Problem 27 - Calculus Single Variable

Question

Determine whether the given improper integral is convergent or divergent. If it converges, then evaluate it. \(\int_{0}^{3} \frac{x}{x^{2}-2} d x\)

Step-by-Step Solution

Answer

The integral is divergent at \( x = \sqrt{2} \).

Step 1: Identify the Type of Integral

The given integral \( \int_{0}^{3} \frac{x}{x^{2}-2} \, dx \) is improper because the integrand \( \frac{x}{x^2-2} \) has a discontinuity. We need to check the denominator \( x^2-2 = 0 \) to identify the discontinuities.

Step 2: Determine Discontinuities

Set \( x^2 - 2 = 0 \) to find \( x = \sqrt{2} \) and \( x = -\sqrt{2} \). Since \( \sqrt{2} \approx 1.414 \) lies within the integration interval \([0, 3]\), the improper integral has an issue at \( x = \sqrt{2} \).

Step 3: Split the Integral

Split the integral at the point of discontinuity \( x = \sqrt{2} \), using a limit to approach the discontinuity:\[ \int_{0}^{3} \frac{x}{x^2-2} \, dx = \lim_{a \to \sqrt{2}^-} \int_{0}^{a} \frac{x}{x^2-2} \, dx + \lim_{b \to \sqrt{2}^+} \int_{b}^{3} \frac{x}{x^2-2} \, dx \].

Step 4: Evaluate Each Part

Evaluate each integral separately:For \( \lim_{a \to \sqrt{2}^-} \int_{0}^{a} \frac{x}{x^2-2} \, dx \), use substitution: let \( u = x^2 - 2 \) giving \( du = 2x \, dx \), so \( \frac{1}{2} du = x \, dx \).\( \int \frac{x}{x^2-2} \, dx = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|x^2-2| + C \).

Step 5: Evaluate the Limits

Compute the evaluated definite integrals:- For \( \lim_{a \to \sqrt{2}^-} \int_{0}^{a} \frac{x}{x^2-2} \, dx \), the integral becomes:\( \lim_{a \to \sqrt{2}^-} \left[ \frac{1}{2} \ln|x^2-2| \right]_{0}^{a} \), which diverges as \( x \to \sqrt{2}^- \) due to \( \ln|x^2-2| \rightarrow -\infty \).Therefore, the integral does not converge because the first part is divergent.

Key Concepts

Convergence and Divergence

When we talk about improper integrals, one of the first questions we ask is whether the integral is convergent or divergent. These are two crucial outcomes:

• Convergence: An improper integral converges if it results in a finite number when evaluated.
• Divergence: Conversely, it diverges if the result is infinite or does not settle to any particular value.

This is particularly important as it tells us whether the integral represents something meaningful or not. For example, in the integral \( \int_{0}^{3} \frac{x}{x^{2}-2} \, dx \), the presence of a discontinuity at \( x = \sqrt{2} \) raises the question of convergence. Because part of the split integral diverges, the entire integral is divergent, meaning it does not settle to a finite sum.

Limit Evaluation

Limits play a key role when dealing with improper integrals. They help handle the behavior of the function near points of discontinuity. In this particular problem, the function \( \frac{x}{x^2-2} \) becomes problematic because it is undefined at \( x = \sqrt{2} \).

To resolve this, we use limits to "approach" the point of discontinuity without actually evaluating it directly. The integral is split at the problematic point into two parts:

• \( \lim_{a \to \sqrt{2}^-} \int_{0}^{a} \frac{x}{x^2-2} \, dx \)
• \( \lim_{b \to \sqrt{2}^+} \int_{b}^{3} \frac{x}{x^2-2} \, dx \)

Even though limits can often be used to coax a result from an improper integral, in this case, trying to evaluate the first part reveals that the limit causes divergence, pointing to the whole integral being divergent.

Integration Techniques

Applying the right integration technique is essential, especially when confronting improper integrals. Here, the technique used is substitution, which simplifies the integral into a more manageable form. For this problem, substitution involves:

• Letting \( u = x^2 - 2 \) so that \( du = 2x \, dx \).
• Rewriting the integral in terms of \( u \): \( \frac{1}{2} \int \frac{1}{u} \, du \).
• This integral simplifies to \( \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|x^2 - 2| + C \).

This method turns the integral into a logarithmic form, which is easier to evaluate. However, the divergence issue arises when applying limits, showcasing that sometimes the divergence is inherent, regardless of the simplification.

Discontinuities in Integrals

Discontinuities are the points at which integrals might "misbehave." They occur when the function is not defined or becomes infinite. The integral \( \int_{0}^{3} \frac{x}{x^2-2} \, dx \) contains a discontinuity at \( x = \sqrt{2} \).

This spot is where the denominator of the integrand becomes zero, leading to an undefined point or a vertical asymptote. Identifying these points is the first step in handling improper integrals.

By breaking the integral at the discontinuity, we can evaluate its behavior on both sides, but if any part diverges, as in this case, then so does the entire integral. Discontinuities require careful handling, often involving limits and sometimes other techniques such as partial fraction decomposition.