Problem 75 - Calculus Single Variable

Question

The given function \(f(x)\) is unbounded as \(x \rightarrow 0^{+} .\) Determine a function \(g(x)=c x^{p}\) such that (a) \(0 \leq f(x) \leq g(x)\) for each \(x\) in \((0,1],\) and (b) \(\int_{0}^{1} g(x) d x\) is convergent. This shows that \(\int_{0}^{1} f(x) d x\) is convergent by the Comparison Theorem. By determining a positive \(\varepsilon\) such that \(\int_{0}^{\varepsilon} g(x) d x<5 \times 10^{-4},\) approximate \(\int_{0}^{1} f(x) d x\) to three decimal places. \(f(x)=\sqrt{x} \csc (x)\)

Step-by-Step Solution

Answer

\( \int_{0}^{1} f(x) \, dx \approx 2 \) when approximated to three decimal places.

Step 1: Identify the behavior of the function

The function is given as \( f(x) = \sqrt{x} \csc(x) \), and \( \csc(x) = \frac{1}{\sin(x)} \). As \( x \to 0^{+} \), the sine function \( \sin(x) \) behaves like \( x \), thus \( \csc(x) = \frac{1}{x} \). Hence, \( f(x) \approx \frac{\sqrt{x}}{x} = x^{-\frac{1}{2}} \). This suggests that the function \( f(x) \) is unbounded as \( x \to 0^{+} \).

Step 2: Propose a suitable bounding function

To choose \( g(x) = cx^p \) such that \( 0 \leq f(x) \leq g(x) \) and \( \int_{0}^{1} g(x) \, dx \) is convergent, note the behavior \( f(x) \approx x^{-\frac{1}{2}} \). We can choose \( g(x) = x^{-\frac{1}{2}} \), since 0.5 is greater than 0, and ensures the integral will converge. The constant \( c = 1 \) fits the required conditions.

Step 3: Verify convergence of the integral

Calculate \( \int_{0}^{1} x^{-\frac{1}{2}} \, dx \). By integration, \( \int x^{-\frac{1}{2}} \, dx = 2x^{\frac{1}{2}} \). Evaluating from 0 to 1 gives \( 2(1) - 2(0) = 2 \). Thus, \( \int_{0}^{1} g(x) \, dx = 2 \), which is finite, confirming the convergence of the integral.

Step 4: Evaluate a small threshold for convergence

We are asked to find \( \varepsilon \) such that \( \int_{0}^{\varepsilon} x^{-\frac{1}{2}} \, dx < 5 \times 10^{-4} \). Calculate: \( 2\varepsilon^{0.5} < 5 \times 10^{-4} \). Solving gives \( \varepsilon^{0.5} < 2.5 \times 10^{-4} \), hence \( \varepsilon < (2.5 \times 10^{-4})^2 = 6.25 \times 10^{-8} \), which is a very small positive number ensuring it meets the condition.

Step 5: Approximate the integral \( \int_{0}^{1} f(x) \, dx \)

Given that \( 0 \leq f(x) \leq x^{-\frac{1}{2}} \) and knowing \( \int_{0}^{1} x^{-\frac{1}{2}} \, dx = 2 \), and after considering the threshold, approximate \( \int_{0}^{1} f(x) \, dx \). Since \( \varepsilon \) is very small and \( f(x) \) remains bounded within \( (0,1] \, by \) the function \( g(x) \), the integral approximately equals the convergent value of \( g(x) \), yielding approximately 2.

Key Concepts

Comparison Theorem

The Comparison Theorem is a powerful tool used in calculus to determine the convergence of an integral. When dealing with improper integrals, it can often be tricky to figure out if they converge or diverge. The Comparison Theorem allows us to compare our problematic integral with another integral whose behavior we already understand.

Here's how it works: If we have two functions, say \( f(x) \) and \( g(x) \), and for all \( x \) in a certain interval, \( 0 \leq f(x) \leq g(x) \), and if the integral of \( g(x) \) converges, then the integral of \( f(x) \) must also converge. Conversely, if \( g(x) \) diverges and \( 0 \leq g(x) \leq f(x) \), then the integral of \( f(x) \) will also diverge.

In the exercise, we used the Comparison Theorem to conclude that \( \int_{0}^{1} \sqrt{x} \csc(x) \, dx \) converges by comparing it to \( \int_{0}^{1} x^{-\frac{1}{2}} \, dx \), which is known to be convergent. Thus, this theorem acts like a logical bridge, helping us reason about integrals that are otherwise quite complex to handle.

Convergence of Integrals

Understanding when an integral converges is crucial, especially in the context of improper integrals that extend over infinite intervals or unbounded functions. Convergence of an integral refers to the integral having a finite value when computed over a given limit.

For an improper integral of the form \( \int_{0}^{1} f(x) \, dx \), convergence is determined by the behavior of \( f(x) \) as it approaches the endpoints of the interval. In our exercise, we were particularly interested in the behavior near \( x = 0 \).

Directly, if \( f(x) \) becomes unbounded as \( x \rightarrow 0^{+} \), it would typically cause the integral to diverge. However, we cleverly bounded \( f(x) \) by another function \( g(x) = x^{-\frac{1}{2}} \) in our example, ensuring that our bounding function \( g(x) \), which is convergent, helps us assert the convergence of the original function's integral. This nuanced approach is key in handling integrals with asymptotic behaviors at their boundaries.

Bounding Functions

Bounding involves enclosing a function within an upper and lower bound, which simplifies analysis and helps in determining convergence or divergence.

In mathematical analysis, we often find it extremely beneficial to bound a complex function by a simpler function whose behavior is already well-understood. For instance, in this exercise, we needed to bound the function \( f(x) = \sqrt{x} \csc(x) \) to prove convergence using the Comparison Theorem.

• We found that near \( x = 0^{+} \), \( f(x) \) behaves approximately like \( x^{-\frac{1}{2}} \), which guided the selection of our bounding function.
• We chose \( g(x) = x^{-\frac{1}{2}} \), which satisfies both \( f(x) \leq g(x) \) for \( x \in (0,1] \) and results in a convergent integral.

The bounding strategy is crucial in calculus as it allows us to use simpler functions to explore the characteristics of more complicated integrals.

Cosecant Function

The cosecant function, denoted as \( \csc(x) \), is an important trigonometric function. It is defined as the reciprocal of the sine function: \( \csc(x) = \frac{1}{\sin(x)} \).

The behavior of \( \csc(x) \) is particularly interesting near zero, where \( \sin(x) \) is also close to zero. As \( x \to 0^{+} \), \( \sin(x) \approx x \), making \( \csc(x) \approx \frac{1}{x} \). This behavior leads to the function becoming unbounded as \( x \to 0^{+} \).

In this exercise, \( \csc(x) \) contributes significantly to the complexity of \( f(x) = \sqrt{x} \csc(x) \). The hyperbolic nature of \( \csc(x) \) as \( x \to 0^{+} \) means that without careful consideration (and the use of bounding functions), the integral of \( f(x) \) could potentially diverge. However, by bounding this complexity, we managed to evaluate the integral successfully.