Problem 66 - Calculus for Biology and Medicine

Question

Evaluate each definite integral. $$ \int_{1}^{2} \ln \left(x^{2} e^{x}\right) d x $$

Step-by-Step Solution

Answer

The result of the integral is \( 4 \ln(2) - \frac{1}{2} \).

Step 1: Break Down the Logarithm

Utilize the logarithmic property \( \ln(ab) = \ln(a) + \ln(b) \). Here, apply it to the integrand \( \ln(x^2 e^x) \), which simplifies to \( \ln(x^2) + \ln(e^x) \). This becomes \( 2\ln(x) + x \), since \( \ln(e^x) = x \).

Step 2: Simplify the Integral

Substitute the simplified integrand back into the integral. This results in \( \int_{1}^{2} (2\ln(x) + x) \ dx \). This can be split into two separate integrals: \( 2\int_{1}^{2} \ln(x) \ dx + \int_{1}^{2} x \ dx \).

Step 3: Integrate \( \ln(x) \)

Use integration by parts: let \( u = \ln(x) \) and \( dv = dx \), which gives \( du = \frac{1}{x} dx \) and \( v = x \). The formula \( \int u \, dv = uv - \int v \, du \) is used: thus, the integration of \( \ln(x) \) becomes \( x\ln(x) - \int x \frac{1}{x} \ dx = x\ln(x) - \int \ dx = x\ln(x) - x \). Evaluate this from 1 to 2.

Step 4: Integrate \( x \)

The antiderivative of \( x \) is \( \frac{x^2}{2} \). Evaluate this from 1 to 2.

Step 5: Calculate the Integrals

Evaluate \( x \ln(x) - x \) from 1 to 2: \([2 \ln(2) - 2] - [1 \ln(1) - 1]\), which simplifies to \( 2 \ln(2) - 2 + 1 \) since \( \ln(1) = 0 \). Also, evaluate \( \frac{x^2}{2} \) from 1 to 2, which gives \( \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \).

Step 6: Combine Results and Simplify

Sum the results from both integrals: \([2(2 \ln(2) - 1)] + \frac{3}{2} \). This yields \( 4 \ln(2) - 2 + \frac{3}{2} \). Combine and simplify: \( 4 \ln(2) - \frac{1}{2} \).

Key Concepts

Logarithmic Integration

Logarithmic integration involves integrating a function that contains a logarithm. These types of integrals often require breaking down the logarithmic expression into simpler components. In our problem, we start with the expression \( \ln(x^2 e^x) \). To simplify, we use the property \( \ln(ab) = \ln(a) + \ln(b) \). By applying this rule:

• We express \( \ln(x^2 e^x) \) as \( \ln(x^2) + \ln(e^x) \).
• Knowing \( \ln(e^x) = x \), the expression simplifies further to \( 2\ln(x) + x \).

With the expression simplified, the integration becomes more straightforward. Logarithmic integration often sets the stage for more complex techniques such as integration by parts. It's a great tool to reinforce understanding of how logarithmic properties can simplify calculus problems.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It is useful for integrals where straightforward integration is too complex. The formula is:\[ \int u \, dv = uv - \int v \, du \]For integrating \( \ln(x) \), we choose:

• \( u = \ln(x) \), hence \( du = \frac{1}{x} \, dx \)
• \( dv = dx \), which leads to \( v = x \)

Applying integration by parts gives us:

• \( uv = x \ln(x) \)
• The integral \( \int v \, du = \int x \frac{1}{x} \, dx = \int dx = x \)

Thus, the integrated form is \( x \ln(x) - x \), evaluated from the limits 1 to 2. This evaluation and simplification yield a specific numerical result which contributes to solving the entire definite integral. Integration by parts is especially helpful when dealing with products that involve logarithmic functions.

Calculus Techniques

Calculus offers a variety of techniques for computing definite integrals, each useful for different types of functions. In this exercise, once we have used logarithmic properties to simplify the integrand, we split the problem into simpler parts.First, the integration of the function \( x \) is straightforward as it is simply \( \frac{x^2}{2} \). Evaluating this from 1 to 2 gives \( \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \).Calculus often involves summing the results of different methods to handle more complex integrals. The technique of simplifying a compound expression with more than one distinct function:

• Ensures that we handle each part of the integral separately.
• Simplifies the final calculations, making the integration process more manageable.

By combining techniques like integration by parts with direct integration, we not only solve the problem but also reinforce the connection between different calculus operations.