📐 Concept diagram

05-03 - The Fundamental Theorem of Calculus

Phase: 5 | Subject: 05-03 Prerequisites: 05-02-the-definite-integral.md Next subject: 05-04-integration-by-substitution.md

Learning Objectives

By the end of this subject, you will be able to:

1. State both parts of the Fundamental Theorem of Calculus (FTC)
2. Use FTC Part 2 to evaluate definite integrals
3. Understand the relationship between derivatives and integrals
4. Apply the Net Change Theorem

Core Content

The Fundamental Theorem of Calculus

FTC connects differentiation and integration — they are inverse operations.

Part 1: The Derivative of an Integral

If F(x) = ∫[a to x] f(t)dt, then F'(x) = f(x).

Meaning: The derivative of the area function gives back the original function.

Example: F(x) = ∫[0 to x] t²dt By FTC: F'(x) = x²

Verification: F(x) = [t³/3][0 to x] = x³/3. F'(x) = x². ✓

Part 2: Evaluating Definite Integrals

If F is an antiderivative of f (i.e., F' = f), then:

$∫[a to b] f(x)dx = F(b) - F(a)
$

This is why we care about antiderivatives! They let us compute definite integrals exactly.

Example: ∫[1 to 3] 2x dx Antiderivative: F(x) = x² ∫[1 to 3] 2x dx = F(3) - F(1) = 9 - 1 = 8

Verification with geometry: y = 2x from x = 1 to x = 3. Area of trapezoid: (1/2)(2 + 6) × 2 = 8. ✓

Net Change Theorem

$∫[a to b] f'(x)dx = f(b) - f(a)
$

The integral of a RATE of change gives the NET CHANGE in the quantity.

Example: If v(t) is velocity, then ∫[0 to 5] v(t)dt = s(5) - s(0) = displacement.

Example: If C'(x) is marginal cost, then ∫[100 to 200] C'(x)dx = C(200) - C(100) = cost to produce items 101-200.

Key Terms

• 05 03 The Fundamental Theorem Of Calculus
• Correct: A)
• Correct: B)
• Correct: C)
• Example 1: Using FTC Part 2
• Example 2: Derivative of an integral
• Example 3: Net change
• Net Change Theorem
• Part 1: The Derivative of an Integral
• Part 2: Evaluating Definite Integrals
• The Fundamental Theorem of Calculus
• This is why we care about antiderivatives!

Worked Examples

Example 1: Using FTC Part 2

∫[0 to π/2] cos(x)dx Antiderivative: sin(x) = [sin(x)][0 to π/2] = sin(π/2) - sin(0) = 1 - 0 = 1

Example 2: Derivative of an integral

d/dx[∫[0 to x] √(t + 1)dt] = √(x + 1) by FTC Part 1.

Example 3: Net change

A tank leaks oil at rate r(t) = 10 - t litres/minute. How much leaks in first 5 minutes?

∫[0 to 5] (10 - t)dt = [10t - t²/2][0 to 5] = 50 - 12.5 = 37.5 litres.

Practice Problems

Problem 1: ∫[0 to 2] 3x²dx

Problem 2: ∫[1 to e] (1/x)dx

Problem 3: ∫[-π to π] sin(x)dx

Problem 4: d/dx[∫[2 to x] t³dt]

Problem 5: If v(t) = 4t - t², find displacement from t = 0 to t = 3.

Summary

Key takeaways:

• FTC Part 1: derivative of integral gives back the function
• FTC Part 2: ∫[a to b] f(x)dx = F(b) - F(a)
• Integration and differentiation are inverse operations
• Net Change Theorem: ∫[a to b] f'(x)dx = f(b) - f(a)

Pitfalls

• Confusing FTC Part 1 and Part 2. Part 1 says d/dx∫[a to x] f(t)dt = f(x) (derivative undoes integral). Part 2 says ∫[a to b] f(x)dx = F(b) - F(a) (antiderivative evaluates integral). They answer opposite questions — know which tool does what.
• Forgetting the chain rule with FTC Part 1. d/dx∫[a to g(x)] f(t)dt = f(g(x))·g'(x). If the upper limit is a function of x (not just x), you must multiply by its derivative.
• Subtracting in the wrong order. FTC Part 2 gives F(b) - F(a), NOT F(a) - F(b). Reversing the order flips the sign. Always evaluate the upper limit minus the lower limit.
• Applying FTC without checking continuity. FTC Part 2 requires f to be continuous on [a,b]. Integrating 1/x² on [-1, 1] using FTC directly gives a wrong (negative) answer because the function has a vertical asymptote at x = 0.
• Confusing net change with total change. The Net Change Theorem gives the net change (end minus start), not the total absolute change. For velocity, ∫v(t)dt gives displacement (net), not total distance travelled.

Quiz

Q1: ∫[0 to 1] 6x dx equals:

A) 3 B) 6 C) 9 D) 12

Q2: If F(x) = ∫[0 to x] t²dt, then F'(x) equals:

A) x² B) 2x C) x³/3 D) 0

Q3: ∫[1 to 2] (1/x)dx equals:

A) 1 B) ln(2) C) 2 D) 1/2

Q4: The Net Change Theorem states that:

A) The integral of a function is its antiderivative B) The integral of a rate of change equals the net change in the quantity C) The derivative of an integral is the original function D) The area under a curve is always positive

Q5: If ∫[0 to 3] f(x)dx = 10, then ∫[0 to 3] 2f(x)dx = ?

A) 5 B) 10 C) 20 D) 40

Next Steps

Next up: 05-04-integration-by-substitution.md