Problem 24
Question
Determine whether each integral is convergent. If the integral is convergent, compute its value. $$ \int_{0}^{\infty} \frac{1}{\sqrt{x+1}} d x $$
Step-by-Step Solution
Verified Answer
The integral \( \int_{0}^{\infty} \frac{1}{\sqrt{x+1}} \, dx \) is divergent.
1Step 1: 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.
2Step 2: 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: \[ \lim_{b \to \infty} \int_{0}^{b} \frac{1}{\sqrt{x+1}} \, dx. \]
3Step 3: 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: \[ \int \frac{1}{\sqrt{x+1}} \, dx = 2 \sqrt{x+1}. \]
4Step 4: Evaluate the Definite Integral
Substitute the antiderivative into the definite integral and evaluate from 0 to \( b \):\[ \int_{0}^{b} \frac{1}{\sqrt{x+1}} \, dx = \left[ 2 \sqrt{x+1} \right]_{0}^{b} = 2 \sqrt{b+1} - 2. \]
5Step 5: 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.
6Step 6: 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.
Key Concepts
Integral ConvergenceImproper IntegralsLimit ProcessAntiderivative
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 — if the integral converges.
Other exercises in this chapter
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