Chapter 7: Problem 96
Evaluate the integrals in Exercises \(93-106.\) $$\int_{1}^{e} x^{(\ln 2)-1} d x$$
Short Answer
Expert verified
The integral evaluates to \( \frac{1}{\ln 2} \).
Step by step solution
01
Identify the integration formula
The given integral \( \int_{1}^{e} x^{(\ln 2)-1} \, dx \) suggests the power rule for integration can be used: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n eq -1 \). Here, \( n = (\ln 2) - 1 \).
02
Apply the power integration rule
Apply the integration rule to the function by increasing the exponent by one and dividing by the new exponent. Thus, we integrate as follows:\[\int x^{(\ln 2)-1} \, dx = \frac{x^{(\ln 2)-1+1}}{(\ln 2)-1+1} = \frac{x^{\ln 2}}{\ln 2}.\]
03
Evaluate the definite integral
Now, evaluate the indefinite integral from \( x = 1 \) to \( x = e \):\[\left[ \frac{x^{\ln 2}}{\ln 2} \right]_{1}^{e} = \frac{e^{\ln 2}}{\ln 2} - \frac{1^{\ln 2}}{\ln 2}.\]
04
Simplify the evaluation
Simplify each term in the evaluated expression:1. \( e^{\ln 2} = 2 \) because raising \( e \) to the power of \( \ln 2 \) results in \( 2 \).2. \( 1^{\ln 2} = 1 \) because any number raised to any power of zero remains 1.Thus, the expression becomes:\[\frac{2}{\ln 2} - \frac{1}{\ln 2} = \frac{2 - 1}{\ln 2} = \frac{1}{\ln 2}.\]
05
Conclusion
The evaluated integral is simplified to \( \frac{1}{\ln 2} \). This is the value of the definite integral from \( 1 \) to \( e \) for the given function.
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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 fundamental technique useful for solving many integration problems. When you encounter an integral of the form \( \int x^n \, dx \), where \( n \) is a real number not equal to -1, the Power Rule can be your go-to method.
The rule itself is straightforward: you increase the power of \( x \) by one and divide by the new power, plus add the integration constant \( C \) if needed. More formally, the rule is expressed as:
The rule itself is straightforward: you increase the power of \( x \) by one and divide by the new power, plus add the integration constant \( C \) if needed. More formally, the rule is expressed as:
- \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)
- \( \left[ \frac{x^{n+1}}{n+1} \right]_{a}^{b} = \frac{b^{n+1}}{n+1} - \frac{a^{n+1}}{n+1} \)
Exponential Functions in Integrals
Dealing with exponential functions in integrals may seem tricky at first, but understanding their properties makes integration much easier. Let's take a closer look at such functions and their integration properties.
In the context of our definite integral \( \int_{1}^{e} x^{(\ln 2)-1} dx \), we encounter an exponent involving \( \ln 2 \). This highlights how integrals often contain functional exponents like natural logarithms, which require careful handling.
In the context of our definite integral \( \int_{1}^{e} x^{(\ln 2)-1} dx \), we encounter an exponent involving \( \ln 2 \). This highlights how integrals often contain functional exponents like natural logarithms, which require careful handling.
- Recognize expressions with natural logs, \( \ln(x) \), usually translate in integrals to powers of exponential bases, \( e \).
- When \( e \) is involved as a base raised to any variable or constant power, it simplifies effectively thanks to properties like \( e^{\ln a} = a \).
Integration Techniques
Integration encompasses a set of methods and techniques tailored to different types of functions and problems. Picking the right technique helps simplify and solve the problem efficiently.
Some common techniques include:
Some common techniques include:
- Power Rule: As reviewed earlier, ideal for polynomials and functions in the power form.
- Substitution: Useful when the integral involves a composition of functions. It simplifies the integrand by changing variables.
- Integration by Parts: Best for products of functions, derived from the product rule of differentiation.
- Partial Fraction Decomposition: Efficient when dealing with rational functions, breaking them down into simpler fractions.