Chapter 7: Problem 60
Evaluate the integrals. \(\int_{1}^{e} x^{(\ln 2)-1} d x\)
Short Answer
Expert verified
The value of the integral is \( \frac{1}{\ln 2} \).
Step by step solution
01
Understanding the Integral
The integral given is \( \int_{1}^{e} x^{(\ln 2)-1} \, dx \). Here, \( x \) is in the base, and \( (\ln 2 - 1) \) is the exponent. We aim to integrate this power function within the limits from 1 to \( e \).
02
Applying the Power Rule for Integration
The power rule for integration states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. For definite integrals, the constant term can be ignored. Here, \( n = \ln 2 - 1 \), so the integral becomes \( \int_{1}^{e} x^{(\ln 2)-1} \, dx = \left[ \frac{x^{\ln 2}}{\ln 2} \right]_{1}^{e} \).
03
Evaluating the Integral at the Upper Limit
Substitute the upper limit \( e \) into the integrated function: \( \frac{e^{\ln 2}}{\ln 2} \). Note that \( e^{\ln 2} = 2 \), because \( e^x \) and \( \ln(x) \) are inverses and \( e^{\ln a} = a \). So the expression evaluates to \( \frac{2}{\ln 2} \).
04
Evaluating the Integral at the Lower Limit
Substitute the lower limit 1 into the integrated function: \( \frac{1^{\ln 2}}{\ln 2} \). Since any number to the power of 0 is 1, the expression simplifies to \( \frac{1}{\ln 2} \).
05
Calculating the Definite Integral
Subtract the result of the lower limit from the result of the upper limit: \( \frac{2}{\ln 2} - \frac{1}{\ln 2} = \frac{1}{\ln 2} \). This computation gives the final value of the definite integral: \( \frac{1}{\ln 2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Rule for Integration
The power rule for integration is a handy tool that helps us integrate functions of the form \( x^n \). This rule allows us to simplify the process of finding integrals by providing a direct formula. To apply this rule, you'll use the formula:
In our example, the integral \( \int_{1}^{e} x^{(\ln 2)-1} \, dx \) uses the power rule. Here, \( n = \ln 2 - 1 \). Plugging this into the formula, the expression becomes:
- \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)
In our example, the integral \( \int_{1}^{e} x^{(\ln 2)-1} \, dx \) uses the power rule. Here, \( n = \ln 2 - 1 \). Plugging this into the formula, the expression becomes:
- \( \left[ \frac{x^{\ln 2}}{\ln 2} \right]_{1}^{e} \)
Exponential Functions
Exponential functions are a crucial part of mathematics, characterized by a constant base raised to a variable exponent. The most well-known base is Euler's number \( e \), approximately equal to 2.71828.
When working with exponential functions that involve the natural base \( e \), remember the fundamental property:
Understanding this concept allows you to break down and evaluate functions involving \( e \), simplifying processes like integration.
When working with exponential functions that involve the natural base \( e \), remember the fundamental property:
- \( e^{\ln a} = a \)
Understanding this concept allows you to break down and evaluate functions involving \( e \), simplifying processes like integration.
Natural Logarithm
The natural logarithm, noted as \( \ln(x) \), is a special type of logarithm with a base of \( e \), the mathematical constant. It serves as the inverse function of exponential functions involving \( e \), which means the two operations cancel each other out.
For instance, the property \( e^{\ln x} = x \) shows this inverse relationship clearly. This understanding assists in simplifying calculations that involve both exponential and logarithmic elements.
In our integral calculation, we faced terms like \( \frac{2}{\ln 2} \) and \( \frac{1}{\ln 2} \), stemming from evaluating exponentials and the inverse properties of the natural logarithm. These concepts underline how intricate operations can be broken down into simpler ones.
For instance, the property \( e^{\ln x} = x \) shows this inverse relationship clearly. This understanding assists in simplifying calculations that involve both exponential and logarithmic elements.
In our integral calculation, we faced terms like \( \frac{2}{\ln 2} \) and \( \frac{1}{\ln 2} \), stemming from evaluating exponentials and the inverse properties of the natural logarithm. These concepts underline how intricate operations can be broken down into simpler ones.
- Recognizing this inverse relationship helps in performing accurate mathematical operations, especially within calculus tasks such as integration and differentiation.