Solved Exercises on Defining Definite Integrals with Riemann Sums

Master defining definite integrals with Riemann sums: left, right, midpoint, and trapezoidal sums through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Left Riemann Sum Definition
Exercise 1
Define the definite integral \(\int_0^2 x^2 dx\) using left Riemann sums. Show that \(\int_0^2 x^2 dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} f(x_i) \Delta x\).
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. Partition the interval \([a,b]\) into \(n\) equal subintervals
  2. Calculate \(\Delta x = \frac{b-a}{n}\)
  3. Find left endpoints: \(x_i = a + i\Delta x\) for \(i = 0, 1, ..., n-1\)
  4. Form the sum: \(L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x\)
  5. Take the limit: \(\int_a^b f(x)dx = \lim_{n \to \infty} L_n\)
Interval
\([0,2]\)
Width
\(\Delta x = \frac{2}{n}\)
Left sum
\(L_n = \frac{2}{n}\sum_{i=0}^{n-1} \left(\frac{2i}{n}\right)^2\)
Step 1: Set up the partition

For \(\int_0^2 x^2 dx\), we have \(a = 0\), \(b = 2\)

Divide \([0,2]\) into \(n\) equal subintervals of width \(\Delta x = \frac{2-0}{n} = \frac{2}{n}\)

Step 2: Find left endpoints

The left endpoints are: \(x_0 = 0, x_1 = \frac{2}{n}, x_2 = \frac{4}{n}, ..., x_{n-1} = \frac{2(n-1)}{n}\)

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

Step 3: Form the left Riemann sum

\(L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x = \sum_{i=0}^{n-1} \left(\frac{2i}{n}\right)^2 \cdot \frac{2}{n}\)

\(L_n = \frac{2}{n} \sum_{i=0}^{n-1} \frac{4i^2}{n^2} = \frac{8}{n^3} \sum_{i=0}^{n-1} i^2\)

Step 4: Use the sum of squares formula

\(\sum_{i=0}^{n-1} i^2 = \sum_{i=1}^{n-1} i^2 = \frac{(n-1)n(2n-1)}{6}\)

(Note: The \(i=0\) term is 0, so we can start from \(i=1\))

Step 5: Substitute and simplify

\(L_n = \frac{8}{n^3} \cdot \frac{(n-1)n(2n-1)}{6} = \frac{8(n-1)(2n-1)}{6n^2}\)

\(L_n = \frac{4(n-1)(2n-1)}{3n^2} = \frac{4(2n^2 - 3n + 1)}{3n^2}\)

\(L_n = \frac{8n^2 - 12n + 4}{3n^2} = \frac{8}{3} - \frac{4}{n} + \frac{4}{3n^2}\)

Step 6: Take the limit

\(\lim_{n \to \infty} L_n = \lim_{n \to \infty} \left(\frac{8}{3} - \frac{4}{n} + \frac{4}{3n^2}\right) = \frac{8}{3}\)

\(\int_0^2 x^2 dx = \frac{8}{3}\)
Final answer:

\(\int_0^2 x^2 dx = \lim_{n \to \infty} \frac{8}{n^3} \sum_{i=0}^{n-1} i^2 = \frac{8}{3}\)

Applied rules:

• Definition of definite integral: As limit of Riemann sums

• Sum of squares: \(\sum_{i=1}^{n-1} i^2 = \frac{(n-1)n(2n-1)}{6}\)

• Limit properties: Terms with \(n\) in denominator approach zero

2 Right Riemann Sum Definition
Exercise 2
Define the definite integral \(\int_1^3 \frac{1}{x} dx\) using right Riemann sums. Show that \(\int_1^3 \frac{1}{x} dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x\).
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
\([1,3]\)
Width
\(\Delta x = \frac{2}{n}\)
Right sum
\(R_n = \frac{2}{n}\sum_{i=1}^{n} \frac{1}{1 + \frac{2i}{n}}\)
Step 1: Set up the partition

For \(\int_1^3 \frac{1}{x} dx\), we have \(a = 1\), \(b = 3\)

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

Step 2: Find right endpoints

The right endpoints are: \(x_1 = 1 + \frac{2}{n}, x_2 = 1 + \frac{4}{n}, ..., x_n = 1 + \frac{2n}{n} = 3\)

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

Step 3: Form the right Riemann sum

\(R_n = \sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} \frac{1}{1 + \frac{2i}{n}} \cdot \frac{2}{n}\)

\(R_n = \frac{2}{n} \sum_{i=1}^{n} \frac{1}{1 + \frac{2i}{n}}\)

Step 4: Recognize the limit

For this specific function \(f(x) = \frac{1}{x}\), we know that:

\(\int_1^3 \frac{1}{x} dx = \ln|x| \Big|_1^3 = \ln(3) - \ln(1) = \ln(3)\)

Step 5: Verify the Riemann sum approaches the correct value

As \(n \to \infty\), the right Riemann sum \(R_n\) approaches \(\ln(3) \approx 1.0986\)

This demonstrates that \(\int_1^3 \frac{1}{x} dx = \lim_{n \to \infty} \frac{2}{n} \sum_{i=1}^{n} \frac{1}{1 + \frac{2i}{n}} = \ln(3)\)

\(\int_1^3 \frac{1}{x} dx = \ln(3)\)
Final answer:

\(\int_1^3 \frac{1}{x} dx = \lim_{n \to \infty} \frac{2}{n} \sum_{i=1}^{n} \frac{1}{1 + \frac{2i}{n}} = \ln(3)\)

Applied rules:

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

• Integral of reciprocal: \(\int \frac{1}{x} dx = \ln|x| + C\)

• Logarithmic properties: \(\ln(a) - \ln(b) = \ln(\frac{a}{b})\)

3 Midpoint Riemann Sum Definition
Exercise 3
Define the definite integral \(\int_0^1 e^x dx\) using midpoint Riemann sums. Show that \(\int_0^1 e^x dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(\bar{x}_i) \Delta x\).
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
\([0,1]\)
Width
\(\Delta x = \frac{1}{n}\)
Midpoint sum
\(M_n = \frac{1}{n}\sum_{i=1}^{n} e^{\frac{2i-1}{2n}}\)
Step 1: Set up the partition

For \(\int_0^1 e^x dx\), we have \(a = 0\), \(b = 1\)

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

Step 2: Find midpoints

The midpoints are: \(\bar{x}_1 = \frac{1}{2n}, \bar{x}_2 = \frac{3}{2n}, ..., \bar{x}_n = \frac{2n-1}{2n}\)

In general: \(\bar{x}_i = 0 + \left(i - \frac{1}{2}\right) \cdot \frac{1}{n} = \frac{2i-1}{2n}\) for \(i = 1, 2, ..., n\)

Step 3: Form the midpoint Riemann sum

\(M_n = \sum_{i=1}^{n} f(\bar{x}_i) \Delta x = \sum_{i=1}^{n} e^{\frac{2i-1}{2n}} \cdot \frac{1}{n}\)

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

Step 4: Evaluate the known integral

We know that: \(\int_0^1 e^x dx = e^x \Big|_0^1 = e^1 - e^0 = e - 1\)

Step 5: Verify the limit

As \(n \to \infty\), the midpoint sum \(M_n\) approaches \(e - 1 \approx 1.7183\)

This demonstrates that \(\int_0^1 e^x dx = \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} e^{\frac{2i-1}{2n}} = e - 1\)

\(\int_0^1 e^x dx = e - 1\)
Final answer:

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

Applied rules:

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

• Exponential integral: \(\int e^x dx = e^x + C\)

• Exponential properties: \(e^0 = 1\)

Summary: Defining Definite Integrals with Riemann Sums
\(\int_a^b f(x)dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x\)
Definition of Definite Integral
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:

Partition: A division of interval \([a,b]\) into \(n\) subintervals

Norm of partition: The length of the largest subinterval

Riemann sum: A sum of the form \(\sum_{i=1}^{n} f(x_i^*) \Delta x_i\) where \(x_i^*\) is in the \(i\)-th subinterval

Definite integral: The limit of Riemann sums as the norm of the partition approaches zero

Complete methodology:
  1. Partition the interval: Divide \([a,b]\) into \(n\) subintervals of width \(\Delta x = \frac{b-a}{n}\)
  2. Choose sample points: Select \(x_i^*\) in each subinterval (left, right, midpoint, or arbitrary)
  3. Form the sum: Calculate \(R_n = \sum_{i=1}^{n} f(x_i^*) \Delta x\)
  4. Take the limit: Evaluate \(\lim_{n \to \infty} R_n\)
Tip 1: For uniform partitions, all subintervals have equal width \(\Delta x = \frac{b-a}{n}\).
Tip 2: As \(n \to \infty\), Riemann sums approach the exact definite integral value.
Tip 3: The definite integral exists if the limit of Riemann sums is the same regardless of sample point choices.
Tip 4: Midpoint and trapezoidal rules typically converge faster than left/right rules.
Integrability condition: A bounded function is Riemann integrable if and only if the set of its discontinuities has measure zero.
Convergence rate: Left/right: O(1/n), Midpoint: O(1/n²), Trapezoidal: O(1/n²).
Solution: Exercises 4 to 5
4 Trapezoidal Rule Definition
Exercise 4
Define the definite integral \(\int_0^1 x^3 dx\) using the trapezoidal rule. Show that \(\int_0^1 x^3 dx = \lim_{n \to \infty} T_n\) where \(T_n\) is the trapezoidal sum.
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: Set up the partition

For \(\int_0^1 x^3 dx\), we have \(a = 0\), \(b = 1\)

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

Step 2: 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 3: Apply trapezoidal rule formula

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

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

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

Step 4: Express in sigma notation

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

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

Step 5: Use sum of cubes formula

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

\(T_n = \frac{1}{2n}\left[\frac{2}{n^3} \cdot \frac{(n-1)^2n^2}{4} + 1\right]\)

\(T_n = \frac{1}{2n}\left[\frac{(n-1)^2}{2n} + 1\right] = \frac{(n-1)^2}{4n^2} + \frac{1}{2n}\)

Step 6: Simplify and take the limit

\(T_n = \frac{n^2 - 2n + 1}{4n^2} + \frac{1}{2n} = \frac{1}{4} - \frac{1}{2n} + \frac{1}{4n^2} + \frac{1}{2n}\)

\(T_n = \frac{1}{4} + \frac{1}{4n^2}\)

\(\lim_{n \to \infty} T_n = \lim_{n \to \infty} \left(\frac{1}{4} + \frac{1}{4n^2}\right) = \frac{1}{4}\)

Step 7: Verify with direct integration

\(\int_0^1 x^3 dx = \left[\frac{x^4}{4}\right]_0^1 = \frac{1}{4} - 0 = \frac{1}{4}\)

\(\int_0^1 x^3 dx = \frac{1}{4}\)
Final answer:

\(\int_0^1 x^3 dx = \lim_{n \to \infty} T_n = \frac{1}{4}\) where \(T_n = \frac{1}{2n}\left[\frac{2}{n^3}\sum_{i=1}^{n-1} i^3 + 1\right]\).

Applied rules:

• Trapezoidal formula: Average of left and right sums

• Sum of cubes: \(\sum_{i=1}^{n} i^3 = \left[\frac{n(n+1)}{2}\right]^2\)

• Limit evaluation: Terms with \(n\) in denominator approach zero

5 Equivalence of Riemann Sums
Exercise 5
Prove that for a continuous function \(f(x)\) on \([a,b]\), all Riemann sums (left, right, midpoint, trapezoidal) converge to the same definite integral as \(n \to \infty\).
Definition:

Integrability: A function \(f(x)\) is Riemann integrable on \([a,b]\) if all Riemann sums converge to the same limit as the partition norm approaches zero.

Left sum
\(L_n \to \int_a^b f(x)dx\)
Right sum
\(R_n \to \int_a^b f(x)dx\)
All converge
\(\to \int_a^b f(x)dx\)
Step 1: Establish the theoretical foundation

For a continuous function \(f(x)\) on the closed interval \([a,b]\), we know that:

• \(f(x)\) is uniformly continuous on \([a,b]\)

• \(f(x)\) is bounded on \([a,b]\)

• \(f(x)\) is Riemann integrable on \([a,b]\)

Step 2: Analyze the difference between sums

Consider the difference between right and left sums:

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

\(= \Delta x[f(x_n) - f(x_0)] = \frac{b-a}{n}[f(b) - f(a)]\)

Step 3: Take the limit of the difference

\(\lim_{n \to \infty} (R_n - L_n) = \lim_{n \to \infty} \frac{b-a}{n}[f(b) - f(a)] = 0\)

This shows that \(R_n\) and \(L_n\) approach the same limit.

Step 4: Extend to midpoint sums

For the midpoint sum \(M_n\), note that each rectangle has the same width as in \(L_n\) and \(R_n\), but the height is taken at the midpoint of each subinterval.

By the Mean Value Theorem, for each subinterval \([x_{i-1}, x_i]\), there exists a point \(c_i\) such that the area of the midpoint rectangle approximates the area under the curve on that subinterval.

Step 5: Use continuity property

Since \(f\) is continuous, as \(\Delta x \to 0\) (which happens as \(n \to \infty\)), the values \(f(x_{i-1})\), \(f(x_i)\), and \(f(\bar{x}_i)\) all approach the same average value for the subinterval.

This ensures that all Riemann sums converge to the same limit.

Step 6: Conclusion

Therefore, for any continuous function \(f(x)\) on \([a,b]\):

\(\lim_{n \to \infty} L_n = \lim_{n \to \infty} R_n = \lim_{n \to \infty} M_n = \lim_{n \to \infty} T_n = \int_a^b f(x)dx\)

All Riemann sums converge to \(\int_a^b f(x)dx\)
Final answer:

For continuous functions on \([a,b]\), all Riemann sums (left, right, midpoint, trapezoidal) converge to the same definite integral as \(n \to \infty\).

Applied rules:

• Continuity implies integrability: Continuous functions on closed intervals are Riemann integrable

• Uniform continuity: Ensures that function values in small intervals become arbitrarily close

• Limit properties: Differences between different Riemann sums approach zero as \(n \to \infty\)

Key Concepts, Laws, Methods, and Formulas
\(\int_a^b f(x)dx = \sup L(f,P) = \inf U(f,P)\)
Darboux Definition
Key definitions:

Darboux sums: Upper sum \(U(f,P) = \sum M_i \Delta x_i\) and lower sum \(L(f,P) = \sum m_i \Delta x_i\) where \(M_i\) and \(m_i\) are the supremum and infimum of \(f\) on the \(i\)-th subinterval.

Integrability criterion: \(f\) is integrable if \(\sup L(f,P) = \inf U(f,P)\) over all partitions \(P\).

Refinement: A partition \(Q\) is a refinement of \(P\) if every point of \(P\) is also in \(Q\).

Complete methodology:
  1. Verify integrability: Check if the function satisfies integrability conditions
  2. Choose partition: Usually uniform partitions with \(\Delta x = \frac{b-a}{n}\)
  3. Select sample points: Based on the type of Riemann sum needed
  4. Formulate sum: Express the Riemann sum using sigma notation
  5. Evaluate limit: Calculate \(\lim_{n \to \infty}\) of the Riemann sum
Tip 1: For smooth functions, the error in left/right sums is approximately \(\pm\frac{(b-a)^2}{2n}f'(\xi)\) for some \(\xi\).
Tip 2: When computing limits of Riemann sums, factor out \(\Delta x\) first.
Tip 3: Use standard summation formulas: \(\sum 1 = n\), \(\sum i = \frac{n(n+1)}{2}\), \(\sum i^2 = \frac{n(n+1)(2n+1)}{6}\), \(\sum i^3 = \left[\frac{n(n+1)}{2}\right]^2\).
Tip 4: The Fundamental Theorem of Calculus provides a shortcut, but Riemann sums establish the theoretical foundation.
Integrability conditions: Bounded functions with finitely many discontinuities are Riemann integrable.
Numerical significance: Riemann sums form the basis for numerical integration methods.
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: \(\sum_{i=1}^{n} i^k\) for \(k = 0, 1, 2, 3\)

Visualization: Convergence of Different 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)
\(T_n = \frac{1}{2n}[f(0) + 2\sum_{i=1}^{n-1} f(x_i) + f(1)]\) (trapezoidal)

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

  • Left sum: Underestimates (since \(f(x) = x^2\) is increasing)
  • Right sum: Overestimates (since \(f(x) = x^2\) is increasing)
  • Midpoint sum: Most accurate, converges quadratically
  • Trapezoidal sum: Also quadratic convergence

Questions & Answers

Question: Why do we need to define definite integrals using Riemann sums? Can't we just use antiderivatives?

Answer: This is a fundamental question about mathematical foundations:

  • Historical order: Riemann sums were developed before the Fundamental Theorem of Calculus
  • Theoretical foundation: We need to define what the definite integral MEANS before we can prove the FTC
  • Existence: Riemann sums establish when integrals exist and what functions are integrable
  • Non-elementary functions: Some functions don't have elementary antiderivatives but still have definite integrals
  • Computation: Riemann sums are the basis for numerical integration methods

The FTC connects antiderivatives to definite integrals, but we must first define definite integrals independently.

Question: How do I know if a function is Riemann integrable?

Answer: A function is Riemann integrable if:

  • Boundedness: The function is bounded on the closed interval \([a,b]\)
  • Discontinuities: The set of discontinuities has Lebesgue measure zero

Practically, this means that functions that are continuous or have only finitely many jump discontinuities are Riemann integrable. Examples of integrable functions include polynomials, sine, cosine, exponential functions, and piecewise continuous functions.

Functions like Dirichlet's function (1 at rationals, 0 at irrationals) are not Riemann integrable.

Question: What's the difference between Darboux and Riemann integrals?

Answer: Both approaches lead to the same concept, but they're defined differently:

  • Riemann integral: Defined using tagged partitions and sample points, taking the limit of Riemann sums
  • Darboux integral: Defined using upper and lower sums (supremum and infimum over subintervals)

The beauty is that they're equivalent: A function is Riemann integrable if and only if it's Darboux integrable, and the values of the integrals are the same. The Darboux approach is often easier for proving theoretical results, while Riemann sums are more intuitive for computation.

Question: Why do different Riemann sums give the same limit for continuous functions?

Answer: For continuous functions, this happens because of uniform continuity:

  • Continuity: Small changes in input produce small changes in output
  • Uniform continuity: On a closed interval, we can make output differences arbitrarily small by making input differences small enough, regardless of where we are in the interval
  • Refinement: As subintervals become narrower, the function values at different sample points within each subinterval become closer together

This means that as \(n \to \infty\), the difference between any two Riemann sums approaches zero, so they all converge to the same limit.

Question: How do Riemann sums connect to real-world applications?

Answer: Riemann sums model many real-world situations:

  • Distance traveled: Summing velocity × time intervals
  • Total accumulation: Adding up rates of change over time
  • Area/volume calculations: Approximating irregular shapes
  • Probability: Riemann sums lead to probability density functions
  • Numerical methods: All computer integration algorithms are based on Riemann sums

Essentially, whenever we need to accumulate quantities that change continuously, Riemann sums provide the mathematical foundation.