Chapter 5: Problem 72
Evaluate each limit by interpreting it as a Riemann sum in which the given interval is divided into \(n\) subintervals of equal width. \(\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{n}{n^{2}+k^{2}} ;[0,1]\)
Short Answer
Expert verified
The limit evaluates to \(\frac{\pi}{4}\).
Step by step solution
01
Identify the Width of Subintervals
The interval given is \([0, 1]\) which has length 1. Divide this interval into \(n\) subintervals of equal width. Therefore, the width of each subinterval (denoted as \(\Delta x\)) is \(\Delta x = \frac{1-0}{n} = \frac{1}{n}\).
02
Determine the Sample Points
For a Riemann sum, we typically use sample points in each subinterval. A simple choice is to use the right endpoints. For the \(k\)-th subinterval, the right endpoint is given by \((k \cdot \Delta x)\), which is \((k \cdot \frac{1}{n})\).
03
Express the Sum as a Riemann Sum
The given sum is \(\sum_{k=1}^{n} \frac{n}{n^{2} + k^{2}}\). Substitute \(k = x_k n\) where \(x_k = \frac{k}{n}\) is the right endpoint of the subinterval. The sum can be approximated by the Riemann sum: \[\sum_{k=1}^{n} \frac{n}{n^{2} + k^{2}} \approx \sum_{k=1}^{n} \frac{1}{1+(\frac{k}{n})^{2}} \cdot \frac{1}{n},\] which is an approximation of the integral of function \(f(x) = \frac{1}{1+x^2}\).
04
Convert the Sum into a Definite Integral
The expression \((\frac{1}{1+(\frac{k}{n})^{2}} \cdot \frac{1}{n})\) represents the Riemann sum for the integral of \(f(x) = \frac{1}{1+x^2}\) over \[0, 1\]. Therefore, the limit \(\lim_{n \to \infty} \sum_{k=1}^{n} \frac{n}{n^{2} + k^{2}} \) can be evaluated as the definite integral \([ \int_{0}^{1} \frac{1}{1+x^2} \, dx]\).
05
Evaluate the Definite Integral
The integral \(\int_{0}^{1} \frac{1}{1+x^2} \, dx\) is a standard integral that equals \[ \left. \tan^{-1}(x) \right|_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4}. \]
06
State the Result
The evaluated limit is \(\frac{\pi}{4}\). Therefore, \(lim_{n \to \infty} \sum_{k=1}^{n} \frac{n}{n^{2} + k^{2}} = \frac{\pi}{4}\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits in Calculus
In calculus, **limits** serve as the foundational building block for more complex concepts such as derivatives and integrals. Limits help us understand the behavior of functions as they approach a certain point. Here, the limit describes what happens to the Riemann sum as the number of subintervals increases infinitely.
When evaluating \( \lim_{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{n}{n^{2}+k^{2}} \), the goal is to grasp the behavior of our sum as \( n \), the number of subintervals, approaches infinity. This process helps us transition from finite sums to the continuous area under a curve.
The critical step is recognizing that as \( n \) becomes very large, the sum begins to resemble a definite integral, providing a powerful method for evaluating such sums efficiently using calculus techniques.
When evaluating \( \lim_{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{n}{n^{2}+k^{2}} \), the goal is to grasp the behavior of our sum as \( n \), the number of subintervals, approaches infinity. This process helps us transition from finite sums to the continuous area under a curve.
The critical step is recognizing that as \( n \) becomes very large, the sum begins to resemble a definite integral, providing a powerful method for evaluating such sums efficiently using calculus techniques.
Definite Integral
The **definite integral** is a central concept in calculus that gives the accumulated total of a function over a specific interval. In our problem, the definite integral is formed as we transition from a Riemann sum.
The evaluated integral results in \( \tan^{-1}(x) \) from \( 0 \) to \( 1 \), demonstrating how easily we can move from sums to precise calculations of areas.
- The given integral \( \int_0^1 \frac{1}{1+x^2} \, dx \) emerges as the limit of the Riemann sum, providing the exact area under the curve of the function \( f(x) = \frac{1}{1+x^2} \).
- The definite integral can be thought of as the sum of infinitely many tiny quantities, offering a precise method to calculate areas and solve various real-world problems.
The evaluated integral results in \( \tan^{-1}(x) \) from \( 0 \) to \( 1 \), demonstrating how easily we can move from sums to precise calculations of areas.
Subintervals
**Subintervals** play a pivotal role in approximating areas under curves through Riemann sums. By dividing an interval \( [0, 1] \) into \( n \) equal parts, each segment's width is determined by \( \Delta x = \frac{1}{n} \).
This process demonstrates how we bridge discrete sums with continuous functions through a limit approach.
- This division allows us to approximate the area under a curve by summing up the areas of rectangles with widths \( \Delta x \) and heights determined by the function's value at specific points.
- In Riemann sums, the choice of sample points, often the right endpoints \, \( k \Delta x \), significantly influences the approximation.
This process demonstrates how we bridge discrete sums with continuous functions through a limit approach.