Problem 96 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(93-106.\) $$\int_{1}^{e} x^{(\ln 2)-1} d x$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{\ln 2} \).

Step 1: 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, \( n = (\ln 2) - 1 \).

Step 2: 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:\[\int x^{(\ln 2)-1} \, dx = \frac{x^{(\ln 2)-1+1}}{(\ln 2)-1+1} = \frac{x^{\ln 2}}{\ln 2}.\]

Step 3: Evaluate the definite integral

Now, evaluate the indefinite integral from \( x = 1 \) to \( x = e \):\[\left[ \frac{x^{\ln 2}}{\ln 2} \right]_{1}^{e} = \frac{e^{\ln 2}}{\ln 2} - \frac{1^{\ln 2}}{\ln 2}.\]

Step 4: 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:\[\frac{2}{\ln 2} - \frac{1}{\ln 2} = \frac{2 - 1}{\ln 2} = \frac{1}{\ln 2}.\]

Step 5: 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.

Key Concepts

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:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

In cases of definite integrals, such as \( \int_{a}^{b} x^n \, dx \), you evaluate the antiderivative at both bounds \( a \) and \( b \) and subtract the results as part of the Fundamental Theorem of Calculus:

• \( \left[ \frac{x^{n+1}}{n+1} \right]_{a}^{b} = \frac{b^{n+1}}{n+1} - \frac{a^{n+1}}{n+1} \)

In the example from the exercise, \( n = (\ln 2) - 1 \). After applying the Power Rule, we find the antiderivative \( \frac{x^{\ln 2}}{\ln 2} \). This shows how effectively the rule reduces the complexity of the problem into manageable parts and leads to the final evaluation.

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.

• 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 \( e^{\ln a} = a \).

In our exercise’s evaluation, we see that since \( e^{\ln 2} = 2 \), it simplifies the integration result. Mastering these logarithmic and exponential rules aids greatly in manipulating and computing integrals effectively.

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:

• 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.

Applying these techniques requires discernment in selecting the method suited to a particular integral. In our exercise, the use of the Power Rule was key due to the structure of the integrand. Learning to match the appropriate strategy to the integral’s form is an essential skill in calculus, allowing for more streamlined solutions.