Problem 73 - 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. \(f(x)=\sqrt{1+1 / x}\)

Step-by-Step Solution

Answer

Use \(g(x) = \frac{1}{\sqrt{x}}\); \(\varepsilon = 6.25 \times 10^{-8}\). \(\int_{0}^{1} f(x) dx \approx 2\).

Step 1: Identify the behavior of f(x) as x approaches 0 from the right

The function given is \( f(x) = \sqrt{1 + \frac{1}{x}} \). As \( x \to 0^{+} \), the \( \frac{1}{x} \to \infty \), which makes \( 1 + \frac{1}{x} \to \infty \). Therefore, \( f(x) \to \infty \) as \( x \to 0^{+} \). This confirms that \( f(x) \) is indeed unbounded as \( x \to 0^{+} \).

Step 2: Choose an appropriate function g(x) for comparison

The task requires finding \( g(x) = c x^{p} \) such that \( 0 \leq f(x) \leq g(x) \). Consider \( g(x) = \frac{c}{x^{p}} \). We need \( \sqrt{1 + \frac{1}{x}} \leq \frac{c}{x^{p}} \) for \( x \in (0,1] \). Simplifying, we have \( 1 + \frac{1}{x} \leq \frac{c^2}{x^{2p}} \). For this inequality to hold near \( x = 0 \), choose \( p = \frac{1}{2} \) and \( c = 1 \), making \( g(x) = \frac{1}{\sqrt{x}} \).

Step 3: Verify if \( g(x) \) is appropriate

Check the convergence condition \( \int_{0}^{1} \frac{1}{\sqrt{x}} dx \). The integral evaluates as follows: \[ \int \frac{1}{\sqrt{x}} dx = 2\sqrt{x} + C \]. Evaluating from 0 to 1 gives \( 2(1) - 2(0) = 2 \), which is finite. Hence, the integral of \( g(x) \) is convergent, and \( f(x) \leq g(x) \) fulfills the Comparison Test conditions.

Step 4: Determine \( \varepsilon \) such that \( \int_{0}^{\varepsilon} g(x) dx < 5 \times 10^{-4} \)

Determine \( \varepsilon \) by solving \( \int_{0}^{\varepsilon} \frac{1}{\sqrt{x}} dx = 2\sqrt{x} \Big|_0^{\varepsilon} < 5 \times 10^{-4} \). This requires \( 2\sqrt{\varepsilon} < 5 \times 10^{-4} \). Solving this inequality gives \( \varepsilon < \left( \frac{5 \times 10^{-4}}{2} \right)^2 = 6.25 \times 10^{-8} \).

Step 5: Approximate \( \int_{0}^{1} f(x) dx \) using \( \varepsilon \)

Now approximate \( \int_{0}^{1} f(x) dx \approx \int_{\varepsilon}^{1} f(x) dx \) where \( \varepsilon = 6.25 \times 10^{-8} \). Since \( f(x)\) closely follows \( g(x) \), for \( x\) near 1, approximate the integral using numerical methods or by approximating the last calculated part \( \int_{\varepsilon}^{1} \sqrt{1 + \frac{1}{x}} \approx \int_{0}^{1} \frac{1}{\sqrt{x}} dx = 2\).

Key Concepts

Comparison Test

The comparison test is a fundamental tool in calculus used to determine the convergence or divergence of improper integrals. The beauty of the comparison test lies in its simplicity: it allows you to compare a difficult-to-integrate function with another function that is easier to manage. The main idea is straightforward:

• If you have two functions, say, \( f(x) \) and \( g(x) \), where \( 0 \leq f(x) \leq g(x) \), and if the integral of \( g(x) \) over a particular interval converges (i.e., gives a finite number), then the integral of \( f(x) \) over that same interval also converges.
• Conversely, if the integral of \( g(x) \) diverges (i.e., tends to infinity), and if \( f(x) \geq g(x) \), then the integral of \( f(x) \) also diverges.

In our example, we compared \( f(x) = \sqrt{1 + \frac{1}{x}} \) to the simpler function \( g(x) = \frac{1}{\sqrt{x}} \), successfully applying the comparison test to confirm convergence. This process simplifies complex problems and aids in evaluating a wide range of improper integrals, providing assurance in determining whether they will reach a finite value.

Convergence of Integrals

Convergence of integrals is crucial in calculus, particularly in improper integrals, where limits approach infinity or where we might encounter discontinuities, such as at points where the function becomes undefined. The concept of convergence tells us whether performing integration over an interval will result in a meaningful, finite outcome or if the result will be infinite.
An improper integral converges if it evaluates to a finite number. This commonly happens if the area under the curve of the function you're integrating is limited. For example, in our case, we investigated the improper integral:

• The integral \( \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx \) converges because it evaluates to 2, a finite number, over the given interval from 0 to 1.

If a function is well behaved near its limits, as in the function \( g(x) = \frac{1}{\sqrt{x}} \) at \( x \rightarrow 0^+ \), the integral's finite result assures us that our function is not out of control, affirming convergence.

Function Behavior Near Singularities

Understanding a function's behavior near singularities is essential when evaluating improper integrals. Singularities are points where a function can't be properly defined—often where it tends to infinity or breaks down in some manner.

• In our exercise, we noted that the function \( f(x) = \sqrt{1+\frac{1}{x}} \) approaches infinity as \( x \to 0^+ \), presenting a classic example of singularity analysis.
• Functions often become unbounded at their singularities, but with intelligent comparison and substitution (like our \( g(x) = \frac{1}{\sqrt{x}} \)), we can manage their behavior.

By carefully picking boundary functions such as \( g(x) \) to compare \( f(x) \) against, we can correctly apply the comparison test even at these singular points. Calculus provides a framework for handling such challenging integrals, ensuring that problematic points in a function don’t derail our calculations, leading to incorrect conclusions. Function analysis near these singularities allows us to confidently state the convergence or divergence of an integral.