Problem 68 - Thomas Calculus

Question

Evaluate the integrals. \(\int_{2}^{3} \frac{2 \log _{2}(x-1)}{x-1} d x\)

Step-by-Step Solution

Answer

The integral evaluates to \(\ln(2)\).

Step 1: Identify the Integral Type

The integral given is \[\int_{2}^{3} \frac{2 \log _{2}(x-1)}{x-1} d x\].This is a definite integral which involves a logarithmic function. Let's simplify the expression inside the integral.

Step 2: Simplify the Integrand

Notice that \(\frac{2 \log_2(x-1)}{x-1} = 2 \cdot \frac{\log_2(x-1)}{x-1}\). Here, we can substitute and simplify using the change of base formula for logarithms: \[\log_2(x-1) = \frac{\ln(x-1)}{\ln(2)}\].Thus, the integral becomes: \[\int_{2}^{3} \frac{2}{\ln(2)} \cdot \frac{\ln(x-1)}{x-1} \, dx\].This is a constant multiple of a standard logarithmic integral form.

Step 3: Integral Substitution Preparation

Recognizing this form, let's make the integral easier to solve. Define \(u = \ln(x-1)\). Then \(\frac{du}{dx} = \frac{1}{x-1}\) or \(du = \frac{1}{x-1} \, dx\).The integral now becomes:\[\int_{2}^{3} \frac{2}{\ln(2)} \, u \, du\],where \(u = \ln(x-1)\) and \(du = \frac{1}{x-1} \, dx\).

Step 4: Evaluate the Integral

Perform the integration with respect to \(u\): \[\int u \, du = \frac{u^2}{2}\].Thus,\[\int_{2}^{3} \frac{2}{\ln(2)} \, u \, du = \frac{2}{\ln(2)} \cdot \frac{u^2}{2} = \frac{u^2}{\ln(2)}\].Convert this back in terms of \(x\), using \(u = \ln(x-1)\):\[= \frac{(\ln(x-1))^2}{\ln(2)}\].

Step 5: Apply the Limits of Integration

Insert the limits from \(x = 2\) to \(x = 3\):\[\left[\frac{(\ln(x-1))^2}{\ln(2)}\right]_2^3\].When \(x = 3\), \(\ln(3-1) = \ln(2)\), and \((\ln(2))^2 = \ln(2)^2\). When \(x = 2\), \(\ln(2-1) = \ln(1) = 0\).The evaluation of the integral becomes:\[\frac{(\ln(2))^2}{\ln(2)} - \frac{0^2}{\ln(2)} = \ln(2)\].

Step 6: Compute the Final Result

The evaluated definite integral is \(\ln(2)\).Thus, the final value of the given integral is \(\ln(2)\).

Key Concepts

Definite Integrals

A definite integral is a way to compute the area under a curve from one specific point to another. This type of integral is important because it gives us the exact value of this area. For the integral of a function over an interval, say from 2 to 3, we write it as \[ \int_{2}^{3} f(x) \, dx \].
Here’s a friendly breakdown of how definite integrals work:

• Definite integrals are represented with limits of integration. These limits tell us the interval over which to evaluate the area.
• Once evaluated, a definite integral gives a numerical value, which represents the total net area between the function and the x-axis within the interval.
• This integrative process can involve simplifying the function or using special techniques to make calculations easier.

In our example, the definite integral was evaluated over the bounds 2 and 3, eventually yielding the value of \(\ln(2)\). This demonstrates how definite integrals can provide precise results for specific function behavior across an interval.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. They help us solve for exponents, especially in equations where the unknown is in exponent form. Understanding logarithmic functions involves knowing their properties and how to convert between bases.
In the given integral \( \int_{2}^{3} \frac{2 \log_2(x-1)}{x-1} dx \), we encountered a logarithmic function. It’s expressed with a base 2 logarithm, \( \log_2(x-1) \).

• Using the change of base formula, we can express any logarithmic function in terms of the natural logarithm (ln), which is more commonly used in calculus.
• The change of base formula states: \[ \log_b(a) = \frac{\ln(a)}{\ln(b)} \]. This conversion simplifies manipulation in calculus, as seen where \( \log_2(x-1) = \frac{\ln(x-1)}{\ln(2)} \).
• Logarithmic properties such as this are essential for simplifying integrands in calculus.

Through understanding and applying such properties, we managed to simplify and solve the expression, demonstrating the power of redefining logarithmic expressions when working with definite integrals.

Substitution Method

The substitution method in calculus is akin to changing lanes smoothly in traffic; it lets you transform complicated integrals into simpler forms. This technique involves substituting a part of the integral with a new variable, making integration easier.
In our integral, \( \int_{2}^{3} \frac{2 \ln(x-1)}{x-1} \, dx \), we applied substitution.

• The substitution simplification requires you to identify a suitable substitution. For instance, \( u = \ln(x-1) \), which transforms the integrand into a simpler form.
• We also derive \( du = \frac{1}{x-1} \, dx \), which allows us to replace all parts of the original variable \( x \) in terms of the new variable \( u \).
• This simplifies the integration process, making it straightforward to integrate \( \int u \, du \), leading eventually to \( \frac{u^2}{2} \).

The substitution method not only streamlines the integration task but also opens up different avenues for solving complex integrals effectively. By converting our integral to one involving \( u \) only, we could calculate and convert it back to \( x \), solving the definite integral easily.