Chapter 7: Problem 20
Evaluate. $$ \int_{1}^{2}(\ln x)^{2} d x $$
Short Answer
Expert verified
The integral evaluates to \(-2 \ln 2 + 2\).
Step by step solution
01
Recognize the Integration Method
The problem requires us to integrate \((\ln x)^2\), which involves an integration technique known as integration by parts. This method is used when the integrand is a product of two functions.
02
Set Up Integration by Parts Formula
The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). We need to choose \( u \) and \( dv \) from the integrand \((\ln x)^2 \). Our choice will be: \( u = (\ln x)^2 \) and \( dv = dx \).
03
Calculate 'du' and 'v'
Differentiate \( u = (\ln x)^2 \) to find \( du \). Using the chain rule: \( du = 2 \ln x \cdot \frac{1}{x} \, dx = \frac{2 \ln x}{x} \, dx \).Integrate \( dv = dx \) to obtain \( v = x \).
04
Apply Integration by Parts Formula
Substitute \( u \), \( du \), \( v \), and \( dv \) into the formula:\[\int (\ln x)^2 \, dx = x(\ln x)^2 - \int x \cdot \frac{2 \ln x}{x} \, dx.\]This simplifies to \[x(\ln x)^2 - 2 \int \ln x \, dx.\]
05
Integrate \( \ln x \) Using Parts Again
To solve \( \int \ln x \, dx \), apply integration by parts once more. Let \( u = \ln x \), \( dv = dx \) then \( du = \frac{1}{x} \, dx \) and \( v = x \). Substitute into the formula:\[ \int \ln x \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C.\]
06
Substitute Back into Original Integration
Substitute the result from Step 5 back into the expression from Step 4:\[x(\ln x)^2 - 2(x \ln x - x + C) = x(\ln x)^2 - 2x \ln x + 2x + C'.\]
07
Evaluate the Definite Integral
Evaluate the expression from Step 6 using the limits from 1 to 2:\[\left[ 2(\ln 2)^2 - 2(2 \ln 2 - 2) \right] - \left[ 1(\ln 1)^2 - 2(1 \ln 1 - 1) \right].\]Since \( \ln 1 = 0 \), the second part simplifies to \( 2 \). Calculate the first part to get:\[2(\ln 2)^2 - 4 \ln 2 + 4.\]
08
Simplify the Expression
Simplify the expression \( 2(\ln 2)^2 - 4 \ln 2 + 4 - 2 \.\) Simplified further, this yields:\[-2 \ln 2 + 2.\]
09
Final Result Interpretation
The solution \(-2 \ln 2 + 2 \) represents the area under the curve \((\ln x)^2\) from 1 to 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, used to compute the area under a curve between two defined points. In this problem, we focused on determining the definite integral of \((\ln x)^2\) from 1 to 2. This process involves solving an integral with specified upper and lower limits, which in this case are 2 and 1.
- The result of a definite integral represents a numerical value, indicating the net area under the curve between the given limits.
- It's vital to understand that evaluating a definite integral can involve the application of several techniques, including integration by parts, to solve more complex integrands like \((\ln x)^2\).
Logarithmic Functions
Logarithmic functions are widely studied in calculus due to their unique properties and wide applications. In our problem, the focus was on the logarithmic function \((\ln x)^2\). Understanding the properties of logarithmic functions was crucial to correctly applying integration techniques.
- Logarithms are the inverse of exponential functions, often arising in calculus problems involving growth and decay processes.
- The natural logarithm, noted as \(\ln x\), is the logarithm to the base \(e\), where \(e\) is Euler's number, approximately 2.718.
Calculus Problem Solving
Successful calculus problem-solving often demands a systematic approach, as evident in solving this integration problem. Our task was to find the integral of \((\ln x)^2\) from 1 to 2, requiring both strategy and technique.
- Identifying the correct method for integration is crucial, particularly for products of functions or complex expressions.
- In this example, integration by parts was the chosen technique, allowing us to breakdown and conquer the integration challenge.
- A deep comprehension of differentiation and integration rules, along with function properties, is essential for successful problem-solving in calculus.