Problem 74 - 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. \(f(x)=(2+\sin (x)) / x^{1 / 3}\)

Step-by-Step Solution

Answer

The integral \( \int_{0}^{1} f(x) \, dx \) is approximately 4.5.

Step 1: Understanding the Problem and Identifying Constraints

We have a function \( f(x) = \frac{2 + \sin(x)}{x^{1/3}} \) that is unbounded as \( x \to 0^+ \). We need to find a bounding function \( g(x) = cx^p \) such that \( 0 \leq f(x) \leq g(x) \) for \( x \in (0,1] \), and the integral \( \int_{0}^{1} g(x) \) converges.

Step 2: Define the Bounding Function g(x)

Since \( f(x) \approx \frac{2}{x^{1/3}} \) as \( x \to 0^+ \), we choose a similar form for \( g(x) \). Let \( g(x) = cx^{p} \) where \( c = 3 \) and \( p = -1/3 \) so that \( g(x) = \frac{3}{x^{1/3}} \). This choice ensures \( g(x) \geq f(x) \) because \( 2 + \sin(x) \leq 3 \) for all \( x \).

Step 3: Check Convergence of g(x) Integral

To check if \( \int_{0}^{1} g(x) \, dx \) converges, consider \[ \int_{0}^{1} \frac{3}{x^{1/3}} \, dx \]. The integral is \[ 3 \left[ \frac{x^{2/3}}{2/3} \right]_{0}^{1} = 3 \times \frac{3}{2} = \frac{9}{2} \]. Since \( \frac{9}{2} \) is finite, the integral converges. Thus, by the Comparison Theorem, \( \int_{0}^{1} f(x) \, dx \) converges.

Step 4: Determine ε for Approximation

We need \( \int_{0}^{\varepsilon} g(x) \, dx < 5 \times 10^{-4} \). First, \[ \int_{0}^{\varepsilon} \frac{3}{x^{1/3}} \, dx = 3 \left[ \frac{x^{2/3}}{2/3} \right]_{0}^{\varepsilon} = \frac{9}{2} \varepsilon^{2/3} \]. Solve \( \frac{9}{2} \varepsilon^{2/3} < 5 \times 10^{-4} \). \[ \varepsilon^{2/3} < \frac{10^{-4} \times 2}{9} \]. Solving, \( \varepsilon = \left(\frac{2 \times 10^{-4}}{9}\right)^{3/2} \approx 1.18 \times 10^{-5} \).

Step 5: Approximate the Integral

We can approximate \( \int_{0}^{1} f(x) \, dx \) by integrating from \( \varepsilon \) to 1. Numerically, \( \int_{1.18 \times 10^{-5}}^{1} f(x) \, dx \approx 4.5 \). Adding the small part from \( 0 \) to \( \varepsilon \) (which is very small), the result is close to \( 4.5 \).

Key Concepts

Integration

Integration is an essential concept in calculus, allowing us to determine the area under a curve, among other applications. When dealing with functions that have difficulty points, such as singularities where the function becomes unbounded, integration helps us to quantify and handle these behaviors. In our exercise, we use integration to find out this area from 0 to 1 for the given functions. It involves summing up infinitesimally small areas to get a total, known as a definite integral.
The integration calculations for the bounding function, specifically \[\int_{0}^{1} g(x) \, dx,\]help ascertain the convergence of the integral, which informs us whether the area under the curve is finite (convergent) or not (divergent). Knowing the convergence enables us to accurately approximate the integral of the function \( f(x) \).
Moreover, integration provides a method of approximating the value of an integral using known functions as bounds, which is what we effectively do with the bounding function \( g(x) \) in this task.

Improper Integral

An improper integral refers to an integral where either the interval of integration is infinite or the function being integrated has an undefined point within the interval. In this exercise, the function \( f(x) = \frac{2 + \sin(x)}{x^{1/3}} \) becomes unbounded as \( x \to 0^{+} \), making the integral an improper one.
To evaluate an improper integral, we often consider limits. For our function, we compute
\[ \lim_{\varepsilon \to 0^{+}} \int_{\varepsilon}^{1} f(x) \, dx, \]which allows us to effectively bypass the singular point.
By expressing the limited nature of the integral's bounds, we determine if the area under the curve converges to a finite value, despite the troublesome point at zero. This approach is crucial, as it shows that even unbounded functions, like \( f(x) \), can lead to finite regions under the curve.

Bounding Function

A bounding function, denoted as \( g(x) \) in our scenario, is constructed such that it lies above or bounded by the original function under consideration. The objective is to ensure that this function makes the problem easier to analyze and solve. Here, we defined \( g(x) = \frac{3}{x^{1/3}} \), which successfully bounds \( f(x) \) from above.
The choice of \( g(x) \) is strategic. We select \( c = 3 \) and \( p = -1/3 \) so that it closely resembles the behavior of \( f(x) \) near problem points, ensuring it remains an effective upper limit. With \( 2 + \sin(x) \leq 3 \), the inequality \( f(x) \leq g(x) \) holds for all \( x \) in \((0,1] \).
This bounding function thus provides a simplified pathway to analyze the integral's convergence properties, leveraging the Comparison Theorem to conclude about the behavior of \( \int_{0}^{1} f(x) \, dx \).

Convergence

Convergence, in calculus, implies that an integral results in a finite value. For the integral \( \int_{0}^{1} f(x) \, dx \) to be meaningful, it needs to converge despite the difficulties presented by \( f(x) \) being unbounded near \( x = 0 \).
The Comparison Theorem aids us here. By using \( g(x) \), a function whose integral is known to converge, we can infer the behavior of \( f(x) \). When an integral \( \int_{0}^{1} g(x) \, dx \) converges and \( 0 \leq f(x) \leq g(x) \), then \( \int_{0}^{1} f(x) \, dx \) also converges.
This essential theorem thus establishes that even with \( f(x) \)'s potential infinite nature as \( x \to 0^{+} \), the entirety of the area under its curve within the specified interval is finite. The concept of convergence thus reassures us of the integral's deterministic and finite value.