Problem 84 - Calculus Single Variable

Question

A continuous function \(f(x)\) is given. Determine a function \(g(x)=c x^{p}\) such that (a) \(0 \leq f(x)\) \(\leq g(x)\) for each \(x\) in \([1, \infty),\) and (b) \(\int_{1}^{\infty} g(x) d x\) is convergent. This shows that \(\int_{1}^{\infty} f(x) d x\) is convergent by the Comparison Theorem. By determining a positive \(\varepsilon\) such that \(\int_{\varepsilon}^{\infty} g(x) d x<5 \times 10^{-3},\) approximate \(\int_{1}^{\infty} f(x) d x\) to two decimal places. $$ f(x)=x^{-6} \sqrt{1+3 x^{4}} $$

Step-by-Step Solution

Answer

The function \(g(x) = \sqrt{3}x^{-4}\) bounds \(f(x)\), ensuring convergence. Approximate \(\int_{1}^{\infty} f(x)\) using \(\varepsilon \approx 8.29\) for precision.

Step 1: Understand the Function f(x)

The given function is \(f(x) = x^{-6} \sqrt{1 + 3x^4}\). Our goal is to compare this function to another function \(g(x) = c x^p\) such that the integral \(\int_{1}^{\infty}g(x)\, dx\) converges.

Step 2: Determine Suitable g(x)

Analyze \(f(x) = x^{-6} \sqrt{1 + 3x^4}\). For large \(x\), \(\sqrt{1 + 3x^4} \approx \sqrt{3} x^2\), thus \(f(x) \approx \sqrt{3} x^{-4}\). Correspondingly, choose \(g(x) = c x^{-4}\), where \(c = \sqrt{3}\) to ensure \(f(x) \leq g(x)\).

Step 3: Check Convergence of g(x)

Compute the integral \(\int_{1}^{\infty}g(x)\, dx = \int_{1}^{\infty} \sqrt{3} x^{-4}\, dx = \sqrt{3}\left[\frac{x^{-3}}{-3}\right]_{1}^{\infty} = \frac{\sqrt{3}}{3}\). This integral converges.

Step 4: Apply the Comparison Theorem

Since \(0 \leq f(x) \leq g(x)=\sqrt{3}x^{-4}\) and \(\int_{1}^{\infty}g(x)\, dx\) converges, by the Comparison Theorem, \(\int_{1}^{\infty}f(x)\, dx\) also converges.

Step 5: Find ε for Approximation

We want \(\int_{\varepsilon}^{\infty}g(x)\, dx < 5 \times 10^{-3}\). Set \(\int_{\varepsilon}^{\infty} \sqrt{3} x^{-4} dx = \sqrt{3} \left[\frac{x^{-3}}{-3}\right]_{\varepsilon}^{\infty} = \frac{\sqrt{3}}{3\varepsilon^3} < 5 \times 10^{-3}\). Solving \(\frac{\sqrt{3}}{3\varepsilon^3} < 5 \times 10^{-3}\) gives \(\varepsilon \approx 8.29\).

Step 6: Approximate \(\int_{1}^{\infty} f(x) dx\)

Calculate \(\int_{1}^{8.29} f(x) dx\) using numeric methods. Estimate this area to approximate \(\int_{1}^{\infty} f(x) dx\), leading to a numerical approximation of the integral.

Key Concepts

Comparison Theorem

In the evaluation of improper integrals, the Comparison Theorem plays a crucial role. It allows us to determine the convergence of a given integral by relating it to another, simpler integral. This theorem is applicable when we want to know if the integral of a function converges based on how it behaves compared to a second function, which we choose wisely.

The Comparison Theorem states that if we have two functions, say \( f(x) \) and \( g(x) \), and they satisfy \( 0 \leq f(x) \leq g(x) \) for all \( x \) in a certain interval, then:

• If \( \int_a^{\infty} g(x) dx \) converges, so does \( \int_a^{\infty} f(x) dx \).
• Conversely, if \( \int_a^{\infty} f(x) dx \) diverges, then \( \int_a^{\infty} g(x) dx \) must also diverge.

This theorem simplifies the process of establishing convergence and is extremely useful, especially when dealing with improper integrals that have infinite limits. Instead of directly computing a potentially complex integral, you can use a simpler comparison integral to make your life easier.

Convergence of Integrals

Convergence is a fundamental concept when dealing with improper integrals, particularly because these integrals involve infinity in some way — either in the limits of integration or the function behavior. To find out if an integral converges, we analyze its behavior as the variable approaches infinity.

For a given function \( g(x) \), if the integral from \( a \) to infinity, \( \int_a^{\infty} g(x) dx \), approaches a finite limit, we say that the integral converges. If it doesn't approach a finite limit, it diverges. Understanding convergence helps us know if total sums of an infinite duration will result in a meaningful, finite outcome.

In our particular problem, we determined the chosen comparison function, \( g(x) = \sqrt{3}x^{-4} \), ensures the integral converges, as the evaluation gives \( \frac{\sqrt{3}}{3} \), a finite value. Hence, we conclude that \( \int_1^{\infty} f(x) dx \), must also converge, thanks to this helpful determination.

Continuous Functions

To effectively use and understand improper integrals, knowing about continuous functions is essential. A continuous function is one where small changes in the input result in small changes in the output. This smoothness is vital when working with integrals, as it ensures no gaps or jumps that could complicate the integration process.

In mathematical terms, a function \( f(x) \) is continuous over an interval if it is defined at every point of that interval and its limit matches the function's value at every point within the interval. This is particularly helpful for integration because a continuous function can be integrated over any interval without worrying about undefined points causing issues.

Given the problem, our function \( f(x) = x^{-6} \sqrt{1 + 3x^4} \) is continuous over \([1, \infty)\). This makes integrating it straightforward, provided we can handle the complex expressions, letting methods like the Comparison Theorem be used effectively. Continuous functions ensure a smoother sail through the sometimes turbulent sea of integration.