/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 191 \(\lim _{n \rightarrow \infty} \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 191

\(\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}}\) (A) \(e\) (B) \(e-1\) (C) \(1-e\) (D) \(e+1\)

Short Answer

Expert verified
(B) \(e-1\)

Step by step solution

01

Recognize the Expression as a Riemann Sum

The given limit and sum expression \( \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}} \) is a Riemann sum. It approximates the integral of a function by dividing the interval into \(n\) sub-intervals. Here, \( \frac{1}{n} \) represents the width of each sub-interval, and \( e^{\frac{r}{n}} \) represents the function value at specific points.
02

Identify the Function and Interval

Observe that this sum can be seen as a Riemann sum for the integral of the function \( f(x) = e^x \) over the interval \([0, 1]\). The terms \( \frac{r}{n} \) are points within the interval, and the function evaluated at these points is \( e^{\frac{r}{n}} \). As \( n \rightarrow \infty \), these points become dense across the interval \[0, 1\].
03

Set up the Integral

The Riemann sum \( \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}} \) is equivalent to the integral \( \int_{0}^{1} e^x \, dx \). This means we will calculate the definite integral of \(e^x\) over the interval from 0 to 1.
04

Compute the Integral

To find \( \int_{0}^{1} e^x \, dx \), we need to find the antiderivative of \(e^x\), which is simply \(e^x\) itself. Using the fundamental theorem of calculus, evaluate \(e^x\) from 0 to 1:\[ \int_{0}^{1} e^x \, dx = \left[ e^x \right]_{0}^{1} = e^1 - e^0 = e - 1. \]
05

Conclusion

The Riemann sum converges to the value of the definite integral \(e-1\) as \(n\) approaches infinity.

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 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite integral
A definite integral is a key mathematical concept used to calculate the total area under a curve within a specific interval on the x-axis. This concept is instrumental in understanding the accumulation of quantities. For example, it can represent the total distance traveled over time given a velocity function.
  • In the context of the original exercise, the definite integral is represented by \int_{0}^{1} e^x \, dx.
  • This integral calculates the area under the exponential function \( e^x \) over the interval from \(0\) to \(1\).
  • Finding the definite integral involves finding the antiderivative of the function, a process which often uses the fundamental theorem of calculus.
The result of the integration provides a precise number, which in this case is \( e-1\). This precise number helps us understand the behavior of functions across a specified domain.
Exponential function
The exponential function, expressed as \(e^x \), is a fundamental mathematical concept that describes growth and decay processes in various scientific fields. It is characterized by a constant base \(e\), a transcendental number approximately equal to \2.718\.
  • In our problem, the function \(e^x\) is being integrated, showing the versatile application of exponentials in calculus.
  • Exponential functions grow significantly faster than polynomial and linear functions, making them crucial for modeling rapid growth or decay.
In the integral from 0 to 1, the exponential function \(e^x \) depicts growth from \(e^0 = 1\) to \(e^1 = e\). This aids in understanding the impact of continuous compounding in various real-life scenarios, like finance and population dynamics.
Limit of a sequence
In mathematics, the limit of a sequence refers to the value that the terms of a sequence approach as the index goes to infinity. Understanding limits is essential for working with sequences and series, particularly in the evaluation of definite integrals and Riemann sums.
  • In this exercise, the limit \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}} is a Riemann sum converter into a definite integral.
  • The result of the limit connects the concept of summing infinitely small quantities to determine the total area under a curve.
As the number of partitions (n) becomes infinite, the sum \(\sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}} \) approaches the definite integral \(\int_{0}^{1} e^x \, dx \), leading to the result \( e-1\). This illustrates how the limit of a sequence forms a bridge to integral calculus.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The values of \(a\) for which the equation \(\int_{0}^{x} \sin ^{2} \frac{t}{2} d t=\) \(a^{2} x^{2}-\frac{1}{2}(3 x-1)+\frac{1}{a^{2}}\) possesses a solution are (A) \(\pm \frac{1}{\sqrt{n \pi+\frac{\pi}{2}}}, n \in N\) (B) \(\pm \frac{1}{\sqrt{2 n \pi+\pi}}, n \in N\) (C) \(\pm \frac{1}{\sqrt{2 n \pi-\frac{\pi}{2}}}, n \in N\) (D) None of these

For \(x>0\), let \(f(x)=\int_{1}^{x} \frac{\ln t}{1+t} d t .\) Then, the value of \(f(e)+f\left(\frac{1}{e}\right)\) is (A) 1 (B) 2 (C) \(\frac{1}{2}\) (D) None of these

The area of the region bounded by the curves \(y=|x-1|\) and \(y=3-|x|\) is (A) 2 sq. units (B) 3 sq. units (C) 4 sq. units (D) 6 sq. units

A function \(f\) defined on \([a, b]\) is bounded above if there exists a real number \(K\) such that \(|f(x)| \leq K, \forall x \in[a, b]\). The real number \(K\) is called an upper bound of \(f . f\) is called bounded below if there exists a real number \(k\) such that \(|f(x)| \geq k, \forall x \in[a, b]\). The real number \(k\) is called a lower bound of \(f\). It may be observed that if \(K\) is an upper bound of \(f\), then every real number greater than or equal to \(K\) is also an upper bound of \(f\). The smallest of all the upper bounds of \(f\) is called the least upper bound or the supremum (sup.) of \(f\). Similarly, the greatest of all the lower bounds of \(f\) is called the greatest lower bound or the infimum (Inf.) of \(f\). If the functions \(f\) and \(g\) are bounded and continuous on \([a, b]\) and \(g\) keeps the same sign on \([a, b]\), then there exists a number \(\mu\) between the infimum and supremum of \(f\) on \([a, b]\) such that $$ \int_{a}^{b} f(x) g(x) d x=\mu \int_{a}^{b} g(x) d x $$. \(\int_{0}^{1} \frac{x^{2}}{\sqrt{1+x^{2}}}\) lies between (A) \(\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{3}}\) (B) \(\frac{1}{\sqrt{2}}\) and 1 (C) \(\frac{1}{3 \sqrt{2}}\) and \(\frac{1}{3}\) (D) None of these

In the following questions an Assertion (A) is given followed by a Reason (R). Mark your responses from the following options: (A) Assertion (A) is True and Reason (R) is True; Reason \((\mathrm{R})\) is a correct explanation for Assertion (A) (B) Assertion (A) is True, Reason \((\mathrm{R})\) is True; Reason (R) is not a correct explanation for Assertion (A) (C) Assertion (A) is True, Reason (R) is False (D) Assertion (A) is False, Reason (R) is True Assertion: \(\int_{0}^{\pi / 2} \sin 2 k x \cot x d x=\frac{\pi}{2}\) Reason: \(\frac{\sin 2 k x}{\sin x}=2(\cos x+\cos 3 x+\ldots+\cos\) \((2 k-1) x)\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.