Problem 64 - Calculus for Biology and Medicine

Question

In Problems 63-68, evaluate each definite integral. $$ \int_{1}^{2} x \ln \left(x^{2}\right) d x $$

Step-by-Step Solution

Answer

The value of the definite integral is \( 2\ln(2) - \frac{3}{4} \).

Step 1: Simplify the Integrand

First, use the property of logarithms to simplify the integrand. Recall that \( \ln(a^b) = b \ln(a) \). In this case, we have \( \ln(x^2) = 2\ln(x) \). Substitute this back into the integral to get:\[ \int_{1}^{2} x \cdot 2 \ln(x) \ dx = 2 \int_{1}^{2} x \ln(x) \ dx \].

Step 2: Use Integration by Parts

Integration by parts is used for integrals of the form \( \int u \, dv = uv - \int v \, du \). Let \( u = \ln(x) \) and \( dv = x \, dx \). Then, \( du = \frac{1}{x} \; dx \) and \( v = \frac{x^2}{2} \). Apply integration by parts:\[ \int x \ln(x) \, dx = \left( \ln(x) \cdot \frac{x^2}{2} \right) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \].

Step 3: Simplify the Second Integral

The second integral simplifies by reducing \( \frac{x^2}{2} \cdot \frac{1}{x} \) to \( \frac{x}{2} \):\[ \int \frac{x}{2} \, dx = \frac{1}{2} \int x \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4} \].

Step 4: Evaluate the Integral

Combining the results from integration by parts, the integral becomes:\[ \frac{x^2}{2} \ln(x) - \frac{x^2}{4} \]. Evaluate this from \( x = 1 \) to \( x = 2 \):- At \( x=2 \): \[ \frac{2^2}{2} \ln(2) - \frac{2^2}{4} = 2\ln(2) - 1 \].- At \( x=1 \): \[ \frac{1^2}{2} \ln(1) - \frac{1^2}{4} = 0 - \frac{1}{4} = -\frac{1}{4} \].

Step 5: Calculate the Definite Integral Value

Subtract the value at \( x=1 \) from the value at \( x=2 \):\[ (2\ln(2) - 1) - (-\frac{1}{4}) = 2\ln(2) - 1 + \frac{1}{4} = 2\ln(2) - \frac{3}{4} \].

Key Concepts

Integration by Parts

Integration by parts is a powerful technique in calculus for solving integrals with products of functions. It's often used when directly integrating a product is difficult. The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

Choosing the right \( u \) and \( dv \) is crucial. Typically, you choose \( u \) to be a function that becomes simpler when differentiated, and \( dv \) to be a function that doesn't get more complicated when integrated.
In the specific problem above, the choice is:

• \( u = \ln(x) \)
• \( dv = x \, dx \)

Differentiating \( u \) gives \( du = \frac{1}{x} \, dx \), and integrating \( dv \) gives \( v = \frac{x^2}{2} \). Plug these into the integration by parts formula to help evaluate the definite integral.

Logarithmic Functions

Logarithmic functions play a crucial role in calculus, especially in problems involving exponential growth and decay. The standard logarithmic function is \( \ln(x) \), the natural logarithm, which is the inverse of the exponential function \( e^x \). Logarithms have unique properties that simplify complex expressions. For example:

• \( \ln(a^b) = b \cdot \ln(a) \)
• \( \ln(1) = 0 \), because any number raised to the power of 0 gives 1

In the given integral, the property \( \ln(x^2) = 2\ln(x) \) is used to simplify the integrand. Recognizing and applying such properties is key when dealing with logarithms in calculus.

Calculus Problem-Solving

Solving calculus problems, such as definite integrals, often combines multiple concepts and techniques. The process involves:

• Simplifying the integrand wherever possible, like transforming \( \ln(x^2) \) to \( 2 \ln(x) \).
• Choosing appropriate methods, such as integration by parts, to solve integrals with products of functions.
• Evaluating the resulting expressions at given bounds if dealing with definite integrals.

In this exercise, the calculated steps end with evaluating the integral between bounds 1 and 2. The result, \( 2\ln(2) - \frac{3}{4} \), is the difference between these calculated limits. Such problems enhance understanding of integration techniques, especially as they apply to functions involving logs and require careful manipulation and substitution to simplify and evaluate.