) and families of functions
- Solve initial value problems to find specific antiderivatives
- Verify antiderivatives by differentiation
- Apply integration to business (marginal cost → total cost)
. . .
Integration is the reverse of differentiation - today we learn to “undo” derivatives!
Part A: The Antiderivative Concept
From Derivatives to Antiderivatives
Remember differentiation?
\[f(x) = x^3 \quad \xrightarrow{\text{differentiate}} \quad f'(x) = 3x^2\]
. . .
Now we reverse the process:
\[F'(x) = 3x^2 \quad \xleftarrow{\text{antidifferentiate}} \quad F(x) = x^3\]
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\(F(x)\) is an antiderivative of \(f(x)\) if \(F'(x) = f(x)\).
In German: \(F\) is the Stammfunktion of \(f\).
The Question We’re Solving
Given a derivative, find the original function
. . .
Example Questions:
- If \(f'(x) = 2x\), what could \(f(x)\) be?
- If the marginal cost is \(C'(x) = 50\), what is the total cost function?
- If velocity is \(v(t) = 3t^2\), what is the position function?
. . .
Business Context: We often know the rate of change (marginal cost, growth rate) and need to find the total quantity (total cost, population).
Finding: First Examples
Question: If \(F'(x) = 2x\), what is \(F(x)\)?
. . .
Think: What function, when differentiated, gives \(2x\)?
- If \(F(x) = x^2\), then \(F'(x) = 2x\) ✓
- But also: \(F(x) = x^2 + 5\) gives \(F'(x) = 2x\) ✓
- And: \(F(x) = x^2 - 100\) gives \(F'(x) = 2x\) ✓
. . .
Problem: There are infinitely many antiderivatives!
The Family of Antiderivatives
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All these curves have the same slope at each x-value!
The Constant of Integration
Solution: We write the general antiderivative with a constant \(C\):
. . .
\[\int 2x \, dx = x^2 + C\]
. . .
- The symbol \(\int\) means “integrate” (find the antiderivative)
- \(dx\) indicates we integrate with respect to \(x\)
- \(+C\) represents any constant (the constant of integration)
- This notation is called an indefinite integral
. . .
Always include +C for indefinite integrals!
Part B: The Power Rule for Integration
Reversing the Power Rule
Recall the power rule for derivatives:
\[\frac{d}{dx}[x^n] = nx^{n-1}\]
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Reverse it for integration:
\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad \text{for } n \neq -1\]
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Memory trick: “Add 1 to the power, divide by the new power”
Power Rule Examples
Example 1: \(\int x^3 \, dx\)
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\[\int x^3 \, dx = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C\]
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Check: \(\frac{d}{dx}\left[\frac{x^4}{4} + C\right] = \frac{4x^3}{4} = x^3\) ✓
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Example 2: \(\int x^5 \, dx\)
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\[\int x^5 \, dx = \frac{x^6}{6} + C\]
Special Cases
Constant function: \(\int 1 \, dx\) (or \(\int dx\))
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Think of \(1 = x^0\): \[\int x^0 \, dx = \frac{x^{0+1}}{0+1} + C = x + C\]
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Linear function: \(\int x \, dx\)
\[\int x^1 \, dx = \frac{x^2}{2} + C\]
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The antiderivative of a constant is a line, the antiderivative of \(x\) is a parabola.
Negative and Fractional Powers
Negative powers: \(\int x^{-2} \, dx\)
. . .
\[\int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} + C = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C\]
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Fractional powers: \(\int x^{1/2} \, dx = \int \sqrt{x} \, dx\)
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\[\int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} + C = \frac{2x^{3/2}}{3} + C = \frac{2\sqrt{x^3}}{3} + C\]
. . .
The power rule fails when \(n = -1\) (we would divide by zero). We’ll handle \(\int \frac{1}{x} \, dx\) later.
Part C: Sum and Constant Multiple Rules
Constant Multiple Rule
If \(k\) is a constant:
\[\int k \cdot f(x) \, dx = k \cdot \int f(x) \, dx\]
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Example: \(\int 5x^3 \, dx\)
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\[\int 5x^3 \, dx = 5 \int x^3 \, dx = 5 \cdot \frac{x^4}{4} + C = \frac{5x^4}{4} + C\]
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Pull constants outside the integral, then integrate!
Sum and Difference Rules
For sums and differences:
\[\int [f(x) \pm g(x)] \, dx = \int f(x) \, dx \pm \int g(x) \, dx\]
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Example: \(\int (x^2 + 3x - 5) \, dx\)
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\[= \int x^2 \, dx + \int 3x \, dx - \int 5 \, dx\]
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\[= \frac{x^3}{3} + 3 \cdot \frac{x^2}{2} - 5x + C = \frac{x^3}{3} + \frac{3x^2}{2} - 5x + C\]
Polynomial Integration
Example: \(\int (4x^3 - 6x^2 + 2x - 7) \, dx\)
- \(\int 4x^3 \, dx = 4 \cdot \frac{x^4}{4} = x^4\)
- \(\int 6x^2 \, dx = 6 \cdot \frac{x^3}{3} = 2x^3\)
- \(\int 2x \, dx = 2 \cdot \frac{x^2}{2} = x^2\)
- \(\int 7 \, dx = 7x\)
. . .
Answer: \(\int (4x^3 - 6x^2 + 2x - 7) \, dx = x^4 - 2x^3 + x^2 - 7x + C\)
. . .
Check: \(\frac{d}{dx}[x^4 - 2x^3 + x^2 - 7x + C] = 4x^3 - 6x^2 + 2x - 7\) ✓
Practice - 10 Minutes
Practice: Basic Integration
Work individually for 3 minutes
Find the following antiderivatives:
- \(\int 6x^2 \, dx\)
- \(\int (x^4 - 3x^2 + 1) \, dx\)
- \(\int \frac{4}{x^3} \, dx\) (rewrite as power first)
- \(\int (2\sqrt{x} + 3) \, dx\)
Integration Drill
Work in pairs
Evaluate these integrals and verify by differentiation:
- \(\int (3x^2 + 4x + 5) \, dx\)
- \(\int (x^3 - 2x + 4) \, dx\)
- \(\int 10x^4 \, dx\)
- \(\int \left(x^2 + \frac{1}{x^2}\right) \, dx\)
Break - 10 Minutes
Part D: Initial Value Problems
Finding a Specific Antiderivative
Problem: The general antiderivative has infinitely many solutions. How do we find a specific one?
. . .
Solution: Use an initial condition, a known point on the function.
- Given \(f'(x)\) and a condition like \(f(a) = b\)
- Find the specific function \(f(x)\).
. . .
Steps:
- Find the general antiderivative with \(+C\)
- Substitute the initial condition to solve for \(C\)
- Write the specific solution
Example: Initial Value Problem
Problem: Find \(f(x)\) if \(f'(x) = 6x^2 - 4\) and \(f(1) = 5\).
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Step 1: Find the general antiderivative
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\[f(x) = \int (6x^2 - 4) \, dx = 2x^3 - 4x + C\]
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Step 2: Use the initial condition \(f(1) = 5\)
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\[f(1) = 2(1)^3 - 4(1) + C = 5; C = 7\]
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Step 3: Write the specific solution
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\[f(x) = 2x^3 - 4x + 7\]
Visualizing the Initial Condition

The initial condition selects one curve from the family!
Total Cost from Marginal Cost
Scenario: A manufacturing company knows its marginal cost function:
\[MC(x) = C'(x) = 0.3x^2 - 2x + 50\]
The fixed costs are €1,000 (cost when producing 0 units).
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Question: Find the total cost function \(C(x)\).
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Step 1: Integrate marginal cost
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\[C(x) = \int (0.3x^2 - 2x + 50) \, dx = 0.1x^3 - x^2 + 50x + C\]
Completing the Business Problem
Step 2: Use the initial condition \(C(0) = 1000\)
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\[C(0) = 0.1(0)^3 - (0)^2 + 50(0) + C = 1000\] \[C = 1000\]
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Step 3: Final answer
\[C(x) = 0.1x^3 - x^2 + 50x + 1000\]
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The constant \(C\) represents the fixed costs, costs that don’t depend on production level.
Visualizing Cost Functions

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The marginal cost (derivative) shows the rate; the total cost (antiderivative) shows the accumulated total.
Part E: Velocity and Position
Application: Motion
Key relationships:
- Position: \(s(t)\) = where the object is
- Velocity: \(v(t) = s'(t)\) = rate of change of position
- Acceleration: \(a(t) = v'(t) = s''(t)\) = rate of change of velocity
. . .
Reversing the relationships:
- If you know velocity, integrate to find position: \(s(t) = \int v(t) \, dt\)
- If you know acceleration, integrate to find velocity: \(v(t) = \int a(t) \, dt\)
Finding Position from Velocity
Problem: A car’s velocity is \(v(t) = 3t^2 + 2t\) m/s, where \(t\) is in seconds. At \(t = 0\), the car is at \(s = 10\) meters. Find the position function \(s(t)\).
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Step 1: Integrate velocity to get position
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\[s(t) = \int (3t^2 + 2t) \, dt = t^3 + t^2 + C\]
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Step 2: Use initial condition \(s(0) = 10\)
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\[s(0) = 0 + 0 + C = 10 \implies C = 10\]
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\[s(t) = t^3 + t^2 + 10\]
Verifying Our Answer
Check: Does \(s'(t) = v(t)\)?
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\[s(t) = t^3 + t^2 + 10\] \[s'(t) = 3t^2 + 2t = v(t) \quad \checkmark\]
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Check: Does \(s(0) = 10\)?
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\[s(0) = 0 + 0 + 10 = 10 \quad \checkmark\]
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Always check your antiderivative by differentiating to recover the original function and substituting the initial condition!
Guided Practice - 20 Minutes
Practice Set A: Basic Antiderivatives
Work individually for 5 minutes
Find the following indefinite integrals:
- \(\int (5x^4 - 3x^2 + 2) \, dx\)
- \(\int (x^3 + x^{-1/2}) \, dx\)
- \(\int \frac{6}{x^4} \, dx\)
- \(\int (4\sqrt{x} - \frac{3}{x^2}) \, dx\)
Practice Set B: Initial Value Problems
Work individually for 7 minutes
Solve these initial value problems:
\(f'(x) = 4x^3 - 6x\), \(f(2) = 10\). Find \(f(x)\).
\(g'(x) = 3x^2 + 4\), \(g(0) = 5\). Find \(g(x)\).
The marginal revenue is \(MR(x) = 100 - 2x\). Revenue is €0 when no units are sold. Find \(R(x)\).
Practice Set C: Business Applications
Work in pairs for 8 minutes
A company’s marginal cost is \(MC(x) = 20 + 0.4x\) euros per unit. Fixed costs are €500. Find the total cost function \(C(x)\) and calculate the total cost of producing 50 units.
A product’s marginal profit is \(MP(x) = 80 - 4x\) euros per unit. The company breaks even (profit = 0) when selling 0 units. Find the profit function \(P(x)\) and determine how many units maximize profit.
Coffee Break - 15 Minutes
Business Applications Deep Dive
From Rates to Totals
Integration connects rates of change to accumulated totals:
| Rate Function (Derivative) | Total Function (Antiderivative) |
|---|---|
| Marginal cost \(C'(x)\) | Total cost \(C(x)\) |
| Marginal revenue \(R'(x)\) | Total revenue \(R(x)\) |
| Marginal profit \(P'(x)\) | Total profit \(P(x)\) |
| Production rate | Total production |
| Growth rate | Total quantity |
. . .
The constant of integration often represents a fixed quantity (fixed costs, initial inventory, starting capital).
Example: Complete Cost Analysis
Scenario: A furniture manufacturer has:
- Marginal cost: \(MC(x) = 0.02x^2 - 2x + 100\) euros per unit
- Fixed costs: €5,000
- Each unit sells for €150
. . .
Tasks:
- Find the total cost function \(C(x)\)
- Find the revenue function \(R(x)\)
- Find the profit function \(P(x)\)
- Determine the break-even points
Solution Part 1: Cost Function
Step 1: Integrate marginal cost
\[C(x) = \int (0.02x^2 - 2x + 100) \, dx\]
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\[C(x) = \frac{0.02x^3}{3} - x^2 + 100x + C\]
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Step 2: Apply initial condition \(C(0) = 5000\)
\[C = 5000\]
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Answer: \(C(x) = \frac{x^3}{150} - x^2 + 100x + 5000\)
Solution Part 2: Revenue and Profit
Revenue: (Price × Quantity)
\[R(x) = 150x\]
. . .
Profit: (Revenue − Cost)
\[P(x) = R(x) - C(x)\]
\[P(x) = 150x - \left(\frac{x^3}{150} - x^2 + 100x + 5000\right)\]
. . .
\[P(x) = -\frac{x^3}{150} + x^2 + 50x - 5000\]
Visualizing the Business Model

Collaborative Problem-Solving
Group Challenge: Startup Analysis
Scenario: A tech startup is launching a new app. Their data analytics team has modeled:
Marginal cost (per user): \(MC(x) = 0.01x + 5\) euros, where \(x\) is thousands of users
Fixed costs: €50,000 (servers, development, etc.)
Revenue per user: €12 per thousand users
Group Challenge: Tasks
Work in groups and then we compare
- Find the total cost function \(C(x)\) for \(x\) thousand users
- Find the revenue function \(R(x)\)
- Find the profit function \(P(x)\)
- Calculate the profit/loss at 1,000 users, 5,000 users, and 10,000 users
- Find the break-even point(s)
- At what user count is profit maximized?
Wrap-Up & Key Takeaways
Today’s Essential Concepts
- Antiderivative: If \(F'(x) = f(x)\), then \(F(x)\) is antiderivative of \(f(x)\)
- Power rule: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) (for \(n \neq -1\))
- Sum/Constant rules: Split integrals, pull out constants
- +C is essential: Represents the family of all antiderivatives
- Initial conditions: Fix the value of \(C\) to find a specific function
- Verification: Always differentiate to check your answer
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Next session: We’ll learn about definite integrals and the Fundamental Theorem of Calculus!
Integration Formulas Summary
| Function | Antiderivative |
|---|---|
| \(x^n\) (\(n \neq -1\)) | \(\frac{x^{n+1}}{n+1} + C\) |
| \(k\) (constant) | \(kx + C\) |
| \(kf(x)\) | \(k \int f(x) \, dx\) |
| \(f(x) \pm g(x)\) | \(\int f(x) \, dx \pm \int g(x) \, dx\) |
. . .
Integration is “undo-ing” differentiation. Ask yourself: “What function, when differentiated, gives me this?”
Final Assessment - 5 Minutes
Quick Check
Work individually, then compare
Find \(\int (2x^3 - 5x + 3) \, dx\)
Given \(f'(x) = 6x - 2\) and \(f(1) = 4\), find \(f(x)\).
A company’s marginal cost is \(MC(x) = 30 + 0.5x\). Fixed costs are €200. What is the cost of producing 20 units?
Next Session Preview
Coming Up: Definite Integrals & FTC
- The definite integral as a signed area
- The Fundamental Theorem of Calculus
- Evaluating integrals using \(F(b) - F(a)\)
- Properties of definite integrals
- Applications to accumulated quantities
. . .
Complete Tasks 06-01
- Practice basic antiderivatives until they’re automatic
- Work through initial value problems
- Focus on business applications (cost, revenue, profit)