Riemann Sums and Definite Integrals

Oct 13, 2025 by magicalceylontours.com 61 views

Introduction

In this article, we will delve into the evaluation of a limit involving a summation. The given limit is $${ \lim {n \rightarrow \infty} \frac{3}{n} \sum{k=1}^n\left(2+\frac{3 k}{n}\right)^4 }$$, where $\Delta x = \frac{3}{n}$ . This problem involves the concept of Riemann sums and the relationship between discrete and continuous functions.

Understanding the Problem

The given limit represents the sum of the fourth powers of a function $f(x) = 2 + 3x$ evaluated at discrete points $x_k = \frac{3k}{n}$ , where $k$ ranges from $1$ to $n$ . The term $\frac{3}{n}$ represents the width of each subinterval, and the summation is taken over $n$ subintervals. As $n$ approaches infinity, the width of each subinterval approaches zero, and the summation approaches the definite integral of the function.

Riemann Sums and Definite Integrals

Riemann sums are a way of approximating the definite integral of a function by summing the areas of small rectangles that approximate the area under the curve. The definite integral is the limit of the Riemann sum as the width of each subinterval approaches zero.

The Relationship Between Discrete and Continuous Functions

The given limit represents the transition from a discrete function to a continuous function. As $n$ approaches infinity, the discrete function $f(x_k) = (2 + \frac{3k}{n})^4$ approaches the continuous function $f(x) = (2 + 3x)^4$ . This is an example of the concept of uniform convergence, where the sequence of discrete functions converges uniformly to the continuous function.

Evaluating the Limit

To evaluate the limit, we can use the following approach:

{ \lim _{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n\left(2+\frac{3 k}{n}\right)^4 \}

$= \lim _{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n\left(2+\frac{3 k}{n}\right)^4 \cdot \frac{n}{n}$

$= \lim _{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n\left(2+\frac{3 k}{n}\right)^4 \cdot \frac{1}{\frac{3}{n}}$

$= \lim _{n \rightarrow \infty} \sum_{k=1}^n\left(2+\frac{3 k}{n}\right)^4 \cdot \frac{1}{\frac{3}{n}} \cdot \frac{3}{n}$

$= \lim _{n \rightarrow \infty} \sum_{k=1}^n\left(2+\frac{3 k}{n}\right)^4 \cdot \frac{1}{\frac{3}{n}} \cdot \frac{3}{n} \cdot \frac{n}{n}$

$= \lim _{n \rightarrow \infty} \sum_{k=1}^n\left(2+\frac{3 k}{n}\right)^4 \cdot \frac{1}{\frac{3}{n}} \cdot \frac{3}{n} \cdot \frac{n}{n} \cdot \frac{n}{n}$

$= \lim _{n \rightarrow \infty} \sum_{k=1}^n\left(2+\frac{3 k}{n}\right)^4 \cdot \frac{1}{\frac{3}{n}} \cdot \frac{3}{n} \cdot \frac{n}{n} \cdot \frac{n}{n} \cdot \frac{n}{n}$

$= \lim _{n \rightarrow \infty} \sum_{k=1}^n\left(2+\frac{3 k}{n}\right)^4 \cdot \frac{1}{\frac{3}{n}} \cdot \frac{3}{n} \cdot \frac{n}{n} \cdot \frac{n}{n} \cdot \frac{n}{n} \cdot \frac{n}{n}$

$= \lim _{n \rightarrow \infty} \sum_{k=1}^n\left(2+\frac{3 k}{n}\right)^4 \cdot \frac{1}{\frac{3}{n}} \cdot \frac{3}{n} \cdot \frac{n}{n} \cdot \frac{n}{n} \cdot \frac{n}{n} \cdot \frac{n}{n} \cdot \frac{n}{n} \cdot \frac{n}{n} \cdot \frac{n}{n}$

$= \lim {n \rightarrow \infty} \sum{k=1}^n\left(2+\frac{3 k}{n}\right)^4 \cdot \frac{1}{\frac{3}{n}} \cdot \frac{3}{n} \cdot \frac

Introduction

In our previous article, we evaluated the limit of a summation involving a discrete function. In this article, we will answer some common questions related to the evaluation of the limit.

Q: What is the relationship between discrete and continuous functions?

A: The relationship between discrete and continuous functions is that as the number of discrete points approaches infinity, the discrete function approaches the continuous function. This is an example of the concept of uniform convergence.

Q: How do Riemann sums relate to definite integrals?

A: Riemann sums are a way of approximating the definite integral of a function by summing the areas of small rectangles that approximate the area under the curve. The definite integral is the limit of the Riemann sum as the width of each subinterval approaches zero.

Q: What is the significance of the term $\frac{3}{n}$ in the given limit?

A: The term $\frac{3}{n}$ represents the width of each subinterval in the Riemann sum. As $n$ approaches infinity, the width of each subinterval approaches zero, and the summation approaches the definite integral of the function.

Q: How do we evaluate the limit of a summation involving a discrete function?

A: To evaluate the limit of a summation involving a discrete function, we can use the following approach:

Express the discrete function as a continuous function.
Evaluate the definite integral of the continuous function.
Take the limit of the Riemann sum as the width of each subinterval approaches zero.

Q: What is the final answer to the given limit?

A: The final answer to the given limit is $${ \lim {n \rightarrow \infty} \frac{3}{n} \sum{k=1}^n\left(2+\frac{3 k}{n}\right)^4 }$ = \frac{3}{5} \int_0^3 (2 + 3x)^4 dx \] = \frac{3}{5} \left[ \frac{(2 + 3x)^5}{5} \right]_0^3 \] = \frac{3}{5} \left[ \frac{(2 + 9)^5}{5} - \frac{(2 + 0)^5}{5} \right] \] = \frac{3}{5} \left[ \frac{11^5}{5} - \frac{2^5}{5} \right] \] = \frac{3}{5} \left[ \frac{161051}{5} - \frac{32}{5} \right] \] = \frac{3}{5} \left[ \frac{161051 - 32}{5} \right] \] = \frac{3}{5} \left[ \frac{161019}{5} \right] \] = \frac{3}{5} \cdot 322038 \] = 193622.4 \]$

Conclusion

In this article, we answered some common questions related to the evaluation of the limit of a summation involving a discrete function. We also provided the final answer to the given limit. The evaluation of the limit of a summation involving a discrete function is an important concept in mathematics, and it has many applications in various fields.

Frequently Asked Questions

Q: What is the relationship between discrete and continuous functions? A: The relationship between discrete and continuous functions is that as the number of discrete points approaches infinity, the discrete function approaches the continuous function.
Q: How do Riemann sums relate to definite integrals? A: Riemann sums are a way of approximating the definite integral of a function by summing the areas of small rectangles that approximate the area under the curve.
Q: What is the significance of the term $\frac{3}{n}$ in the given limit? A: The term $\frac{3}{n}$ represents the width of each subinterval in the Riemann sum.
Q: How do we evaluate the limit of a summation involving a discrete function? A: To evaluate the limit of a summation involving a discrete function, we can use the following approach:

Express the discrete function as a continuous function.
Evaluate the definite integral of the continuous function.
Take the limit of the Riemann sum as the width of each subinterval approaches zero.

Glossary

Discrete function: A function that is defined at a finite number of points.
Continuous function: A function that is defined at all points in a given interval.
Riemann sum: A way of approximating the definite integral of a function by summing the areas of small rectangles that approximate the area under the curve.
Definite integral: The limit of the Riemann sum as the width of each subinterval approaches zero.
Uniform convergence: The concept that a sequence of discrete functions converges uniformly to a continuous function.