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Evaluate the integrals. $$\int_{2}^{4} \frac{d x}{x(\ln x)^{2}}$$

Short Answer

Expert verified
\( \frac{1}{2 \ln 2} \)

Step by step solution

01

Recognize the Integral's Form

The given integral is \( \int_{2}^{4} \frac{dx}{x (\ln x)^2} \). This form hints that a u-substitution could simplify the integration process.
02

Choose a Suitable Substitution

To make the integration simpler, let's perform the substitution \( u = \ln x \), which implies \( du = \frac{1}{x} \, dx \), or equivalently, \( dx = x \, du \). Thus, \( x = e^u \).
03

Change the Limits of Integration

Using the substitution \( u = \ln x \), change the limits: when \( x = 2 \), \( u = \ln 2 \); when \( x = 4 \), \( u = \ln 4 \). Hence, the limits change from \( x = 2 \) to \( x = 4 \) into \( u = \ln 2 \) to \( u = \ln 4 \).
04

Substitute and Simplify the Integral

Under the substitution, the integral becomes:\[\int_{\ln 2}^{\ln 4} \frac{1}{(\ln x)^2} \, dx = \int_{\ln 2}^{\ln 4} \frac{x \, du}{x u^2} = \int_{\ln 2}^{\ln 4} \frac{du}{u^2}.\]The \( x \) terms cancel out, reducing the integral to a standard form \( \int \frac{1}{u^2} \, du \).
05

Evaluate the Simplified Integral

To integrate \( \int \frac{1}{u^2} \, du \), rewrite it as \( \int u^{-2} \, du \). The antiderivative is \( -u^{-1} = -\frac{1}{u} \). Therefore, evaluate:\[-\frac{1}{u} \Bigg|_{\ln 2}^{\ln 4}.\]
06

Apply the Limits of Integration

Now substitute the limits \( u = \ln 2 \) and \( u = \ln 4 \) into the antiderivative:\[-\frac{1}{u} \Bigg|_{\ln 2}^{\ln 4} = \left(-\frac{1}{\ln 4}\right) - \left(-\frac{1}{\ln 2}\right) = \frac{1}{\ln 2} - \frac{1}{\ln 4}.\]
07

Simplify the Result

Recognize that \( \ln 4 = 2 \ln 2 \). Therefore:\[\frac{1}{\ln 2} - \frac{1}{\ln 4} = \frac{1}{\ln 2} - \frac{1}{2 \ln 2} = \frac{2}{2 \ln 2} - \frac{1}{2 \ln 2} = \frac{1}{2 \ln 2}.\]This simplifies to \( \frac{1}{2 \ln 2} \).

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Key Concepts

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

Definite Integrals
Definite integrals are a fundamental concept in calculus, providing a way to calculate the area under a curve between two specific points (often referred to as limits). In more intuitive terms, they give us the net accumulation of a quantity over an interval. A definite integral is expressed as \(int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the limits of integration and \(f(x)\) is the integrand.
  • The value of a definite integral is a number that represents the total area under the curve of the graph of the function between the limits.
  • The process of finding a definite integral involves two main steps: finding the antiderivative (indefinite integral) and then applying the limits of integration.

In our exercise, after substitution and simplification, we evaluated the definite integral on the interval from \( \ln 2 \) to \( \ln 4 \). By calculating the antiderivative and then applying the limits, we found a concise expression for the area under the transformed function.
Natural Logarithm Integration
Integrating functions that involve the natural logarithm \( \ln(x) \) often requires some clever substitution to simplify the process. The natural logarithm, a logarithm with base \(e\), presents unique properties that, when exploited correctly, can streamline integration.
  • When you encounter an integral involving \( \ln(x) \), consider using substitution to simplify the expression. For instance, using \( u = \ln x \) can often reduce the complexity.
  • In our case, this substitution changed the integrand into the more manageable form \( \int \frac{du}{u^2} \), making it simpler to find the antiderivative.

The substitution transformed a challenging natural logarithm expression into a straightforward integral, allowing us to leverage the elegance of substitution in definite integrals.
Antiderivative
The antiderivative, also known as the indefinite integral, is essentially the reverse of differentiation. It seeks to find a function whose derivative is the given function. This concept is crucial when evaluating integrals because it provides the base for calculating definite integrals.
  • In mathematical notation, finding an antiderivative involves the operation \( int f(x) \, dx = F(x) + C \), where \(F(x)\) is the antiderivative and \(C\) is the constant of integration.
  • In our example, we found the antiderivative of \( u^{-2} \) to be \( -\frac{1}{u} \).

This antiderivative helped us define the net change represented by the definite integral over its specified limits. Understanding antiderivatives is a core concept to mastering both indefinite and definite integrals.

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Most popular questions from this chapter

Evaluate the integrals in Exercises \(93-106.\) $$\int \frac{d x}{x\left(\log _{8} x\right)^{2}}$$

True, or false? As \(x \rightarrow \infty\) a. \(\frac{1}{x+3}=O\left(\frac{1}{x}\right)\) b. \(\frac{1}{x}+\frac{1}{x^{2}}=O\left(\frac{1}{x}\right)\) c. \(\frac{1}{x}-\frac{1}{x^{2}}=o\left(\frac{1}{x}\right)\) d. \(2+\cos x=O(2)\) g. \(\ln (\ln x)=O(\ln x)\) h. \(\ln (x)=o\left(\ln \left(x^{2}+1\right)\right)\)

Use your graphing utility. Graph \(f(x)=\tan ^{-1} x\) together with its first two derivatives. Comment on the behavior of \(f\) and the shape of its graph in relation to the signs and values of \(f^{\prime}\) and \(f^{\prime \prime}\).

L'Hópital's Rule does not help with the limits. Try it-you just keep on cycling. Find the limits some other way. $$\lim _{x \rightarrow \infty} \frac{e^{x^{2}}}{x e^{x}}$$

Hanging cables Imagine a cable, like a telephone line or TV cable, strung from one support to another and hanging freely. The cable's weight per unit length is a constant \(w\) and the horizontal tension at its lowest point is a vector of length \(H\). If we choose a coordinate system for the plane of the cable in which the \(x\) -axis is horizontal, the force of gravity is straight down, the positive \(y\) -axis points straight up, and the lowest point of the cable lies at the point \(y=H / w\) on the \(y\) -axis (see accompanying figure), then it can be shown that the cable lies along the graph of the hyperbolic cosine Such a curve is sometimes called a chain curve or a catenary, the latter deriving from the Latin catena, meaning "chain." a. Let \(P(x, y)\) denote an arbitrary point on the cable. The next accompanying figure displays the tension at \(P\) as a vector of length (magnitude) \(T\), as well as the tension \(H\) at the lowest point \(A .\) Show that the cable's slope at \(P\) is b. Using the result from part (a) and the fact that the horizontal tension at \(P\) must equal \(H\) (the cable is not moving), show that \(T=w y .\) Hence, the magnitude of the tension at \(P(x, y)\) is exactly equal to the weight of \(y\) units of cable.

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