Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Riemann Sum
Riemann Sums is a special type of approximation of the area under a curve by dividing it into multiple simple shapes like rectangles or trapezoids and is used in integrals when finite sums are involved. Figuring out the area of a curve is complex hence this method makes it simple. Usually, we take the help of different integration methods for this purpose. This is one of the major parts of integral calculus.
Riemann Integral
Bernhard Riemann's integral was the first systematic description of the integral of a function on an interval in the branch of mathematics known as real analysis.
Question
![### Expressing the Integral as a Limit of Riemann Sums
In this section, we will express the given integral as a limit of Riemann sums using the right endpoints. We are not required to evaluate the limit.
Consider the integral:
\[ \int_{6}^{9} \left( x^2 + \frac{1}{x} \right) dx \]
To express this as a limit of Riemann sums, we will use the right endpoints.
A Riemann sum is represented as:
\[ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x \]
where:
- \( \Delta x = \frac{b - a}{n} \) is the width of each subinterval.
- \( x_i^* \) is a sample point in the \(i\)-th subinterval.
For right endpoints, \( x_i^* \) is the right endpoint of the \(i\)-th subinterval.
Let's identify the components in our specific case:
- \( a = 6 \)
- \( b = 9 \)
- The function \( f(x) = x^2 + \frac{1}{x} \)
The width of each subinterval:
\[ \Delta x = \frac{9 - 6}{n} = \frac{3}{n} \]
The right endpoint \( x_i^* \) for the \(i\)-th subinterval is:
\[ x_i = 6 + i \Delta x = 6 + i \left(\frac{3}{n}\right) = 6 + \frac{3i}{n} \]
Thus, the Riemann sum using right endpoints is:
\[ \lim_{n \to \infty} \sum_{i=1}^{n} \left( \left( 6 + \frac{3i}{n} \right)^2 + \frac{1}{6 + \frac{3i}{n}} \right) \left( \frac{3}{n} \right) \]
This represents the integral:
\[ \int_{6}^{9} \left( x^2 + \frac{1}{x} \right) dx \]
Therefore, the integral is expressed as a limit of Riemann sums using the right endpoints. It can be written as:
\[ \lim](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6432aae-8b0b-47bc-a89f-5ec45f34f71a%2F3e1daad3-56e2-40f8-962a-61369e13e8cf%2Fcwal01e_processed.png&w=3840&q=75)
Transcribed Image Text:### Expressing the Integral as a Limit of Riemann Sums
In this section, we will express the given integral as a limit of Riemann sums using the right endpoints. We are not required to evaluate the limit.
Consider the integral:
\[ \int_{6}^{9} \left( x^2 + \frac{1}{x} \right) dx \]
To express this as a limit of Riemann sums, we will use the right endpoints.
A Riemann sum is represented as:
\[ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x \]
where:
- \( \Delta x = \frac{b - a}{n} \) is the width of each subinterval.
- \( x_i^* \) is a sample point in the \(i\)-th subinterval.
For right endpoints, \( x_i^* \) is the right endpoint of the \(i\)-th subinterval.
Let's identify the components in our specific case:
- \( a = 6 \)
- \( b = 9 \)
- The function \( f(x) = x^2 + \frac{1}{x} \)
The width of each subinterval:
\[ \Delta x = \frac{9 - 6}{n} = \frac{3}{n} \]
The right endpoint \( x_i^* \) for the \(i\)-th subinterval is:
\[ x_i = 6 + i \Delta x = 6 + i \left(\frac{3}{n}\right) = 6 + \frac{3i}{n} \]
Thus, the Riemann sum using right endpoints is:
\[ \lim_{n \to \infty} \sum_{i=1}^{n} \left( \left( 6 + \frac{3i}{n} \right)^2 + \frac{1}{6 + \frac{3i}{n}} \right) \left( \frac{3}{n} \right) \]
This represents the integral:
\[ \int_{6}^{9} \left( x^2 + \frac{1}{x} \right) dx \]
Therefore, the integral is expressed as a limit of Riemann sums using the right endpoints. It can be written as:
\[ \lim
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