91Ó°ÊÓ

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.

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.

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.