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Chapter 28: Problem 30

In Exercises \(27-42,\) solve the given problems by integration. Find the area bounded by \(y=2(\ln x) / x^{2}, y=0\) and \(x=3\)

Short Answer

Expert verified
The area is \(-\frac{2 \ln 3}{3} - \frac{8}{9}\).

Step by step solution

01

Identify the Integral

We are tasked with finding the area bounded by the curve described by the function \(y = \frac{2 \ln x}{x^2}\), the x-axis (\(y=0\)), and a vertical line at \(x=3\). To do this, we need to evaluate the definite integral of \(y\) from the initial x-bound (you need to determine this bound where the function intersects the x-axis) to \(x=3\).
02

Find the Lower Limit of Integration

First, determine the lower limit of integration by finding where \(y = 0\). Solving \(\frac{2 \ln x}{x^2} = 0\) implies \(\ln x = 0\) because \(x^2 > 0\). Thus, \(x = e^0 = 1\). Hence, our lower limit is \(x=1\).
03

Write the Definite Integral

The definite integral to find the area is set up as follows: \[ \int_{1}^{3} \frac{2 \ln x}{x^2} \, dx \]
04

Integrate the Function

We perform integration by parts for \(\int \frac{2 \ln x}{x^2} \, dx\), with \(u = \ln x\) and \(dv = \frac{2}{x^2} \, dx\).1. Differentiate \(u\): \(du = \frac{1}{x} \, dx\)2. Integrate \(dv\): \(v = -\frac{2}{x}\)3. Integration by parts formula: \(\int u \, dv = uv - \int v \, du\).Thus, we have:\(\int \frac{2 \ln x}{x^2} \, dx = -\frac{2 \ln x}{x} + \int \frac{2}{x^2} \cdot \frac{1}{x} \, dx \)Simplify the second integral:\(= -\frac{2}{x} + \int \frac{2}{x^3} \, dx \)= \(-\frac{2 \ln x}{x} + \int -\frac{2}{x^3} \, dx = -\frac{2 \ln x}{x} + \frac{1}{x^2} + C\) (where \(C\) is the constant of integration).
05

Evaluate the Definite Integral

Evaluate the definite integral using the antiderivative calculated in Step 4. Calculate:\[\left[-\frac{2 \ln x}{x} + \frac{1}{x^2}\right]_{1}^{3} = \left(-\frac{2 \ln 3}{3} + \frac{1}{9}\right) - \left(0 + 1\right) \]Simplify:\[= \left(-\frac{2 \ln 3}{3} + \frac{1}{9}\right) - 1 \]This gives the area under the curve from 1 to 3.
06

Simplify the Result

Calculate \(-\frac{2 \ln 3}{3} + \frac{1}{9} - 1\). This leads to:\(= -\frac{2 \ln 3}{3} + \frac{1}{9} - \frac{9}{9} \)Further simplify to get the known precise area:\(= -\frac{2 \ln 3}{3} - \frac{8}{9} \)

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Key Concepts

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

Integration by Parts
Integration by Parts is a useful technique in calculus for integrating products of functions. It stems from the formula \( \int u \, dv = uv - \int v \, du \). By applying this technique, you can often turn a difficult integral into a more manageable one.
  • For the problem \( \int \frac{2 \ln x}{x^2} \, dx \), we identify \( u = \ln x \) and its differential as \( du = \frac{1}{x} \, dx \).
  • We'll let \( dv = \frac{2}{x^2} \, dx \) which integrates to \( v = -\frac{2}{x} \).
Plugging into the formula, we solve the integral to simplify our results. Understanding the choice of \( u \) and \( dv \) is key, as choosing \( u \) to be a component that simplifies when differentiated, like \( \ln x \), often makes the subsequent integral easier to handle.
Bounded Area
Calculating the area bounded by curves involves solving a definite integral between two limits. These limits are where the curve interacts with either another function or a defined axis.
In our case, we are calculating the area of the space between the curve defined by \( y = \frac{2 \ln x}{x^2} \) and the x-axis, from \( x=1 \) to \( x=3 \). To find the lower limit, we set the function equal to zero:
  • Solving \( \frac{2 \ln x}{x^2} = 0 \) leads to \( \ln x = 0 \), giving us \( x = 1 \) since the natural logarithm of 1 is zero.
This is a classic example of how a function's interaction with the x-axis, alongside defined constraints (like \( x=3 \)), defines the region of the area we need to calculate.
Antiderivative Calculation
Finding an antiderivative is a central step in solving definite integrals, representing the function whose derivative is the original integrand. In our exercise, the antiderivative of \( \frac{2 \ln x}{x^2} \) is found using integration by parts.
Following that method, the antiderivative unfolds as:
  • \( -\frac{2 \ln x}{x} \)
  • \( + \frac{1}{x^2} \)
Each term is derived from breaking down the integral and finding simpler components that neatly align to form the antiderivative. Remember, the constants from antiderivatives cancel out in definite integrals, emphasizing the definite integral's dependency on the specific bounds only, rather than any arbitrary constants.
Calculus Problem Solving
Calculus problem solving requires breaking down problems into smaller, actionable steps. For this exercise, follow these steps to ensure accuracy and understanding:
  • Clearly identify the problem and organize your thoughts about what needs solving. This involves understanding which areas the problem encompasses (such as bounded area).
  • Identify boundaries and apply the appropriate calculus technique. Here, it involved integration by parts to handle a product of a logarithm and a power of x.
  • Verify each integral and simplify results. Calculate your specific values and ensure each arithmetic step leads you closer to the result.
Practice will enhance your ability to recognize when particular techniques like integration by parts are necessary and refining each step ensures that you're not just finding results, but also truly understanding the calculus principles applied.

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Most popular questions from this chapter

Solve the given problems by integration. An oil-storage tank can be described as the volume generated by revolving the region bounded by \(y=24 / \sqrt{16+x^{2}}, x=0\) \(y=0,\) and \(x=3\) about the \(x\) -axis. Find the volume (in \(\mathrm{m}^{3}\) ) of the tank.

Solve the given problems by integration. If \(x>0,\) find \(f(x)\) if \(f^{\prime \prime}(x)=x^{-2}, f(1)=0,\) and \(f(2)=0\)

Solve the given problems by integration. Find the volume generated by revolving the first-quadrant region bounded by \(y=4 /\left(x^{4}+6 x^{2}+5\right)\) and \(x=2\) about the \(y\) -axis.

identify the form of each integral as being inverse sine, inverse tangent, logarithmic, or general power, as in Examples 6 and 7. Do not integrate. In each part (a), explain how the choice was made. (a) \(\int \frac{2 d x}{9 x^{2}+4}\) (b) \(\int \frac{2 x d x}{\sqrt{9 x^{2}-4}}\) (c) \(\int \frac{2 x d x}{9 x^{2}-4}\)

Solve the given problems by integration. The force \(F\) (in \(\mathrm{N}\) ) applied by a stamping machine in making a certain computer part is \(F=4 x /\left(x^{2}+3 x+2\right),\) where \(x\) is the distance (in \(\mathrm{cm}\) ) through which the force acts. Find the work done by the force from \(x=0\) to \(x=0.500 \mathrm{cm}\)
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