Problem 60 - Calculus Single Variable

Question

In each of Exercises \(58-69\) use the Comparison Theorem to determine whether the given improper integral is convergent or divergent. In some cases, you may have to break up the integration before applying the Comparison Theorem. \(\int_{2}^{6} \sqrt{\frac{3+\sin (x)}{x-2}} d x\)

Step-by-Step Solution

Answer

The integral \( \int_{2}^{6} \sqrt{\frac{3+\sin(x)}{x-2}} \, dx \) is convergent.

Step 1: Choosing a Function for Comparison

To apply the Comparison Theorem, find a function that behaves similarly to \( \sqrt{\frac{3+\sin(x)}{x-2}} \). Since \(-1 \leq \sin(x) \leq 1\), we have \(2 \leq 3+\sin(x) \leq 4\). Start with \( \sqrt{\frac{2}{x-2}} \leq \sqrt{\frac{3+\sin(x)}{x-2}} \leq \sqrt{\frac{4}{x-2}} \). Use these inequalities for comparison.

Step 2: Checking Convergence of Simplified Integrals

First, check the convergent nature of the integrals \( \int_{2}^{6} \sqrt{\frac{2}{x-2}} \, dx \) and \( \int_{2}^{6} \sqrt{\frac{4}{x-2}} \, dx \). Start by evaluating \( \int_{2}^{6} \frac{1}{\sqrt{x-2}} \, dx \), which is known to be convergent since its antiderivative, \( 2\sqrt{x-2} \), exists and is bounded over the interval [2, 6].

Step 3: Applying the Comparison Theorem

Since \( \sqrt{\frac{2}{x-2}} \leq \sqrt{\frac{3+\sin(x)}{x-2}} \leq \sqrt{\frac{4}{x-2}} \) and both bounds \( \sqrt{\frac{2}{x-2}} \) and \( \sqrt{\frac{4}{x-2}} \) provide convergent integrals based on \( \int_{2}^{6} \frac{1}{\sqrt{x-2}} \, dx \), the integral \( \int_{2}^{6} \sqrt{\frac{3+\sin(x)}{x-2}} \, dx \) is also convergent by the Comparison Theorem.

Key Concepts

Improper Integral

An improper integral occurs when we're dealing with infinite limits of integration or when the function we're integrating has some form of an infinite discontinuity over the interval. In the given problem, when considering the function \( \sqrt{\frac{3+\sin(x)}{x-2}} \), the denominator \( x-2 \) could potentially cause an issue at the point \( x = 2 \). The reason for this is because when \( x \) approaches 2, the denominator approaches zero, which could make the value of the function approach infinity. This is a typical scenario where improper integrals come into play as we handle potential infinite values.When approaching improper integrals like these, a common method is to split the integral at the problematic point, analyze each portion, and use the Comparison Theorem as demonstrated in the original solution, to see if the issue makes the integral converge or diverge.

Convergent Integral

A convergent integral is one where the limit of the integral approaches a finite number. In our example, we solve \( \int_{2}^{6} \sqrt{\frac{2}{x-2}} \, dx \) to understand the behavior of our main integral function. If both the lower-bound simplified integral, referred to as the comparison function, and the upper-bound integral yield finite values, this indicates those integrals are convergent.The approach involves finding a simpler function that bounds our original integral function but still mimics its behavior sufficiently closely. By evaluating these simpler integrals, such as \( \int_{2}^{6} \frac{1}{\sqrt{x-2}} \, dx \), we see it's convergent because it results in a finite answer after integration. If both the upper and lower bounds are convergent, we conclude through the Comparison Theorem that the original integral is also convergent.

Inequalities

Inequalities are a cornerstone when applying the Comparison Theorem, especially for improper integrals. Here, inequalities relate to bounding the function \( \sqrt{\frac{3+\sin(x)}{x-2}} \). Knowing the range of \( \sin(x) \), which lies between \(-1\) and \(1\), helps reshape the expression to form practical bounds for comparison: \( \sqrt{\frac{2}{x-2}} \leq \sqrt{\frac{3+\sin(x)}{x-2}} \leq \sqrt{\frac{4}{x-2}} \).By establishing these inequalities, we create a way to simplify and verify the behavior of complex integrals. Using these bounds, one can work with simpler integrals on either side to understand the behavior of the original integral. If both bounding integrals, which are easier to evaluate, are convergent, we confirm through inequality comparison that the original more complex integral will converge as well.