Lecture 5.1.2: Limits of Riemann Sums

Nov 1, 2024

Lecture Notes: Finding the Area Under a Curve

Introduction

  • Objective: Finding the area under a curve by approximating it with rectangles.
  • Approach: Increasing the number of rectangles improves approximation.
  • Key Concept: All sums (left, right, midpoint) converge to the same limit.

Riemann Sum

  • Definition: A Riemann sum is the sum of the form (\sum_{i=1}^{n} f(x_i^*) \Delta x_i).
    • (x_i^*) is an arbitrary point in the i-th subinterval.
    • (\Delta x_i) is the width of the i-th subinterval.
    • Multiplying height and width gives the area.
  • Convergence: Any Riemann sum converges to the same value if partition is uniform and as the number of rectangles goes to infinity.

Uniform vs. Non-Uniform Partitions

  • Uniform Partition: (\Delta x) is constant ((\Delta x = \frac{b-a}{n})).
  • Non-Uniform Partition: Define the norm of the partition as the maximum width of subintervals. The norm must go to zero for convergence.

Limit of Riemann Sum

  • Area Calculation: Limit of Riemann sum as (n) goes to infinity.
    • Use right-hand endpoint for simplification.
    • Integral form: (\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x).

Example Calculations

Area of a Triangle

  • Setup: Triangle with base (B) and height (H).
  • Function: Line segment (f(x) = \frac{H}{B}x).
  • Riemann Sum Limit: Showed the limit matches (\frac{1}{2} \text{Base} \times \text{Height}).

Area Under (f(x) = x^2 + 1) on ([0, 2])

  • Target Area: 14/3.
  • Partition: (\Delta x = \frac{2}{n}).
  • Result: Confirmed area is (\frac{14}{3}).

Area Under (f(x) = x^3 + 2x) on ([0, \frac{3}{2}])

  • Partition: (\Delta x = \frac{3}{2n}).
  • Result: Area calculated to be (\frac{225}{64}).

Conclusion

  • Current Method: Use Riemann sums with uniform partitions and right-hand endpoints to find areas.
  • Future Improvements: Connect this method to antiderivatives for quicker calculations.
  • Summary: Demonstrated through examples how to calculate exact area under curves using Riemann sums.