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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...
Question
Use the p-test to decide if the following integrals converge or diverge. If it converges finds its convergence area.
Certainly! Here is the transcription of the text for an educational website:

---

The integral shown in the image is:

$$\int_{5}^{\infty} \frac{1}{\sqrt{x}} \, dx$$

This is an improper integral with an infinite upper limit. The integrand is \( \frac{1}{\sqrt{x}} \).

To solve this integral, we can use the following steps:

1. **Rewrite the integrand** using exponent notation:
   \[ \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}} \]

2. **Integrate** using the power rule for integration:
   \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad \text{for} \ n \neq -1 \]

   For our integrand, the exponent \( n = -\frac{1}{2} \):
   \[ \int x^{-\frac{1}{2}} \, dx = \int x^{-\frac{1}{2}} \, dx = \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + C = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C = 2x^{\frac{1}{2}} + C = 2\sqrt{x} + C \]

3. **Evaluate the definite integral** from 5 to \( \infty \):
   \[ \int_{5}^{\infty} x^{-\frac{1}{2}} \, dx = \left[ 2\sqrt{x} \right]_{5}^{\infty} \]

   Taking the limit as \( x \) approaches infinity and evaluating at the lower bound:
   \[ \lim_{t \to \infty} \left[ 2\sqrt{t} \right] - 2\sqrt{5} \]
   \[ = \lim_{t \to \infty} 2\sqrt{t} - 2\sqrt{5} \]

   Since \( \lim_{t \to \infty} 2\sqrt{t} \) approaches \( \infty \):
   \[ \int_{5}^{\infty} \frac{1}{\
Transcribed Image Text:Certainly! Here is the transcription of the text for an educational website: --- The integral shown in the image is: $$\int_{5}^{\infty} \frac{1}{\sqrt{x}} \, dx$$ This is an improper integral with an infinite upper limit. The integrand is \( \frac{1}{\sqrt{x}} \). To solve this integral, we can use the following steps: 1. **Rewrite the integrand** using exponent notation: \[ \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}} \] 2. **Integrate** using the power rule for integration: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad \text{for} \ n \neq -1 \] For our integrand, the exponent \( n = -\frac{1}{2} \): \[ \int x^{-\frac{1}{2}} \, dx = \int x^{-\frac{1}{2}} \, dx = \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + C = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C = 2x^{\frac{1}{2}} + C = 2\sqrt{x} + C \] 3. **Evaluate the definite integral** from 5 to \( \infty \): \[ \int_{5}^{\infty} x^{-\frac{1}{2}} \, dx = \left[ 2\sqrt{x} \right]_{5}^{\infty} \] Taking the limit as \( x \) approaches infinity and evaluating at the lower bound: \[ \lim_{t \to \infty} \left[ 2\sqrt{t} \right] - 2\sqrt{5} \] \[ = \lim_{t \to \infty} 2\sqrt{t} - 2\sqrt{5} \] Since \( \lim_{t \to \infty} 2\sqrt{t} \) approaches \( \infty \): \[ \int_{5}^{\infty} \frac{1}{\
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