91Ó°ÊÓ

In improper integrals, the limits of integration often stretch towards infinity or negative infinity. These limits tell us about the extent to which the integral is being evaluated. For \(\int_{-\infty}^{-1} \frac{1}{\sqrt{2-w}} \, dw\), the limits are \(-\infty\) to \(-1\).
The lower limit \(-\infty\) suggests that we need to analyze the behavior of the integrand as it stretches towards negative infinity. The idea is to examine if the values approach a finite limit or become undefined. The upper limit \(-1\) is a finite number, making the evaluation at this point potentially easier. By checking both limits, we ensure that the function behaves well across the entire range of integration.

When dealing with improper integrals, especially those with infinite limits, it's crucial to check how the function behaves as it approaches \(-\infty\). The integrand \(\frac{1}{\sqrt{2-w}}\) becomes important here.
As \(w\) goes to \(-\infty\), \(2-w\) increases significantly. This increase causes the square root in the denominator to grow, making the entire fraction approach zero. This indicates that the integrand diminishes as \(w\) decreases without bound, which is generally a good sign for convergence.
However, we must also consider the behavior at any upper limits or boundaries. For example, at \(-1\), to ensure there are no discontinuities.

Finding the antiderivative is essential for evaluating definite integrals. For \(\int \frac{1}{u^{1/2}} \, du\), the antiderivative is \(2u^{1/2}\). This function helps determine the result over the range given in the problem.
Antiderivatives allow us to calculate the area under the curve defined by the original function. In an improper integral, the antiderivative is used in conjunction with limit evaluation to get meaningful results. Through substitution, the problem simplifies, making antiderivatives a powerful tool.
By substituting \(u = 2 - w\), integration becomes more straightforward, \(\int \frac{1}{u^{1/2}} \, du\) standardizes the approach.

In problems where infinity is involved, evaluating the limit becomes crucial. Since the antiderivative of \(u^{-1/2}\) gives us \(-2u^{1/2}\), we need to assess the behavior of this expression as \(u\) goes to infinity.
Performing \(\lim_{M \to \infty} -2(\sqrt{M} - \sqrt{3})\) helps determine if the integral converges or diverges. In this case, \(\sqrt{M}\) grows indefinitely, showing that the function does not approach a particular finite value.

• For convergence, the limit must reach a finite number, confirming that the area under the curve is finite.
• If diverging, as here with the limit approaching negative infinity, it indicates that the area extends infinitely.

Understanding this limit tells us that the given integral diverges.