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Chapter 6: Problem 19

Evaluate the following integrals. Include absolute values only when needed. \(\int_{e^{2}}^{e^{3}} \frac{d x}{x \ln x \ln ^{2}(\ln x)}\)

Short Answer

Expert verified
Answer: \(\frac{1}{\ln(2)} - \frac{1}{\ln(3)}\)

Step by step solution

01

Introduce substitution

Let's introduce a substitution \(u = \ln x\). To find \(du\), we'll take the derivative of \(u\) with respect to \(x\) and then express it in terms of dx: \[du = \frac{d (\ln x)}{d x} dx = \frac{1}{x} dx\] Now, we need to change the limits of integration. When \(x = e^2\), \(u = \ln e^2 = 2\). When \(x = e^3\), \(u = \ln e^3 = 3\). Thus, the new limits of integration are \(2\) and \(3\), respectively. Finally, we can substitute \(u\) and \(du\) into the integral: \[\int_{e^2}^{e^3} \frac{1}{x \ln x \ln^2(\ln x)} dx = \int_{2}^{3} \frac{1}{u \ln^2 u} du\]
02

Integrate with respect to u

Now we integrate with respect to \(u\): \[\int_{2}^{3} \frac{1}{u \ln^2 u} du\] This integral is considered an elementary integral (although not as simple as most), and can be recognized to directly have an antiderivative of : \[-\frac{1}{\ln u}\] We can verify this indirectly by observing that the derivative of \(-\frac{1}{\ln u}\) would yield \(\frac{1}{u \ln^2 u}\). Now let's apply the fundamental theorem of calculus to find the value of the definite integral.
03

Use the fundamental theorem of calculus

Now that we have the antiderivative, \(-\frac{1}{\ln u}\), let's apply the fundamental theorem of calculus to find the value of the definite integral: \[-\frac{1}{\ln(3)} - \left(-\frac{1}{\ln(2)}\right) = \frac{1}{\ln(2)} - \frac{1}{\ln(3)}\]
04

Conclusion

Thus, the value of the definite integral \(\int_{e^2}^{e^3} \frac{1}{x \ln x \ln^2(\ln x)} dx\) is: \[\frac{1}{\ln(2)} - \frac{1}{\ln(3)}\]

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Key Concepts

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

Substitution Method
The substitution method is a handy tool to simplify integration by changing variables. It's similar to making a detour in math when the direct path seems complicated. Here, we have used the substitution \(u = \ln x\), which simplifies the given integral into a more manageable form.
  • First, identify a part of the integral that can be substituted. For our problem, choosing \(u = \ln x\) helps because it simplifies the natural logarithm terms.
  • Next, find the differential \(du\). Since \(u = \ln x\), differentiating both sides gives \(du = \frac{1}{x} dx\).
  • Change the limits of integration if it's a definite integral. Here, we replace the original limits \(x = e^2\) and \(x = e^3\) with \(u = 2\) and \(u = 3\), respectively.
  • After substitution, the integral becomes \(\int_{2}^{3} \frac{1}{u \ln^2 u} du\).
Understanding these steps will allow you to apply the substitution method with confidence in similar problems.
Fundamental Theorem of Calculus
The fundamental theorem of calculus connects the concepts of differentiation and integration. It shows how finding areas under curves (integration) relates to finding slopes of curves (differentiation). This theorem is divided into two parts.
  • The first part tells us if we have an antiderivative \(F(x)\) of a function \(f(x)\), the definite integral \(\int_{a}^{b} f(x) \, dx\) is \(F(b) - F(a)\).
  • The second part ensures that every continuous function has an antiderivative, and it can be used to evaluate definite integrals directly.
In our exercise, after finding the antiderivative \(-\frac{1}{\ln u}\), we used the fundamental theorem of calculus to calculate the value at the new limits \(u = 2\) and \(u = 3\). This process ultimately gives us the result \(\frac{1}{\ln(2)} - \frac{1}{\ln(3)}\). The theorem turns a potentially long computation into a simple subtraction.
Elementary Integrals
Elementary integrals are basic integrals that have known solutions. These solutions can be found in standard integral tables or can be derived from common functions.
  • In integration, recognizing these elementary forms is beneficial because it allows us to skip detailed calculations.
  • For instance, the integral \(\int \frac{1}{u \ln^2 u} du\) may not seem simple, but it's known to have the antiderivative \(-\frac{1}{\ln u}\).
Using elementary integrals saves time and reduces errors. Once identified, these integrals can be computed directly, as demonstrated in this problem. Practicing their recognition is crucial for efficiency in solving integrals.

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Most popular questions from this chapter

Power and energy Power and energy are often used interchangeably, but they are quite different. Energy is what makes matter move or heat up and is measured in units of joules (J) or Calories (Cal), where 1 Cal = 4184 J. One hour of walking consumes roughly \(10^{6} \mathrm{J},\) or \(250 \mathrm{Cal} .\) On the other hand, power is the rate at which energy is used and is measured in watts \((\mathrm{W} ; 1 \mathrm{W}=1 \mathrm{J} / \mathrm{s})\) Other useful units of power are kilowatts ( \(1 \mathrm{kW}=10^{3} \mathrm{W}\) ) and megawatts ( \(1 \mathrm{MW}=10^{6} \mathrm{W}\) ). If energy is used at a rate of \(1 \mathrm{kW}\) for 1 hr, the total amount of energy used is 1 kilowatt-hour \((\mathrm{kWh})=\) which is \(3.6 \times 10^{6} \mathrm{J}\) Suppose the power function of a large city over a \(24-\mathrm{hr}\) period is given by $$P(t)=E^{\prime}(t)=300-200 \sin \frac{\pi t}{12}$$ where \(P\) is measured in megawatts and \(t=0\) corresponds to 6:00 P.M. (see figure). a. How much energy is consumed by this city in a typical \(24-\mathrm{hr}\) period? Express the answer in megawatt-hours and in joules. b. Burning 1 kg of coal produces about 450 kWh of energy. How many kg of coal are required to meet the energy needs of the city for 1 day? For 1 year? c. Fission of 1 g of uranium- 235 (U-235) produces about \(16,000 \mathrm{kWh}\) of energy. How many grams of uranium are needed to meet the energy needs of the city for 1 day? For 1 year? d. A typical wind turbine can generate electrical power at a rate of about \(200 \mathrm{kW}\). Approximately how many wind turbines are needed to meet the average energy needs of the city?

Find the volume of the solid of revolution. Sketch the region in question. The region bounded by \(y=1 / \sqrt{x}, y=0, x=2,\) and \(x=6\) revolved about the \(x\) -axis

Theo and Sasha start at the same place on a straight road, riding bikes with the following velocities (measured in \(\mathrm{mi} / \mathrm{hr}\) ). Assume \(t\) is measured in hours. Theo: \(v_{T}(t)=10,\) for \(t \geq 0\) Sasha: \(v_{S}(t)=15 t,\) for \(0 \leq t \leq 1\) and \(v_{S}(t)=15,\) for \(t>1\) a. Graph the velocity functions for both riders. b. If the riders ride for 1 hr, who rides farther? Interpret your answer geometrically using the graphs of part (a). c. If the riders ride for 2 hr, who rides farther? Interpret your answer geometrically using the graphs of part (a). d. Which rider arrives first at the \(10-, 15-\), and 20 -mile markers of the race? Interpret your answer geometrically using the graphs of part (a). e. Suppose Sasha gives Theo a head start of \(0.2 \mathrm{mi}\) and the riders ride for 20 mi. Who wins the race? f. Suppose Sasha gives Theo a head start of \(0.2 \mathrm{hr}\) and the riders ride for 20 mi. Who wins the race?

Consider the cubic polynomial \(f(x)=x(x-a)(x-b),\) where \(0 \leq a \leq b\) a. For a fixed value of \(b,\) find the function \(F(a)=\int_{0}^{b} f(x) d x\) For what value of \(a\) (which depends on \(b\) ) is \(F(a)=0 ?\) b. For a fixed value of \(b\), find the function \(A(a)\) that gives the area of the region bounded by the graph of \(f\) and the \(x\) -axis between \(x=0\) and \(x=b\). Graph this function and show that it has a minimum at \(a=b / 2\). What is the maximum value of \(A(a),\) and where does it occur (in terms of \(b\) )?

Find the volume of the torus formed when the circle of radius 2 centered at (3,0) is revolved about the \(y\) -axis. Use geometry to evaluate the integral.
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