Problem 108
Question
Evaluate the integrals. \begin{equation}\int_{2}^{3} \frac{2 \log _{2}(x-1)}{x-1} d x\end{equation}
Step-by-Step Solution
Verified Answer
The integral evaluates to \( \ln 2 \).
1Step 1: Recognize the Integral Form
The given integral is \( \int_{2}^{3} \frac{2 \log_{2}(x-1)}{x-1} \, dx \). Notice that it resembles the structure suitable for substitution method because of the \( \frac{1}{x-1} \) term, which is the derivative of the natural logarithm, and the \( \log_{2}(x-1) \) expression.
2Step 2: Simplify the Logarithmic Expression
Convert \( \log_{2}(x-1) \) to a natural logarithm using the change of base formula: \( \log_{2}(x-1) = \frac{\ln(x-1)}{\ln 2} \). Thus, the integral becomes \( \int_{2}^{3} \frac{2 \ln(x-1)}{(x-1) \ln 2} \, dx \). Simplifying gives \( \frac{2}{\ln 2} \int_{2}^{3} \frac{\ln(x-1)}{x-1} \, dx \).
3Step 3: Set Up the Substitution
Let \( u = \ln(x-1) \), then \( du = \frac{1}{x-1} \, dx \). Substitute in these variables so the integral transforms to \( \frac{2}{\ln 2} \int \! u \, du \), with limits changing from \( x = 2 \) to \( u = \ln(1) = 0 \) and from \( x = 3 \) to \( u = \ln(2) \).
4Step 4: Integrate with Respect to the New Variable
Now integrate with respect to \( u \): \( \frac{2}{\ln 2} \int_{0}^{\ln(2)} u \, du = \frac{2}{\ln 2} \left[ \frac{u^2}{2} \right]_0^{\ln(2)} \).
5Step 5: Calculate the Definite Integral
Evaluate the definite integral: \( \frac{2}{\ln 2} \left[ \frac{(\ln(2))^2}{2} - \frac{0^2}{2} \right] = \frac{2}{\ln 2} \cdot \frac{(\ln 2)^2}{2} = \ln 2 \).
6Step 6: Final Step: State the Result
The value of the integral \( \int_{2}^{3} \frac{2 \log_{2}(x-1)}{x-1} \, dx \) is \( \ln 2 \).
Key Concepts
Logarithm TransformationSubstitution MethodDefinite Integral EvaluationChange of Base Formula
Logarithm Transformation
When dealing with integrals involving logarithms, it's often useful to transform the logarithmic expression to make integration more manageable. Specifically, in this exercise, we deal with \( \log_{2}(x-1) \). Transforming it using the change of base formula eases the integration process.
\( \log_{2}(x-1) \) can be rewritten in terms of the natural logarithm as \( \log_{2}(x-1) = \frac{\ln(x-1)}{\ln 2} \).
Why do we do this? Because natural logarithms (\( \ln \)) are more straightforward to handle in calculus, particularly when applying substitution methods. This conversion assists in simplifying the integral structure and aligns it more closely with integrals we know how to solve.
\( \log_{2}(x-1) \) can be rewritten in terms of the natural logarithm as \( \log_{2}(x-1) = \frac{\ln(x-1)}{\ln 2} \).
Why do we do this? Because natural logarithms (\( \ln \)) are more straightforward to handle in calculus, particularly when applying substitution methods. This conversion assists in simplifying the integral structure and aligns it more closely with integrals we know how to solve.
Substitution Method
The substitution method is a powerful technique in integral calculus used to simplify and evaluate integrals. It essentially involves replacing a part of the integral with a single variable to transform a complex integral into one that's easier to work with.
The given integral \( \int_{2}^{3} \frac{2 \ln(x-1)}{(x-1) \ln 2} \, dx \) seems complex at first. By choosing an appropriate substitution, the task becomes more manageable. Here, we let \( u = \ln(x-1) \) which simplifies the integral significantly. Consequently, the differential \( du = \frac{1}{x-1} \, dx \) perfectly matches part of the integral's structure, allowing us to rewrite and solve it as \( \int u \, du \).
After substituting, remember to also change the limits of integration to the new variable \( u \). This completes the substitution process and simplifies the definite integral into a simple quadratic integral \( \int u \, du \), convenient for evaluation.
The given integral \( \int_{2}^{3} \frac{2 \ln(x-1)}{(x-1) \ln 2} \, dx \) seems complex at first. By choosing an appropriate substitution, the task becomes more manageable. Here, we let \( u = \ln(x-1) \) which simplifies the integral significantly. Consequently, the differential \( du = \frac{1}{x-1} \, dx \) perfectly matches part of the integral's structure, allowing us to rewrite and solve it as \( \int u \, du \).
After substituting, remember to also change the limits of integration to the new variable \( u \). This completes the substitution process and simplifies the definite integral into a simple quadratic integral \( \int u \, du \), convenient for evaluation.
Definite Integral Evaluation
Evaluating a definite integral involves solving the integral within specific bounds. Once we've simplified the integral using substitution, we need to integrate and evaluate it between the new bounds.
The rewritten integral becomes \( \frac{2}{\ln 2} \int_{0}^{\ln(2)} u \, du \).
Integrating \( u \) with respect to \( u \) yields \( \frac{u^2}{2} \). We then evaluate this expression from the lower bound \( u = 0 \) to the upper bound \( u = \ln(2) \).
To do this, substitute the limits into the antiderivative: \[ \left[ \frac{u^2}{2} \right]_0^{\ln(2)} = \frac{(\ln 2)^2}{2} - \frac{0^2}{2} \].
This calculation provides the evaluated definite integral, leading us to the final result, \( \ln 2 \), for the original problem.
The rewritten integral becomes \( \frac{2}{\ln 2} \int_{0}^{\ln(2)} u \, du \).
Integrating \( u \) with respect to \( u \) yields \( \frac{u^2}{2} \). We then evaluate this expression from the lower bound \( u = 0 \) to the upper bound \( u = \ln(2) \).
To do this, substitute the limits into the antiderivative: \[ \left[ \frac{u^2}{2} \right]_0^{\ln(2)} = \frac{(\ln 2)^2}{2} - \frac{0^2}{2} \].
This calculation provides the evaluated definite integral, leading us to the final result, \( \ln 2 \), for the original problem.
Change of Base Formula
The change of base formula is crucial in transforming logarithmic expressions to a more usable form in calculus operations. It states that for any logarithm, \( \log_b(a) \), it can be re-expressed as \( \log_b(a) = \frac{\ln(a)}{\ln(b)} \).
This transformation is highly useful when solving integrals involving unusual bases, as it converts them into a natural logarithm, which we are more adept at handling in calculus.
In our exercise, the expression \( \log_{2}(x-1) \) becomes \( \frac{\ln(x-1)}{\ln 2} \), simplifying the integral's structure and making it compatible with substitution and further integral calculus techniques.
This formula not only aids in simplifying the integral but also turns the evaluation process into a series of steps involving familiar natural logarithms, which can be solved effectively using standard calculus methods.
This transformation is highly useful when solving integrals involving unusual bases, as it converts them into a natural logarithm, which we are more adept at handling in calculus.
In our exercise, the expression \( \log_{2}(x-1) \) becomes \( \frac{\ln(x-1)}{\ln 2} \), simplifying the integral's structure and making it compatible with substitution and further integral calculus techniques.
This formula not only aids in simplifying the integral but also turns the evaluation process into a series of steps involving familiar natural logarithms, which can be solved effectively using standard calculus methods.
Other exercises in this chapter
Problem 107
Verify the integration formulas in Exercises \(105-108\) $$ \int\left(\sin ^{-1} x\right)^{2} d x=x\left(\sin ^{-1} x\right)^{2}-2 x+2 \sqrt{1-x^{2}} \sin ^{-1}
View solution Problem 107
Evaluate the integrals. \begin{equation}\int_{0}^{9} \frac{2 \log _{10}(x+1)}{x+1} d x\end{equation}
View solution Problem 109
Solve the initial value problems in Exercises \(109-114\) $$ \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}, \quad y(0)=0 $$
View solution Problem 109
Evaluate the integrals. \begin{equation}\int \frac{d x}{x \log _{10} x}\end{equation}
View solution