Problem 44 - Calculus and its applications

Question

Determine whether each improper integral is convergent or divergent, and find its value if it is convergent. $$ \int_{1}^{\infty} \frac{d x}{\sqrt{x}} $$

Step-by-Step Solution

Answer

The integral is divergent.

Step 1: Analyze the Type of Integral

The given integral is \( \int_{1}^{\infty} \frac{dx}{\sqrt{x}} \). This is an improper integral because the upper limit is infinity. We'll first determine whether it converges or diverges.

Step 2: Rewrite the Integral

The integral can be rewritten in a more standard form for integration. We have \( \frac{1}{\sqrt{x}} = x^{-1/2} \). Thus, the integral becomes \( \int_{1}^{\infty} x^{-1/2} \, dx \).

Step 3: Find the Antiderivative

Determine the antiderivative of \( x^{-1/2} \). The antiderivative is \( \frac{x^{1/2}}{1/2} = 2x^{1/2} \).

Step 4: Evaluate the Indefinite Integral

Using the antiderivative, we evaluate the definite integral as \( \lim_{b \to \infty} \left[ 2x^{1/2} \right]_1^b \).

Step 5: Solve the Limit

Evaluate the integral using the limits: \[ \lim_{b \to \infty} \left( 2\sqrt{b} - 2\sqrt{1} \right) = \lim_{b \to \infty} \left( 2\sqrt{b} - 2 \right) \]. As \( b \to \infty \), \( 2\sqrt{b} \to \infty \).

Step 6: Conclude on Convergence or Divergence

Since the result of the limit is infinite, the integral \( \int_{1}^{\infty} \frac{dx}{\sqrt{x}} \) is divergent.

Key Concepts

Convergence and Divergence of Integrals

When we encounter an improper integral, it's crucial to determine if it converges or diverges. An integral converges if it approaches a specific, finite value as the variable approaches infinity or a point of discontinuity. Conversely, it diverges if it does not settle on such a value, often growing without bound.
\(\int_{1}^{\infty} \frac{dx}{\sqrt{x}}\) is a classic example of an improper integral with infinity as its upper limit. To assess convergence or divergence, we evaluate the limit of the integral as the upper bound heads towards infinity. If we find that it results in a finite number, the integral is convergent. However, if it extends to infinity, the integral diverges.
For \(\int_{1}^{\infty} x^{-1/2} \, dx\), the integral diverges as seen by its evaluation, where the result approaches infinity as the integral limit extends to infinity.

Antiderivatives

Identifying the antiderivative is a pivotal step in solving an improper integral. The antiderivative, or primitive, is a function whose derivative is the original function we are integrating.
For our integral, the function \(x^{-1/2}\) requires finding an antiderivative which is a function \(F(x)\) such that \(F'(x) = x^{-1/2}\). By applying the reverse power rule, where the antiderivative of \(x^n\) is \(\frac{x^{n+1}}{n+1}\) (provided \(n eq -1\)), we get that the antiderivative of \(x^{-1/2}\) is \(2x^{1/2}\).
This result allows us to move forward with evaluating the integral, as the integral from 1 to infinity of a function can be represented by the antiderivative evaluated at these bounds.

Limit Evaluation

After identifying the antiderivative, limit evaluation is the crucial step to deduce the behavior of the improper integral's value as the variable approaches infinity.
In the problem \(\int_{1}^{\infty} \frac{dx}{\sqrt{x}}\), we apply the antiderivative obtained, \(2x^{1/2}\), to evaluate the limit from 1 to "b" as \(b\) approaches infinity.
The evaluation results in \[ \lim_{b \to \infty} \left( 2\sqrt{b} - 2 \right) \]. Since \(2\sqrt{b}\) grows indefinitely as \(b\) increases to infinity, the limit becomes infinite. Thus, the evaluation of this limit shows that the integral diverges.
This illustrates why competency in limit evaluation is fundamental: it directly informs us of the integral's behavior and reveals its divergence.