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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![**Problem Statement:**
Evaluate the improper integral from 1 to ∞ of the function:
\[ \int_{1}^{\infty} \frac{1}{(x+2)(2x+3)} \, dx. \]
**Explanation:**
The integral given above is an example of an improper integral because it involves an infinite limit of integration. To solve this integral, we need to find the antiderivative of the integrand \(\frac{1}{(x+2)(2x+3)}\) and then evaluate the limit as the upper bound of integration approaches infinity.
**Steps to Evaluate:**
1. Simplify the integrand by partial fraction decomposition:
\[ \frac{1}{(x+2)(2x+3)} = \frac{A}{x+2} + \frac{B}{2x+3} \]
2. Solve for the constants \(A\) and \(B\):
\[ \frac{1}{(x+2)(2x+3)} = \frac{A(2x+3) + B(x+2)}{(x+2)(2x+3)} \]
Setting the numerators equal, we get:
\[ 1 = A(2x+3) + B(x+2) \]
3. Find appropriate values of \(x\) to solve for \(A\) and \(B\).
4. Integrate each partial fraction separately.
5. Evaluate the integral from 1 to \(b\) and then take the limit as \(b\) approaches infinity.
The detailed explanation and computation can be carried out step by step in a written format or using computational tools for accurate results.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa9e8573f-552b-4431-b931-fec80f931989%2F5b87b96d-c914-498c-a511-863952060e73%2Fiw0lmrh_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Evaluate the improper integral from 1 to ∞ of the function:
\[ \int_{1}^{\infty} \frac{1}{(x+2)(2x+3)} \, dx. \]
**Explanation:**
The integral given above is an example of an improper integral because it involves an infinite limit of integration. To solve this integral, we need to find the antiderivative of the integrand \(\frac{1}{(x+2)(2x+3)}\) and then evaluate the limit as the upper bound of integration approaches infinity.
**Steps to Evaluate:**
1. Simplify the integrand by partial fraction decomposition:
\[ \frac{1}{(x+2)(2x+3)} = \frac{A}{x+2} + \frac{B}{2x+3} \]
2. Solve for the constants \(A\) and \(B\):
\[ \frac{1}{(x+2)(2x+3)} = \frac{A(2x+3) + B(x+2)}{(x+2)(2x+3)} \]
Setting the numerators equal, we get:
\[ 1 = A(2x+3) + B(x+2) \]
3. Find appropriate values of \(x\) to solve for \(A\) and \(B\).
4. Integrate each partial fraction separately.
5. Evaluate the integral from 1 to \(b\) and then take the limit as \(b\) approaches infinity.
The detailed explanation and computation can be carried out step by step in a written format or using computational tools for accurate results.
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