Solved Exercises on Riemann Sums in Summation Notation

Master Riemann sums in summation notation: left, right, midpoint, and trapezoidal sums expressed using sigma notation through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Left Riemann Sum in Sigma Notation
Exercise 1
Express the left Riemann sum for \(\int_1^3 x^2 dx\) using \(n\) subintervals in sigma notation.
Definition:

Left Riemann Sum: \(L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x\) where \(x_i = a + i\Delta x\) and \(\Delta x = \frac{b-a}{n}\).

Left Riemann Sum Method:
  1. Identify the interval bounds (\(a = 1\), \(b = 3\)) and number of subintervals (\(n\))
  2. Calculate \(\Delta x = \frac{b-a}{n} = \frac{3-1}{n} = \frac{2}{n}\)
  3. Find left endpoints: \(x_i = a + i\Delta x = 1 + i \cdot \frac{2}{n}\)
  4. Express the sum: \(L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x\)
Interval
\([1,3]\)
Width
\(\Delta x = \frac{2}{n}\)
Left sum
\(L_n = \frac{2}{n}\sum_{i=0}^{n-1} \left(1 + \frac{2i}{n}\right)^2\)
Step 1: Identify interval and subintervals

Given: \(\int_1^3 x^2 dx\) with \(n\) subintervals

So \(a = 1\), \(b = 3\), and number of subintervals = \(n\)

Step 2: Calculate subinterval width

\(\Delta x = \frac{b-a}{n} = \frac{3-1}{n} = \frac{2}{n}\)

Step 3: Find left endpoints

Left endpoints are: \(x_0, x_1, x_2, ..., x_{n-1}\)

Where \(x_i = a + i\Delta x = 1 + i \cdot \frac{2}{n} = 1 + \frac{2i}{n}\)

Step 4: Set up the left Riemann sum

\(L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x\)

\(L_n = \sum_{i=0}^{n-1} \left(1 + \frac{2i}{n}\right)^2 \cdot \frac{2}{n}\)

\(L_n = \frac{2}{n} \sum_{i=0}^{n-1} \left(1 + \frac{2i}{n}\right)^2\)

\(L_n = \frac{2}{n} \sum_{i=0}^{n-1} \left(1 + \frac{2i}{n}\right)^2\)
Final answer:

The left Riemann sum in sigma notation is \(L_n = \frac{2}{n} \sum_{i=0}^{n-1} \left(1 + \frac{2i}{n}\right)^2\).

Applied rules:

• Left endpoint formula: \(x_i = a + i\Delta x\) for \(i = 0, 1, ..., n-1\)

• General form: \(L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x\)

• Sigma properties: Constants can be factored out of summations

2 Right Riemann Sum in Sigma Notation
Exercise 2
Express the right Riemann sum for \(\int_0^4 \sqrt{x} dx\) using \(n\) subintervals in sigma notation.
Definition:

Right Riemann Sum: \(R_n = \sum_{i=1}^{n} f(x_i) \Delta x\) where \(x_i = a + i\Delta x\) and \(\Delta x = \frac{b-a}{n}\).

Interval
\([0,4]\)
Width
\(\Delta x = \frac{4}{n}\)
Right sum
\(R_n = \frac{4}{n}\sum_{i=1}^{n} \sqrt{\frac{4i}{n}}\)
Step 1: Identify interval and subintervals

Given: \(\int_0^4 \sqrt{x} dx\) with \(n\) subintervals

So \(a = 0\), \(b = 4\), and number of subintervals = \(n\)

Step 2: Calculate subinterval width

\(\Delta x = \frac{b-a}{n} = \frac{4-0}{n} = \frac{4}{n}\)

Step 3: Find right endpoints

Right endpoints are: \(x_1, x_2, x_3, ..., x_n\)

Where \(x_i = a + i\Delta x = 0 + i \cdot \frac{4}{n} = \frac{4i}{n}\)

Step 4: Set up the right Riemann sum

\(R_n = \sum_{i=1}^{n} f(x_i) \Delta x\)

\(R_n = \sum_{i=1}^{n} \sqrt{\frac{4i}{n}} \cdot \frac{4}{n}\)

\(R_n = \frac{4}{n} \sum_{i=1}^{n} \sqrt{\frac{4i}{n}}\)

Step 5: Simplify the expression

\(R_n = \frac{4}{n} \sum_{i=1}^{n} \frac{\sqrt{4i}}{\sqrt{n}} = \frac{4}{n} \cdot \frac{1}{\sqrt{n}} \sum_{i=1}^{n} \sqrt{4i}\)

\(R_n = \frac{4}{n^{3/2}} \sum_{i=1}^{n} \sqrt{4i} = \frac{4}{n^{3/2}} \sum_{i=1}^{n} 2\sqrt{i}\)

\(R_n = \frac{8}{n^{3/2}} \sum_{i=1}^{n} \sqrt{i}\)

\(R_n = \frac{4}{n} \sum_{i=1}^{n} \sqrt{\frac{4i}{n}}\)
Final answer:

The right Riemann sum in sigma notation is \(R_n = \frac{4}{n} \sum_{i=1}^{n} \sqrt{\frac{4i}{n}}\).

Applied rules:

• Right endpoint formula: \(x_i = a + i\Delta x\) for \(i = 1, 2, ..., n\)

• General form: \(R_n = \sum_{i=1}^{n} f(x_i) \Delta x\)

• Radical simplification: \(\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\)

3 Midpoint Riemann Sum in Sigma Notation
Exercise 3
Express the midpoint Riemann sum for \(\int_0^2 e^x dx\) using \(n\) subintervals in sigma notation.
Definition:

Midpoint Riemann Sum: \(M_n = \sum_{i=1}^{n} f(\bar{x}_i) \Delta x\) where \(\bar{x}_i = a + \left(i - \frac{1}{2}\right)\Delta x\).

Interval
\([0,2]\)
Width
\(\Delta x = \frac{2}{n}\)
Midpoint sum
\(M_n = \frac{2}{n}\sum_{i=1}^{n} e^{\frac{2i-1}{n}}\)
Step 1: Identify interval and subintervals

Given: \(\int_0^2 e^x dx\) with \(n\) subintervals

So \(a = 0\), \(b = 2\), and number of subintervals = \(n\)

Step 2: Calculate subinterval width

\(\Delta x = \frac{b-a}{n} = \frac{2-0}{n} = \frac{2}{n}\)

Step 3: Find midpoints

Midpoints are: \(\bar{x}_1, \bar{x}_2, ..., \bar{x}_n\)

Where \(\bar{x}_i = a + \left(i - \frac{1}{2}\right)\Delta x = 0 + \left(i - \frac{1}{2}\right) \cdot \frac{2}{n}\)

\(\bar{x}_i = \frac{2(i - \frac{1}{2})}{n} = \frac{2i - 1}{n}\)

Step 4: Set up the midpoint Riemann sum

\(M_n = \sum_{i=1}^{n} f(\bar{x}_i) \Delta x\)

\(M_n = \sum_{i=1}^{n} e^{\frac{2i-1}{n}} \cdot \frac{2}{n}\)

\(M_n = \frac{2}{n} \sum_{i=1}^{n} e^{\frac{2i-1}{n}}\)

\(M_n = \frac{2}{n} \sum_{i=1}^{n} e^{\frac{2i-1}{n}}\)
Final answer:

The midpoint Riemann sum in sigma notation is \(M_n = \frac{2}{n} \sum_{i=1}^{n} e^{\frac{2i-1}{n}}\).

Applied rules:

• Midpoint formula: \(\bar{x}_i = a + \left(i - \frac{1}{2}\right)\Delta x\)

• General form: \(M_n = \sum_{i=1}^{n} f(\bar{x}_i) \Delta x\)

• Algebraic manipulation: \(\left(i - \frac{1}{2}\right) \cdot \frac{2}{n} = \frac{2i-1}{n}\)

Summary: Riemann Sums in Summation Notation
\(\int_a^b f(x)dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x\)
Connection Between Integrals and Riemann Sums
Left Sum
\(L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x\)
Uses left endpoints
Right Sum
\(R_n = \sum_{i=1}^{n} f(x_i) \Delta x\)
Uses right endpoints
Midpoint Sum
\(M_n = \sum_{i=1}^{n} f(\bar{x}_i) \Delta x\)
Uses midpoints
Key definitions:

Uniform partition: Dividing \([a,b]\) into \(n\) equal subintervals of width \(\Delta x = \frac{b-a}{n}\)

Sample points: Specific points within each subinterval where the function is evaluated

General Riemann sum: \(R_n = \sum_{i=1}^{n} f(x_i^*) \Delta x\) where \(x_i^*\) is any point in the \(i\)-th subinterval

Complete methodology:
  1. Identify interval: Determine \([a,b]\) and number of subintervals \(n\)
  2. Calculate width: Find \(\Delta x = \frac{b-a}{n}\)
  3. Find sample points: Determine the location of sample points for the chosen method
  4. Set up sum: Express the Riemann sum using sigma notation
Tip 1: For left sums, use indices from 0 to n-1; for right sums, use 1 to n.
Tip 2: Midpoint formula: \(\bar{x}_i = a + \left(i - \frac{1}{2}\right)\Delta x\).
Tip 3: Always verify that your sample points are within the correct subintervals.
Tip 4: Factor out constants like \(\Delta x\) from the sigma notation when possible.
Convergence: As \(n \to \infty\), all Riemann sums approach the definite integral value.
Accuracy: Midpoint and trapezoidal rules are typically more accurate than left/right rules.
Solution: Exercises 4 to 5
4 Trapezoidal Rule in Sigma Notation
Exercise 4
Express the trapezoidal rule for \(\int_0^1 x^3 dx\) using \(n\) subintervals in sigma notation.
Definition:

Trapezoidal Rule: \(T_n = \frac{\Delta x}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)]\)

Interval
\([0,1]\)
Width
\(\Delta x = \frac{1}{n}\)
Trapezoidal sum
\(T_n = \frac{1}{2n}[f(0) + 2\sum_{i=1}^{n-1} f(x_i) + f(1)]\)
Step 1: Identify interval and subintervals

Given: \(\int_0^1 x^3 dx\) with \(n\) subintervals

So \(a = 0\), \(b = 1\), and number of subintervals = \(n\)

Step 2: Calculate subinterval width

\(\Delta x = \frac{b-a}{n} = \frac{1-0}{n} = \frac{1}{n}\)

Step 3: Find all endpoints

All endpoints: \(x_0 = 0, x_1 = \frac{1}{n}, x_2 = \frac{2}{n}, ..., x_n = 1\)

In general: \(x_i = \frac{i}{n}\) for \(i = 0, 1, 2, ..., n\)

Step 4: Apply trapezoidal rule formula

\(T_n = \frac{\Delta x}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)]\)

\(T_n = \frac{1/n}{2}[f(0) + 2f(1/n) + 2f(2/n) + ... + 2f((n-1)/n) + f(1)]\)

Step 5: Express in sigma notation

\(T_n = \frac{1}{2n}[f(0) + 2\sum_{i=1}^{n-1} f(x_i) + f(1)]\)

For \(f(x) = x^3\):

\(T_n = \frac{1}{2n}[0^3 + 2\sum_{i=1}^{n-1} \left(\frac{i}{n}\right)^3 + 1^3]\)

\(T_n = \frac{1}{2n}[0 + \frac{2}{n^3}\sum_{i=1}^{n-1} i^3 + 1]\)

\(T_n = \frac{1}{2n}\left[2\sum_{i=1}^{n-1} \left(\frac{i}{n}\right)^3 + 1\right]\)
Final answer:

The trapezoidal rule in sigma notation is \(T_n = \frac{1}{2n}\left[2\sum_{i=1}^{n-1} \left(\frac{i}{n}\right)^3 + 1\right]\).

Applied rules:

• Trapezoidal formula: Average of left and right sums

• Endpoint weights: First and last terms get weight 1, middle terms get weight 2

• Sigma manipulation: Factor out constants from summations

5 Limit Connection to Integrals
Exercise 5
Show that \(\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2 = \int_0^1 x^2 dx\) and evaluate both sides.
Definition:

Riemann sum limit: \(\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x = \int_a^b f(x)dx\) when the mesh approaches zero.

Riemann sum
\(\frac{1}{n} \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2\)
Integral
\(\int_0^1 x^2 dx\)
Evaluation
\(\frac{1}{3}\)
Step 1: Recognize the Riemann sum structure

We have: \(\frac{1}{n} \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2 = \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2 \cdot \frac{1}{n}\)

This is in the form \(\sum_{i=1}^{n} f(x_i) \Delta x\) where:

\(\Delta x = \frac{1}{n}\), \(x_i = \frac{i}{n}\), and \(f(x) = x^2\)

Step 2: Identify the interval

As \(i\) goes from 1 to \(n\), \(x_i = \frac{i}{n}\) goes from \(\frac{1}{n}\) to 1.

As \(n \to \infty\), \(\frac{1}{n} \to 0\), so the interval is \([0,1]\).

Step 3: Apply the limit theorem

By the definition of definite integrals as limits of Riemann sums:

\(\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2 = \int_0^1 x^2 dx\)

Step 4: Evaluate the Riemann sum directly

\(\frac{1}{n} \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2 = \frac{1}{n} \cdot \frac{1}{n^2} \sum_{i=1}^{n} i^2 = \frac{1}{n^3} \sum_{i=1}^{n} i^2\)

Using \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\):

\(\frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2}\)

\(= \frac{2n^2 + 3n + 1}{6n^2} = \frac{2}{6} + \frac{3}{6n} + \frac{1}{6n^2} = \frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}\)

Step 5: Take the limit

\(\lim_{n \to \infty} \left(\frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}\right) = \frac{1}{3}\)

Step 6: Evaluate the integral

\(\int_0^1 x^2 dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3} - 0 = \frac{1}{3}\)

Both equal \(\frac{1}{3}\)
Final answer:

\(\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2 = \int_0^1 x^2 dx = \frac{1}{3}\)

This demonstrates the fundamental connection between Riemann sums and definite integrals.

Applied rules:

• Limit definition: Definite integrals as limits of Riemann sums

• Standard formula: \(\sum i^2 = \frac{n(n+1)(2n+1)}{6}\)

• Algebraic limit: Terms with \(n\) in denominator approach zero

Key Concepts, Laws, Methods, and Formulas
\(R_n = \sum_{i=1}^{n} f(x_i^*) \Delta x \to \int_a^b f(x)dx \text{ as } n \to \infty\)
Fundamental Connection
Key definitions:

Mesh: The length of the largest subinterval in a partition

Norm of partition: Maximum subinterval width, denoted \(\|\Delta\|\)

Refinement: A partition with smaller mesh than another partition

Complete methodology:
  1. Partition setup: Divide interval \([a,b]\) into \(n\) equal subintervals
  2. Sample point selection: Choose appropriate points within each subinterval
  3. Sigma expression: Write the Riemann sum using summation notation
  4. Limit analysis: Study behavior as \(n \to \infty\)
Tip 1: For uniform partitions, \(\Delta x = \frac{b-a}{n}\) is constant across all subintervals.
Tip 2: The sample point index ranges from 0 to n-1 for left sums, 1 to n for right sums.
Tip 3: When evaluating limits of Riemann sums, use standard summation formulas.
Tip 4: The accuracy of different methods follows: Midpoint ≈ Trapezoid > Left/Right (for smooth functions).
Error analysis: For smooth functions, midpoint error ≈ -(1/2) × trapezoid error.
Convergence rate: Left/right: O(1/n), Midpoint/Trapezoid: O(1/n²).
Formulas to know by heart:

• Left sum: \(L_n = \sum_{i=0}^{n-1} f(a + i\Delta x) \Delta x\)

• Right sum: \(R_n = \sum_{i=1}^{n} f(a + i\Delta x) \Delta x\)

• Midpoint: \(M_n = \sum_{i=1}^{n} f(a + (i-\frac{1}{2})\Delta x) \Delta x\)

• Trapezoid: \(T_n = \frac{\Delta x}{2}[f(x_0) + 2\sum_{i=1}^{n-1} f(x_i) + f(x_n)]\)

• Standard sums: \(\sum 1 = n\), \(\sum i = \frac{n(n+1)}{2}\), \(\sum i^2 = \frac{n(n+1)(2n+1)}{6}\)

Visualization: Convergence of Riemann Sums
Exercise 6: Accuracy Comparison
Compare convergence rates of different Riemann sums for \(\int_0^1 x^2 dx = \frac{1}{3}\):
\(L_n = \frac{1}{n}\sum_{i=0}^{n-1} \left(\frac{i}{n}\right)^2\) (left)
\(R_n = \frac{1}{n}\sum_{i=1}^{n} \left(\frac{i}{n}\right)^2\) (right)
\(M_n = \frac{1}{n}\sum_{i=1}^{n} \left(\frac{i-0.5}{n}\right)^2\) (midpoint)

Analysis: The chart shows how different methods converge to the exact value \(\frac{1}{3} \approx 0.333\).

  • Left sum: Underestimates (decreasing function)
  • Right sum: Overestimates (increasing function)
  • Midpoint sum: Most accurate, converges quadratically

Questions & Answers

Question: Why do we use sigma notation for Riemann sums instead of just calculating the rectangular areas directly?

Answer: Sigma notation is essential for several reasons:

  • Compact representation: Instead of writing out dozens or hundreds of terms, we use a concise formula
  • Mathematical analysis: It allows us to manipulate and analyze the sums algebraically
  • Limit calculations: We can study what happens as \(n \to \infty\) using sigma notation
  • Generalization: The same notation works for any number of subintervals
  • Algorithm development: Computer programs can easily implement sigma notation

Without sigma notation, working with large numbers of subintervals would be impractical.

Question: How do I know which Riemann sum method to use in practice?

Answer: The choice depends on your needs:

  • For accuracy: Use midpoint or trapezoidal rules (converge faster)
  • For simplicity: Left or right sums are computationally easier
  • For theoretical purposes: Any method works for defining integrals
  • For specific applications: Sometimes the physical context suggests a natural choice

In practice, many applications use more sophisticated methods like Simpson's rule, but Riemann sums provide the foundation for understanding numerical integration.

Question: What happens if the function is not continuous? Do Riemann sums still work?

Answer: It depends on the type of discontinuities:

  • Finitely many jump discontinuities: Riemann sums still converge to the integral if the function is bounded
  • Infinite discontinuities: The function may not be Riemann integrable
  • Dense set of discontinuities: The function may fail to be Riemann integrable

A function is Riemann integrable on \([a,b]\) if and only if it is bounded and the set of discontinuities has measure zero. Continuous functions are always Riemann integrable.

Question: How do I verify that my sigma notation expression is correct?

Answer: Use these verification techniques:

  • Small n test: Try \(n = 2\) or \(n = 3\) and manually calculate both the sigma expression and the actual sum
  • Endpoint check: Verify that your sample points are within the correct subintervals
  • Count terms: Make sure you have exactly \(n\) terms in your sum
  • Boundary conditions: Check that your first and last sample points are correct

For example, if using left sums with \(n=2\) on \([0,2]\), you should have sample points at \(x_0=0\) and \(x_1=1\), not \(x_1=1\) and \(x_2=2\).

Question: Are there practical applications where I'd need to work with the sigma notation directly?

Answer: Yes, there are several important applications:

  • Error analysis: Understanding how approximation errors behave
  • Algorithm design: Implementing numerical integration in software
  • Convergence studies: Determining how many subintervals are needed for desired accuracy
  • Theoretical proofs: Establishing properties of definite integrals
  • Monte Carlo methods: Probabilistic integration techniques

Even though we often use software packages for numerical integration, understanding the underlying sigma notation is crucial for validating results and understanding limitations.