Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 4: Problem 2
In Exercises 1 and 2, use Example 1 as a model to evaluate the limit $$\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(c_{i}\right) \Delta x_{i}$$ over the region bounded by the graphs of the equations. $$ \begin{array}{l} f(x)=2 \sqrt[3]{x}, \quad y=0, \quad x=0, \quad x=1 \\ \text { (Hint: Let } \left.c_{i}=i^{3} / n^{3} .\right) \end{array} $$
Short Answer
Expert verified
The value of the limit is 3.
Step by step solution
01
Define The Variables
First of all, it is important to understand what each variable means. \(f(x)\) is the function to be integrated. \(x\) represents the x-values inside the region. Negotiable \(x\) values range between 0 and 1 which also defines the limits of integration. Here, the task states \(c_{i} = i^3/n^3\) which will be the average values of \(x\) in the interval we will consider while the term \(\Delta x_{i}\) represents the width of each interval.
02
Apply the Definitions
Next, we apply the definitions of the given variables and perform substitutions into the given limit. As a result, we get: \[\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f\left(\dfrac{i^3}{n^3}\right) \Delta x_{i} = \lim_{n \rightarrow \infty} \dfrac{1}{n} \sum_{i=1}^{n} f\left(\dfrac{i^3}{n^3}\right)\]Here, we assumed the width of small intervals, \(\Delta x_{i}\), approaches 1/n as \(n\) tends to infinity.
03
Implement the Function
The function \(f(x)\) is defined as \(2\sqrt[3]{x}\), so replace \(f\left(\dfrac{i^3}{n^3}\right)\) with \(2\sqrt[3]{\dfrac{i^3}{n^3}}\):\[\lim_{n \rightarrow \infty} \dfrac{1}{n} \sum_{i=1}^{n} 2\sqrt[3]{\dfrac{i^3}{n^3}}\]
04
Simplify the Expression
The expression inside the square root simplifies to \(\dfrac{i}{n}\):\[\lim_{n \rightarrow \infty} \dfrac{1}{n} \sum_{i=1}^{n} 2\sqrt[3]{\dfrac{i}{n}}\]
05
Finding The Limit
Now, by summing over all \(i\), we multiply by \(2/n\) (since there are \(n\) terms in the sum) and take the limit as \(n\) approaches infinity. This converts the sum into an integral from 0 to 1 of \(2\sqrt[3]{x}\) dx. Thus, we get: \[\lim_{n \rightarrow \infty} \dfrac{1}{n} \sum_{i=1}^{n} 2\sqrt[3]{\dfrac{i}{n}} = \int_{0}^{1} 2\sqrt[3]{x}\, dx\]
06
Calculating The Integral
Finally, calculate the integral, which gives the final result:\[\int_{0}^{1} 2\sqrt[3]{x}\, dx = \left[3x^{4/3}\right]_0^1 = 3 - 0 = 3\]
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.
Limit Evaluation
Limit evaluation is a fundamental aspect of calculus. It involves finding the value that a function approaches as the input approaches a particular value. In the context of integration using Riemann sums, the limit helps us to approximate the area under a curve by increasing the number of subintervals to infinity.
In this problem, the limit evaluation is critical to translating a Riemann sum into a definite integral. As the number of divisions, represented by \(n\), approaches infinity, each subinterval becomes infinitesimally small. This is expressed in our formula as \(\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f\big(c_{i}\big) \Delta x_{i}\).
In this problem, the limit evaluation is critical to translating a Riemann sum into a definite integral. As the number of divisions, represented by \(n\), approaches infinity, each subinterval becomes infinitesimally small. This is expressed in our formula as \(\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f\big(c_{i}\big) \Delta x_{i}\).
- Increasing \(n\) refines the approximation of the area under the curve.
- As \(n\) tends to infinity, the approximation becomes exact, converting the sum into an integral.
- This is a core step in calculus for moving from discrete sums to continuous integrals.
Riemann Sum
The Riemann sum is a method of approximating integrals, which is essentially the area under a curve on a graph. It involves dividing the domain into small subintervals and summing up the product of function values at sample points and the widths of these intervals.
In our exercise, the sum \(\sum_{i=1}^{n} f\left(\frac{i^3}{n^3}\right) \Delta x_{i}\) represents the Riemann sum:
In our exercise, the sum \(\sum_{i=1}^{n} f\left(\frac{i^3}{n^3}\right) \Delta x_{i}\) represents the Riemann sum:
- Each term in the sum corresponds to an area element in the approximation.
- \(c_i = i^3/n^3\) is chosen to represent the sample point in each subinterval.
- The term \(\Delta x_{i}\) denotes the width of intervals, approaching \(1/n\) as \(n\) increases.
Definite Integral
A definite integral is an important concept in calculus used to calculate the exact area under a curve between two points on the x-axis. In this problem, the transition from a Riemann sum to a definite integral involves taking the limit of the sum expression.
Our aim is to compute \(\int_{0}^{1} 2\sqrt[3]{x}\, dx\), which represents the area under the function \(f(x) = 2\sqrt[3]{x}\) from \(x = 0\) to \(x = 1\):
Our aim is to compute \(\int_{0}^{1} 2\sqrt[3]{x}\, dx\), which represents the area under the function \(f(x) = 2\sqrt[3]{x}\) from \(x = 0\) to \(x = 1\):
- Representing an infinite sum as an integral simplifies computation and provides exact results.
- The limits of integration, 0 and 1 in this case, define the interval over which we seek the area.
- Understanding how to convert sums to integrals is key to applying calculus to real-world scenarios.
Integral Evaluation
Integral evaluation refers to the process of calculating the value of a definite integral, which gives the exact area under a curve within specified limits. In this exercise, the integration of \(2\sqrt[3]{x}\) from 0 to 1 insightfully demonstrates the culmination of translating a Riemann sum into a meaningful area.
The integration process entails finding the antiderivative of the function \(2\sqrt[3]{x}\) which is \(3x^{4/3}\), then evaluating this at the given limits:
The integration process entails finding the antiderivative of the function \(2\sqrt[3]{x}\) which is \(3x^{4/3}\), then evaluating this at the given limits:
- The antiderivative simplifies calculations: \([3x^{4/3}]_0^1 = 3(1) - 3(0) = 3\).
- This result confirms the area under the curve is exactly 3, as computed from the integral.
- Integral evaluation is vital for making sense of real-world phenomena described by functions.
