Use the Definition to find an expression for the area under the graph of f as a limit. Do not evaluate the limit. f(x) = x² + √1 + 2x, 3 ≤ x ≤ 5 n lim Σ 718 /= 1
Use the Definition to find an expression for the area under the graph of f as a limit. Do not evaluate the limit. f(x) = x² + √1 + 2x, 3 ≤ x ≤ 5 n lim Σ 718 /= 1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
 to find an expression for the area under the graph of \( f \) as a limit. Do not evaluate the limit.
### Given Function
\[ f(x) = x^2 + \sqrt{1 + 2x}, \quad 3 \leq x \leq 5 \]
### Expression
\[ \lim_{n \to \infty} \sum_{i=1}^{n} \]
## Explanation
To find the expression for the area under the graph of the function \( f(x) \) on the interval [3, 5], we need to use the definition of the definite integral as a limit of Riemann sums. According to the definition:
\[ \int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x \]
Where:
- \( a = 3 \) and \( b = 5 \) are the lower and upper bounds, respectively.
- \( \Delta x = \frac{b - a}{n} \) is the width of each subinterval.
- \( x_i^* \) is a sample point in the \( i \)-th subinterval.
To construct the Riemann sum:
- \( \Delta x = \frac{5 - 3}{n} = \frac{2}{n} \)
- Let \( x_i = 3 + i\Delta x \)
Then,
\[ \sum_{i=1}^{n} f(x_i) \Delta x \]
\[ = \sum_{i=1}^{n} \left( (3 + i \frac{2}{n})^2 + \sqrt{1 + 2(3 + i \frac{2}{n})} \right) \frac{2}{n} \]
Thus, the expression becomes:
\[ \lim_{n \to \infty} \sum_{i=1}^{n} \left[ \left(3 + i \frac{2}{n}\right)^2 + \sqrt{1 + 2 \left(3 + i \frac{2}{n}\right)} \right] \frac{2}{n} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa80fd0a7-9ae8-4bc5-b926-d69bfc10c0d7%2Fb08a05a2-9982-4876-8a24-695c12711e72%2Ft6clymm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Problem Statement
Use the [Definition](https://en.wikipedia.org/wiki/Integral) to find an expression for the area under the graph of \( f \) as a limit. Do not evaluate the limit.
### Given Function
\[ f(x) = x^2 + \sqrt{1 + 2x}, \quad 3 \leq x \leq 5 \]
### Expression
\[ \lim_{n \to \infty} \sum_{i=1}^{n} \]
## Explanation
To find the expression for the area under the graph of the function \( f(x) \) on the interval [3, 5], we need to use the definition of the definite integral as a limit of Riemann sums. According to the definition:
\[ \int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x \]
Where:
- \( a = 3 \) and \( b = 5 \) are the lower and upper bounds, respectively.
- \( \Delta x = \frac{b - a}{n} \) is the width of each subinterval.
- \( x_i^* \) is a sample point in the \( i \)-th subinterval.
To construct the Riemann sum:
- \( \Delta x = \frac{5 - 3}{n} = \frac{2}{n} \)
- Let \( x_i = 3 + i\Delta x \)
Then,
\[ \sum_{i=1}^{n} f(x_i) \Delta x \]
\[ = \sum_{i=1}^{n} \left( (3 + i \frac{2}{n})^2 + \sqrt{1 + 2(3 + i \frac{2}{n})} \right) \frac{2}{n} \]
Thus, the expression becomes:
\[ \lim_{n \to \infty} \sum_{i=1}^{n} \left[ \left(3 + i \frac{2}{n}\right)^2 + \sqrt{1 + 2 \left(3 + i \frac{2}{n}\right)} \right] \frac{2}{n} \]
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