Session 06-02 - Definite Integrals & The Fundamental Theorem

Section 06: Integral Calculus

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Entry Quiz

Quick Review from Session 06-01

Test your understanding of Antiderivatives

  1. Find \(\int (4x^3 - 6x + 2) \, dx\)

  2. Find \(\int \frac{3}{x^2} \, dx\)

  3. If \(f'(x) = 2x + 1\) and \(f(0) = 3\), find \(f(x)\).

  4. Why do we write \(+C\) when finding indefinite integrals?

Homework Discussion

Your questions from Session 06-01

Focus on antiderivatives and indefinite integrals

  • Challenges with the power rule
  • Initial value problems
  • Business applications (marginal cost → total cost)
  • Verification by differentiation

Today we connect integration to area and discover the Fundamental Theorem of Calculus!

Learning Objectives

What You’ll Master Today

  • Understand the definite integral as a limit of sums
  • Interpret definite integrals as signed area under curves
  • Apply the Fundamental Theorem to evaluate integrals
  • Use the evaluation formula \(\int_a^b f(x) \, dx = F(b) - F(a)\)
  • Apply properties of definite integrals
  • Interpret net change using definite integrals
  • Connect integration to accumulation in business contexts

The Fundamental Theorem connects the area problem to antiderivatives!

The Area Problem

Motivating Question

How do we find the area under a curve?

Curve f(x) = 0.5x² + 1 with the region beneath it between the dashed lines x = 1 and x = 3 shaded and marked with a large question mark: the area under a curved graph that we cannot yet compute.

Question: What is the area of the shaded region between \(x = 1\) and \(x = 3\)?

The Rectangle Approximation Idea

Strategy: Approximate the curved region with rectangles!

Three panels approximating the area under f(x) = 0.5x² + 1 between x = 1 and x = 3 with 4, 8 and 16 rectangles. The rectangles overshoot the rising curve, and the overshoot shrinks as the rectangles get narrower.

More rectangles → Better approximation!

The Riemann Sum

Definition: A Riemann sum approximates area using rectangles:

\[\sum_{i=1}^{n} f(x_i^*) \cdot \Delta x\]

  • \(\Delta x = \frac{b-a}{n}\) is the width of each rectangle
  • \(x_i^*\) is a sample point in the \(i\)-th subinterval
  • \(f(x_i^*)\) is the height of the \(i\)-th rectangle
  • We sum the areas of all \(n\) rectangles
  • As \(n \to \infty\), the Riemann sum approaches the exact area!

From Sum to Integral

The Definite Integral:

\[\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \cdot \Delta x\]

  • \(\int_a^b\) means “integrate from \(a\) to \(b\)”
  • \(a\) is the lower limit of integration
  • \(b\) is the upper limit of integration
  • The result is a number (not a function!)

Signed Area

Area Above vs. Below the x-axis

Key insight: Definite integrals measure signed area!

Parabola f(x) = x² − 4. The region between curve and x-axis is shaded as negative area from x = 0 to x = 2, where the curve runs below the axis, and as positive area from x = 2 to x = 3, where it runs above.

Signed Area Rules

The definite integral gives signed area:

  • Above x-axis: Area counts as positive
  • Below x-axis: Area counts as negative
  • Total: Positive area minus negative area

Signed area ≠ Total area

If you want total (unsigned) area, you need to handle regions above and below separately!

Example: Signed vs. Total Area

Line f(x) = x between the dashed bounds x = −2 and x = 3. The small triangle below the x-axis from −2 to 0 is labelled −2, the larger triangle above the axis from 0 to 3 is labelled +4.5: the two areas count with opposite signs.
  • Signed area: \(\int_{-2}^{3} x \, dx = 4.5 - 2 = 2.5\)
  • Total area: \(|{-2}| + 4.5 = 6.5\)

Break

Fundamental Theorem

The Big Connection

The Fundamental Theorem of Calculus (FTC):

If \(F(x)\) is an antiderivative of \(f(x)\), then:

\[\int_a^b f(x) \, dx = F(b) - F(a)\]

This is one of the most important theorems in mathematics! It connects:

  • Differentiation (finding rates of change)
  • Integration (finding accumulated quantities/areas)

Why the FTC Works

Intuition: The integral accumulates the rate of change.

  • \(f(x)\) represents a rate of change
  • \(F(x)\) represents the total accumulated quantity
  • \(F(b) - F(a)\) = total change from \(x = a\) to \(x = b\)

Imagine filling a water tank where \(f(x)\) is the flow rate (liters per minute).

  • \(F(x)\) is the water level at time \(x\)
  • So \(F(b) - F(a)\) = (final level) − (initial level) = total liters added!
  • Each tiny change \(f(x) \cdot \Delta x\) adds up to \(F(b) - F(a)\).

Notation: The Evaluation Bar

We write:

\[\int_a^b f(x) \, dx = F(x) \Big|_a^b = F(b) - F(a)\]

The notation \(F(x) \Big|_a^b\) means “evaluate \(F\) at \(b\) and subtract \(F\) at \(a\)”

Also written as:

\[\left[ F(x) \right]_a^b = F(b) - F(a)\]

Example 1: Basic Evaluation

Evaluate: \(\int_1^4 2x \, dx\)

Step 1: Find an antiderivative

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

Step 2: Apply the FTC

\[\int_1^4 2x \, dx = x^2 \Big|_1^4 = 4^2 - 1^2 = 16 - 1 = 15\]

No \(+C\) needed for definite integrals! The constants cancel: \((F(b) + C) - (F(a) + C) = F(b) - F(a)\)

Example 2: Polynomial

Evaluate: \(\int_0^3 (x^2 - 2x + 1) \, dx\)

Step 1: Find an antiderivative

\[F(x) = \frac{x^3}{3} - x^2 + x\]

Step 2: Evaluate at bounds

\[= \left[\frac{x^3}{3} - x^2 + x\right]_0^3 = \left(\frac{27}{3} - 9 + 3\right) - \left(0 - 0 + 0\right)\]

\[= 9 - 9 + 3 = 3\]

Example 3: With Negative Values

Evaluate: \(\int_{-1}^{2} 3x^2 \, dx\)

Step 1: Antiderivative is \(F(x) = x^3\)

Step 2: Evaluate

\[\int_{-1}^{2} 3x^2 \, dx = x^3 \Big|_{-1}^{2} = 2^3 - (-1)^3 = 8 - (-1) = 9\]

Be careful with negative lower limits! \((-1)^3 = -1\), not \(1\).

Integral Properties

Key Properties

Property 1: Reversing Limits

\[\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx\]

Property 2: Splitting Intervals

\[\int_a^b f(x) \, dx + \int_b^c f(x) \, dx = \int_a^c f(x) \, dx\]

More Properties

Property 3: Constant Multiple

\[\int_a^b k \cdot f(x) \, dx = k \int_a^b f(x) \, dx\]

Property 4: Sum/Difference

\[\int_a^b [f(x) \pm g(x)] \, dx = \int_a^b f(x) \, dx \pm \int_a^b g(x) \, dx\]

These properties are the same as for indefinite integrals, just applied to definite integrals!

Using Properties: Example

Given: \(\int_0^5 f(x) \, dx = 12\) and \(\int_0^3 f(x) \, dx = 7\), find: \(\int_3^5 f(x) \, dx\)

Solution: Using the splitting property:

\[\int_0^3 f(x) \, dx + \int_3^5 f(x) \, dx = \int_0^5 f(x) \, dx\]

\[7 + \int_3^5 f(x) \, dx = 12\]

\[\int_3^5 f(x) \, dx = 5\]

Guided Practice

Practice Set A: Basic Evaluation

Work individually for 5 minutes

Evaluate these definite integrals:

  1. \(\int_0^2 3x^2 \, dx\)
  2. \(\int_1^4 (2x - 1) \, dx\)
  3. \(\int_{-2}^{1} x^3 \, dx\)
  4. \(\int_0^1 (x^2 + x + 1) \, dx\)

Practice Set B: More Complex

Work individually for 7 minutes

  1. \(\int_1^9 \frac{2}{\sqrt{x}} \, dx\)

  2. \(\int_{-1}^{3} (4 - x^2) \, dx\)

  3. Given \(\int_0^6 g(x) \, dx = 15\) and \(\int_4^6 g(x) \, dx = 8\), find \(\int_0^4 g(x) \, dx\)

  4. \(\int_2^5 (3x^2 - 4x + 2) \, dx\)

Practice Set C: Area Interpretation

Work in pairs for 8 minutes

  1. Calculate the signed area: \(\int_{-2}^{2} x^3 \, dx\). Explain why this result makes geometric sense.

  2. For \(f(x) = x - 1\) from \(x = 0\) to \(x = 3\):

    1. Calculate the signed area \(\int_0^3 (x-1) \, dx\)
    2. Find the total (unsigned) area between the curve and x-axis

Coffee Break

Net Change Theorem

Net Change Interpretation

The Fundamental Theorem as Net Change:

\[\int_a^b f'(x) \, dx = f(b) - f(a)\]

In words: The integral of a rate of change gives the net change in the original quantity.

Rate (Derivative) Integral Gives
Marginal cost \(C'(x)\) Change in cost \(C(b) - C(a)\)
Population growth rate Change in population
Production rate Total production

Example: Total Cost Change

Scenario: A company’s marginal cost is \(MC(x) = C'(x) = 20 + 0.5x\) EUR per unit. What is the total cost of increasing production from 10 to 50 units?

Solution:

\[\int_{10}^{50} (20 + 0.5x) \, dx = \left[20x + 0.25x^2\right]_{10}^{50}\]

\[= (1000 + 625) - (200 + 25) = 1625 - 225 = 1400 \text{ EUR}\]

The total additional cost of producing 40 more units is 1,400 EUR.

Example: Monthly Profit Rate

Scenario: A startup’s monthly profit rate (in thousand EUR) is:

\[P'(t) = t^2 - 4t + 3 \text{ (thousand EUR/month)}\]

where \(t\) is months since launch.

Question: What is the net profit over the first 4 months?

Solution:

\[\int_0^4 (t^2 - 4t + 3) \, dt = \left[\frac{t^3}{3} - 2t^2 + 3t\right]_0^4 = \frac{4}{3} \text{ thousand} = 1{,}333 \text{ EUR}\]

Net Profit vs. Total Cash Flow

Net profit = \(\int_a^b P'(t) \, dt\) (profits minus losses) - total cash movement = \(\int_a^b |P'(t)| \, dt\) (all money that moved)

Profit rate P'(t) = t² − 4t + 3 over months 0 to 4, a parabola that crosses zero at t = 1 and t = 3. The areas above the axis in months 0 to 1 and 3 to 4 are shaded as profit periods, the area below the axis in months 1 to 3 as a loss period.

Why Both Measures Matter

Business interpretation:

  • Months 0-1: Profitable (early adopters)
  • Months 1-3: Losses (scaling costs exceed revenue)
  • Months 3-4: Profitable again (economies of scale kick in)

For investors: Net profit of 1,333 EUR sounds good!

For cash management: The company needed reserves to survive the loss period (months 1-3), even though it ended up profitable overall.

Team Problem-Solving

Challenge: Production Analysis

Scenario: A manufacturing plant’s production rate varies during an 8-hour shift:

\[P(t) = 50 + 30t - 3t^2 \text{ units per hour}\]

where \(t\) is hours since the shift started.

Group Tasks

Work in groups of 3-4

  1. Graph \(P(t)\) for \(0 \leq t \leq 8\). At what time is production rate highest?

  2. Calculate the total production during the first 4 hours: \(\int_0^4 P(t) \, dt\)

  3. Calculate total production for the entire 8-hour shift.

Group Tasks II

  1. When does production rate drop to zero? What does this mean?

  2. If workers are paid 0.50 EUR per unit, calculate total labor cost for:

    1. The first half of the shift (0-4 hours)
    2. The second half (4-8 hours)

Wrap-Up & Takeaways

Today’s Essential Concepts

  • Definite integral = limit of Riemann sums = signed area
  • Fundamental Theorem: \(\int_a^b f(x) \, dx = F(b) - F(a)\)
  • Signed area: Above x-axis positive, below negative
  • Properties: Reversing limits, splitting intervals, linearity
  • Net change: \(\int_a^b f'(x) \, dx = f(b) - f(a)\)
  • Applications: Total cost, displacement, accumulated quantities

Next session: Area under curves and between curves with applications!

Comparison: Indefinite vs. Definite

Indefinite Integral Definite Integral
\(\int f(x) \, dx\) \(\int_a^b f(x) \, dx\)
Result is a function + C Result is a number
Family of antiderivatives Signed area / net change
Need \(+C\) No \(+C\) needed

Final Assessment

Quick Check

Work individually, then compare

  1. Evaluate \(\int_1^3 (2x + 4) \, dx\)

  2. If \(\int_0^5 f(x) \, dx = 20\) and \(\int_0^2 f(x) \, dx = 6\), find \(\int_2^5 f(x) \, dx\)

  3. A company’s marginal profit is \(MP(x) = 100 - 2x\). Find the total profit gained by increasing production from 20 to 40 units.

Next Session Preview

Up: Area Problems & Applications

  • Area under a single curve
  • Area between a curve and the x-axis
  • Handling regions where \(f(x) < 0\)
  • Exponential and logarithmic integrals
  • Business applications: total profit, accumulated production

Complete Tasks 06-02

  • Practice evaluating definite integrals
  • Focus on the FTC and its applications
  • Work on signed vs. total area problems