Problem 54 - Calculus Single Variable

Question

In each of Exercises \(41-54,\) determine whether the given improper integral is convergent or divergent. If it converges, then evaluate it. \(\int_{0}^{1} \frac{1}{\sqrt{\sqrt{x+1}-1} \sqrt{x+1}} d x\)

Step-by-Step Solution

Answer

The integral converges, and its value is \( 4\sqrt{\sqrt{2}-1} \).

Step 1: Analyze the Integral

First, note that the integral is improper because the integrand becomes undefined or infinite at the lower limit of integration (as \( x \to 0 \)). We need to determine if the integral converges or diverges.

Step 2: Substitution to Simplify

Let us use the substitution \( u = \sqrt{x+1} \) so that \( du = \frac{1}{2\sqrt{x+1}} dx \) and hence, \( dx = 2u \, du \). When \( x = 0, \) then \( u = 1 \) and when \( x = 1, \) then \( u = \sqrt{2} \). Substitute into the integral, transforming it to \( \int_{1}^{\sqrt{2}} \frac{1}{\sqrt{u-1} \cdot u} \, 2u \, du \). This simplifies to \( 2 \int_{1}^{\sqrt{2}} \frac{1}{\sqrt{u-1}} \, du \).

Step 3: Evaluate Simplified Integral

Examine the integral \( \int \frac{1}{\sqrt{u-1}} \, du \). The antiderivative of \( \frac{1}{\sqrt{u-1}} \) is \( 2\sqrt{u-1} + C \). Therefore, evaluate the integral: \[2 \int_{1}^{\sqrt{2}} \frac{1}{\sqrt{u-1}} \, du = 2 \left[ 2\sqrt{u-1} \right]_{1}^{\sqrt{2}} = 4[\sqrt{\sqrt{2}-1} - \sqrt{0}].\]

Step 4: Calculate the Final Value

The expression simplifies to \( 4 \sqrt{\sqrt{2}-1} \). This value is finite, thus the integral converges and evaluates to \( 4 \sqrt{\sqrt{2}-1} \).

Step 5: Conclusion on Convergence

Since the value calculated is finite, we conclude that the improper integral is convergent. The exact evaluation of this integral is \( 4 \sqrt{\sqrt{2}-1} \).

Key Concepts

Convergence of Integrals

Improper integrals often bring confusion regarding whether they converge or diverge. This means we have to determine if their value is finite or infinite as we approach a certain limit. When dealing with such integrals, it is crucial to first identify any points where the function becomes undefined or exhibits unbounded behavior. In improper integrals, these points typically occur at the limits of integration.
To decide convergence:

• Find if the integrand blows up or becomes infinite at some point.
• Two common methods to check for convergence are comparison tests and limit tests.
• If the value of the definite integral is finite, the integral converges; otherwise, it diverges.

In the given exercise, the investigation of the point where the integrand becomes undefined at the lower limit helps ascertain that we are dealing with an improper integral. Ultimately, since applying suitable manipulations and evaluating the integral yields a finite number, it is concluded to converge.

Substitution Method

Substitution is a powerful technique in integration, especially when simplifying complex integrals. By changing variables, we can transform a difficult integral into a more manageable form, making the integration process simpler.
In the original exercise, the substitution \( u = \sqrt{x+1} \) was employed:

• Calculate the differential, \( du = \frac{1}{2\sqrt{x+1}} dx \), to express \( dx \) in terms of \( du \).


• Transform the limits of integration based on \( u \); from \( x = 0 \) to \( u = 1 \), and \( x = 1 \) to \( u = \sqrt{2} \).


• Substitute back into the integral to simplify the calculation, transforming it into an integral with respect to \( u \), which is often easier to solve.

The clever choice of substitution allows us to simplify the expression and proceed with the calculation, which would otherwise be more complicated to manage.

Integration Techniques

Integration techniques are problem-solving tactics that help evaluate integrals. Each technique needs to be chosen skillfully based on the integral encountered.
For the given problem, after the substitution, the core technique involves finding the antiderivative of a square root function. The integral \( \int \frac{1}{\sqrt{u-1}} \, du \) appears, and solving it requires understanding:

• The antiderivative of \( \frac{1}{\sqrt{u-1}} \) is known to be \( 2\sqrt{u-1} \).
• Evaluating this definite integral involves plugging in the limits of the transformed integral, thus ensuring we accurately compute its value.

By leveraging known antiderivatives and systematic techniques, integrals that initially seem daunting can be unraveled into more straightforward tasks.

Limit Analysis in Calculus

Limit analysis is a fundamental concept in calculus, crucial for understanding improper integrals. When dealing with functions that approach infinity or become undefined at certain points, limits help assess behavior at those critical points.
Let's break down its role:

• Limits help evaluate the behavior of the integral as it approaches a potentially problematic boundary or point of discontinuity.


• In the provided exercise, although the substitution simplifies the integral, the original task involves ensuring the limits do not lead to undefined or infinite values.


• By evaluating the exact limit at the bounds, especially at potential points of infinity, we ascertain whether the integral converges to a specific finite value.

In essence, limits provide the framework to ensure that the evaluation of improper integrals delivers definitive results, signifying convergence or divergence reliably.