Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 7: Problem 24
Determine whether each integral is convergent. If the integral is convergent, compute its value. $$ \int_{0}^{\infty} \frac{1}{\sqrt{x+1}} d x $$
Short Answer
Expert verified
The integral \( \int_{0}^{\infty} \frac{1}{\sqrt{x+1}} \, dx \) is divergent.
Step by step solution
01
Analyze the Integral's Limits
The integral given is an improper integral because one of the limits is infinity: \( \int_{0}^{\infty} \frac{1}{\sqrt{x+1}} \, dx \). We need to evaluate if the integral is convergent, which means finding out if it approaches a finite limit.
02
Set Up the Limit Process
To evaluate the convergence of the integral, we'll use a limit process with a variable \( b \) as the upper limit of integration: \[ \lim_{b \to \infty} \int_{0}^{b} \frac{1}{\sqrt{x+1}} \, dx. \]
03
Find the Antiderivative
The antiderivative of \( \frac{1}{\sqrt{x+1}} \) is \( 2 \sqrt{x+1} \). We'll use this to evaluate the integral. Find the antiderivative: \[ \int \frac{1}{\sqrt{x+1}} \, dx = 2 \sqrt{x+1}. \]
04
Evaluate the Definite Integral
Substitute the antiderivative into the definite integral and evaluate from 0 to \( b \):\[ \int_{0}^{b} \frac{1}{\sqrt{x+1}} \, dx = \left[ 2 \sqrt{x+1} \right]_{0}^{b} = 2 \sqrt{b+1} - 2. \]
05
Apply the Limit Process
Now, apply the limit as \( b \) approaches infinity:\[ \lim_{b \to \infty} (2 \sqrt{b+1} - 2). \]As \( b \to \infty \), \( \sqrt{b+1} \to \infty \), therefore \( 2 \sqrt{b+1} \) also goes to infinity, making the entire expression diverge.
06
Conclude the Analysis
Since the limit goes to infinity, the improper integral\[ \int_{0}^{\infty} \frac{1}{\sqrt{x+1}} \, dx \]does not converge to a finite number.
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.
Integral Convergence
Understanding whether an integral converges is crucial in determining if it has a finite value. Convergence means that as the integral is evaluated over a specific range, the sum of the areas under the curve approaches a fixed, finite number. For an integral to converge, particularly when involving infinitely long intervals, the function within the integral must decrease rapidly enough as it approaches infinity.
- If an integral converges, it settles on a finite number, indicating the area bounded by the curve is measurable.
- If it diverges, it means the area is infinitely large and unbounded.
Improper Integrals
Improper integrals frequently occur in real-world applications and calculus problems. These integrals include one or more infinite limits or integrands that become infinite within the interval of integration.
- An integral like \( \int_{0}^{\infty} \frac{1}{\sqrt{x+1}} \, dx \) is improper because it evaluates from zero to infinity.
- To handle improper integrals, a convergence test is often required to determine if they have a finite area under the curve.
Limit Process
The limit process is a mathematical technique used to evaluate integrals with infinite limits. It involves replacing an infinite limit with a variable, solving the integral as if the variable were finite, and then finding the limit as the variable approaches infinity.
- In our problem, the integral \( \int_{0}^{\infty} \frac{1}{\sqrt{x+1}} \, dx \) is assessed by first evaluating \( \int_{0}^{b} \frac{1}{\sqrt{x+1}} \, dx \).
- The variable \( b \) serves as a placeholder for the upper limit, replacing infinity temporarily.
- After solving the integral with \( b \), the limit \( \lim_{b \to \infty} \) is applied to determine convergence.
Antiderivative
An antiderivative is key in solving integrals. It is a function whose derivative yields the original function in the integrand. By finding the antiderivative of a function, it becomes possible to evaluate definite integrals.
- For the integral \( \int \frac{1}{\sqrt{x+1}} \, dx \), the antiderivative is \( 2 \sqrt{x+1} \).
- This antiderivative allows us to express the integral in terms of the variable \( x \), from its lower limit to its upper limit.
- By substituting these limits into the antiderivative, we obtain the definite integral's value 鈥� if the integral converges.
