Problem 90 - Essential Calculus

Question

For what value of \(c\) does the integral \(\int_{1}^{\infty}\left(\frac{c x}{x^{2}+2}-\frac{1}{3 x}\right) d x\) converge? Evaluate the integral for this value of \(c\)

Step-by-Step Solution

Answer

The integral will converge for \(c ≤ 1\). The value of the integral for the range of convergence is \(c / 2\) ln 2.

Step 1: Determine The Value of \(c\)

Apply the comparison test to determine the condition under which the integral will converge. In general, if \(0 ≤ g(x) ≤ f(x)\) for \(a ≤ x < ∞\) and \(\int_{a}^{\infty} g(x) dx\) converges, then \(\int_{a}^{\infty} f(x) dx\) also converges. Comparing the function \(f(x) = \frac{c x}{x^{2}+2}-\frac{1}{3 x}\) with \(g(x) = \frac{c}{x} + \frac{1}{3x} \), \(c > 0\).Hence, \(\int_{1}^{\infty} \frac{c}{x} dx\) + \(\int_{1}^{\infty} \frac{1}{3x} dx\) is going to be finite, obviously implies that \(c ≤ 1\). So, the integral will converge only for \(c ≤ 1\).

Step 2: Perform Integration

Now that we've established the condition under which the integral will converge (when \(c ≤ 1\)), we find it relevant to evaluate the integral with this value of \(c\). To do that, we split the integral into two parts and solve each one separately.\(\int_{1}^{\infty} \frac{c x}{x^{2}+2} dx\) - \(\int_{1}^{\infty} \frac{1}{3 x} dx\)The first integral can simplify to \(c / 2\) ln\((x^2+2)\) from 1 to infinity which simplifies to \(c / 2\) ln 2.The second integral results in \(-1 / 3\) ln x from 1 to infinity which is 0.Combining the two values, we can write the solution of the integral as \(c / 2\) ln 2.

Key Concepts

Comparison Test

The Comparison Test is a handy tool when we want to determine the convergence of an improper integral. It allows us to compare a complex integral with another one that we can easily analyze. In simple terms, if we have two functions, say, \(f(x)\) and \(g(x)\), and we know that \(0 \leq g(x) \leq f(x)\) for all \(x\) in our interval, and if \(\int g(x)\ dx\) converges, then \(\int f(x)\ dx\) will also converge. In the given exercise, we applied the Comparison Test to deduce the range of the constant \(c\) where the integral \(\int_{1}^{\infty} \left(\frac{c x}{x^2+2} - \frac{1}{3 x}\right) dx\) remains convergent. By comparing with simpler integrals like \(\frac{c}{x}\) and ensuring that \(c \, \leq \, 1\), we found the threshold at which convergence occurs. Using comparison with well-known functions often provides a clearer picture of the integral's behavior on its domain.

Integration

Integration is a fundamental operation in calculus that aggregates small areas into a total value. It effectively computes accumulated values, often represented as the area under a curve. When dealing with improper integrals such as \(\int_{1}^{\infty} \left( \frac{c x}{x^2+2} - \frac{1}{3 x} \right) dx\), the process involves an assessment of limits as the variable approaches infinity.In solving the problem, we broke down the integral into more manageable parts. The first component considered was \(\int_{1}^{\infty} \frac{c x}{x^2+2} dx\), which simulates the behavior by approximating the analysis through logarithmic properties. The outcome of this integration was expressed in terms of \(\frac{c}{2}\ln(x^2 + 2)\) evaluated from 1 to infinity, leading us to \(\frac{c}{2}\ln 2\). The other part \(\int_{1}^{\infty} \frac{1}{3x} dx\) results in 0 due to straightforward logarithmic properties. Separating and solving each part with their limits helps evaluate complex expressions faultlessly.

Convergence of Integrals

The concept of convergence in the context of integrals is fascinated with whether the result of an integral approaches a finite number as we extend its bounds to infinity. Improper integrals, such as \(\int_{1}^{\infty} \left( \frac{c x}{x^2+2} - \frac{1}{3 x} \right) dx\), require careful analysis to ensure they do not diverge, i.e., become infinitely large or small.In the provided exercise, determining the convergence entailed applying limits at infinity. For the main expression to be convergent, the integrals of its components across their range must individually converge. The integral \(\int_{1}^{\infty} \frac{c}{x} dx\) converges because it resembles the known \(\frac{1}{x} \) which converges logarithmically. Ensuring \(c \leq 1\) keeps the overall evaluation finite.Understanding whether or not an integral converges is crucial as it informs whether the function described approaches a sum that is determinable. Convergence assures that the results are meaningful and applicable within real-world contexts.