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Exercises \(17-32,\) determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$\int_{4}^{\infty} \frac{1}{x(\ln x)^{3}} d x$$

Short Answer

Expert verified
The given integral converges and the value of the integral is approximately 0.036.

Step by step solution

01

Setting up the Improper Integral

The integral \(\int_{4}^{\infty} \frac{1}{x(\ln x)^{3}} dx\) needs to be worked out as a limit because it is an improper integral. Write it as a limit: \(\lim_{t \to \infty} \int_{4}^{t} \frac{1}{x(\ln x)^{3}} dx\).
02

Applying Direct Comparison Test

The next step is to apply the direct comparison test to see whether the integral converges or diverges. For the integral to be convergent, the function to be integrated should be lesser than or equal to a function which is a known convergent. Evaluate the limit \(\lim_{x \to \infty} \frac{\frac{1}{x(\ln x)^{3}}}{\frac{1}{x}} = \lim_{x \to \infty} \frac{1}{(\ln x)^{3}}\) which is \(0\). Since this is \(0\), the original integral is convergent.
03

Evaluating the Integral

In order to evaluate this integral, use substitution method. Let \(u=\ln x\) and hence \(du = \frac{1}{x} dx\). The integral changes to \(\int_{\ln 4}^{\infty} \frac{1}{u^{3}} du\). Evaluating this, we get \(-\frac{1}{2u^{2}}\)
04

Evaluating the Limit

Substitute back the limit of \(u\), we get \(-\frac{1}{2} \lim_{u \to \infty} \frac{1}{u^{2}} - \left(-\frac{1}{2} \right) \left( \frac{1}{( \ln 4 )^{2}} \right)\). The first term goes to 0, so the final answer will be \(\frac{1}{2( \ln 4 )^{2}}\) which approximately equals to 0.036.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direct Comparison Test
When we tackle an improper integral, one of the essential tools at our disposal is the Direct Comparison Test. This test is a way to deduce the convergence or divergence of an integral by comparing it to another integral whose behavior we already know. The crucial factor here is to find a comparison function that is easier to integrate and has a known convergence or divergence.

Imagine you are at a sports event trying to judge how fast an athlete is running; you might not know their exact speed but if you compare it with someone whose speed you know, you can estimate whether they are faster or slower. Similarly, in calculus, if our intended function to integrate, say f(x), is less or equal to a known function that converges, and greater or equal to a known function that diverges, we can infer the behavior of our integral accordingly.
In the given problem, by comparing \(\frac{1}{x(\ln x)^{3}}\) with \(\frac{1}{x}\), which is a p-series with p = 1 that diverges, and finding the limit of the ratio to be zero, we conclude that the original function converges because it is less than a function (\textbackslash\textbackslash(\frac{1}{x}\textbackslash\textbackslash)) that diverges as x approaches infinity.

The power of this test lies in its ability to categorize the behavior of complex integrals without going into the intricate process of evaluating them fully.
Limit of a Function
Grasping the concept of the limit of a function is like peering through a telescope; it allows us to observe the behavior of functions as they approach a particular point. Limits are foundational in calculus and are used to understand the behavior of functions at points where they may not be directly evaluated.

When dealing with the improper integral \(\int_{4}^{\infty} \frac{1}{x(\ln x)^{3}} dx\), we cannot simply plug in infinity to evaluate it. Instead, we approach infinity more subtly by seeing what value the function_tends_ toward as x gets larger and larger. This method involves setting a boundary, t, that approaches infinity and watching what happens to our integral within those bounds.

The limit is significant because it can tell us if the area under the curve up to infinity is finite (convergent) or infinite (divergent). In our exercise, by evaluating the limit of a simplified ratio, we establish that the function indeed approaches a value, thus confirming that the integral converges.
Substitution Method in Integration
The substitution method is akin to changing the lens on a camera to get a better view. In integration, it involves transforming the integral into a simpler form that we are capable of evaluating. This method is especially useful when facing complex expressions.

Imagine trying to solve a puzzle; sometimes, reorganizing the pieces can make the picture much clearer. Likewise, with the substitution method, we choose a part of our original function to be 'u', which is a new variable that simplifies the integral. For instance, in our problem, by setting \(u=\ln x\), the integral \(\int_{4}^{\infty} \frac{1}{x(\ln x)^{3}} dx\) becomes \(\int_{\ln 4}^{\infty} \frac{1}{u^{3}} du\), which is a much simpler function to handle.

Substituting back our limits and evaluating the integral becomes a straightforward task. The substitution method not only simplifies calculations but also can reveal connections between different kinds of functions, such as algebraic and transcendental, by bridging the gap between complex and simple expressions.

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Most popular questions from this chapter

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