Problem 41 - Thomas Calculus

Question

In Exercises \(35-64\) , use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$ \int_{0}^{\pi} \frac{d t}{\sqrt{t}+\sin t} $$

Step-by-Step Solution

Answer

The integral \( \int_{0}^{\pi} \frac{dt}{\sqrt{t} + \sin t} \) converges.

Step 1: Analyze the Integrand

The integrand is \( \frac{1}{\sqrt{t} + \sin t} \). Since \( 0 \leq \sin t \leq 1 \) for all \( t \in [0, \pi] \), we have \( \sqrt{t} \leq \sqrt{t} + \sin t \leq \sqrt{t} + 1 \).

Step 2: Apply the Direct Comparison Test

We compare \( \int_{0}^{\pi} \frac{dt}{\sqrt{t} + \sin t} \) to simpler integrals that bound it. Consider \( \int_{0}^{\pi} \frac{dt}{\sqrt{t} + 1} \) and \( \int_{0}^{\pi} \frac{1}{\sqrt{t}} \, dt \).

Step 3: Evaluate Comparison Integrals

Evaluate \( \int_{0}^{\pi} \frac{1}{\sqrt{t}} \, dt \), which can be recognized as an improper integral at \( t = 0 \). It is equivalent to \( \lim_{a \to 0^+} \int_{a}^{\pi} t^{-1/2} \, dt = 2(\pi^{1/2}) - 2(0^{1/2}) \), a convergent value.

Step 4: Justify Comparison and Conclusion

Since \( \frac{1}{\sqrt{t} + \sin t} \leq \frac{1}{\sqrt{t}} \), and \( \int_{0}^{\pi} \frac{1}{\sqrt{t}} \, dt \) converges, the original integral \( \int_{0}^{\pi} \frac{dt}{\sqrt{t} + \sin t} \) also converges by the Direct Comparison Test.

Key Concepts

Direct Comparison Test

The Direct Comparison Test is a method used in calculus to assess whether an integral converges or diverges. This technique involves comparing the integral of interest to a simpler integral that is known to converge or diverge. To effectively perform the Direct Comparison Test, follow these steps:

• Choose a comparable function that is simpler and bounds the original function.
• Determine if this simpler integral converges or diverges.
• Use the known behavior of the simpler integral to make conclusions about the original integral.

If the function you are considering is always less than or equal to a function with a convergent integral, the original integral also converges. For example, in our exercise, the function \(\frac{1}{\sqrt{t} + \sin t}\)is compared to\(\frac{1}{\sqrt{t}}\).Since\(\int_{0}^{\pi} \frac{1}{\sqrt{t}} \, dt\)converges, this suggests that the original integral also converges because the function is bounded above by this simpler, convergent function.

Improper Integrals

Improper integrals arise when the interval of integration is infinite, or when the integrand is undefined or becomes infinite within the interval. They often involve limits to cope with these challenges.To evaluate improper integrals, especially when the integrand has a singularity or discontinuity:

• Identify where the function becomes undefined, either at endpoints or within the interval.
• Rewrite the integral as a limit, replacing the problematic point with a variable approaching the point.
• Evaluate the rewritten limit.

In the example,\(\int_{0}^{\pi} \frac{1}{\sqrt{t}} \, dt\),the function becomes problematic at\(t = 0\).Thus, the improper integral is rewritten as\(\lim_{a \to 0^+} \int_{a}^{\pi} \frac{1}{\sqrt{t}} \, dt\),which evaluates to a finite value, indicating convergence.

Trigonometric Functions

Trigonometric functions like sine and cosine often appear in integrals. Understanding their properties can greatly assist in evaluating integrals.Key properties of the sine function, as used in calculations, include:

• Its periodicity with a period of\(2\pi\).
• Its range, which is limited between\(-1\)and\(1\).
• Special points, such as\(\sin(0)=0\)and\(\sin(\frac{\pi}{2})=1\).

In the integral\(\frac{1}{\sqrt{t} + \sin t}\),the sine function contributes to the range and behavior of the integrand:\(\sin t\)remains between\(0\)and\(1\)while\(t\)is between\(0\)and\(\pi\).This ensures the function adds a small positive value to\(\sqrt{t} \),a fact that is used in the Direct Comparison Test by keeping the boundedness in check.

Convergence of Integrals

The convergence of integrals involves determining whether the integral results in a finite number. Convergence can depend on the behavior of the function as well as the integration limits. In general, finding whether an integral converges requires:

• Analyzing any points where the integrand may become infinite or undefined.
• Applying comparison tests or limit processes to address these points.
• Using known integrated results or properties of similar functions to draw conclusions.

In this exercise, the convergence is established via comparison with a simpler function that is known to converge:\(\int_{0}^{\pi} \frac{1}{\sqrt{t}} \, dt \).The integral\(\frac{1}{\sqrt{t} + \sin t}\)is managed similarly and deduced to converge by ensuring it's bounded by the comparison's convergent behavior.