Problem 17 - Calculus Single Variable

Question

Integrate by parts to evaluate the given definite integral. $$ \int_{1}^{e} x \ln (x) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{e^2 + 1}{4} \).

Step 1: Understanding the Method

Integration by parts is useful for integrals of products of functions. It is based on the formula: \( \int u \, dv = uv - \int v \, du \), where \( u \) and \( dv \) are parts of the integral to be chosen strategically.

Step 2: Choosing Functions for Integration by Parts

In the integral \( \int_{1}^{e} x \ln(x) \, dx \), let \( u = \ln(x) \) and \( dv = x \, dx \). The derivative \( du = \frac{1}{x} \, dx \), and the integral of \( dv \) gives \( v = \frac{x^2}{2} \).

Step 3: Substituting into Integration by Parts Formula

By applying the integration by parts formula, we get: \[ \int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) \bigg|_1^e - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \].

Step 4: Simplifying the Integral

The integral becomes: \[ \frac{x^2}{2} \ln(x) \bigg|_1^e - \int \frac{x}{2} \, dx \], so let's evaluate \( \int \frac{x}{2} \, dx \), which simplifies to \( \frac{x^2}{4} \).

Step 5: Evaluating the Antiderivative

Substitute back into the expression to get: \[ \frac{x^2}{2} \ln(x) \bigg|_1^e - \frac{x^2}{4} \bigg|_1^e \] and evaluate at the bounds \( x = e \) and \( x = 1 \).

Step 6: Calculating Boundary Values

Calculate each term: \( \left[ \frac{e^2}{2} \ln(e) \right] - \left[ \frac{1}{2} \ln(1) \right] = \frac{e^2}{2} \cdot 1 - 0 = \frac{e^2}{2} \). Evaluate \( \frac{x^2}{4} \bigg|_1^e \): \( \left[ \frac{e^2}{4} - \frac{1}{4} \right] = \frac{e^2-1}{4} \).

Step 7: Final Calculation

The integral evaluates to: \[ \left( \frac{e^2}{2} \right) - \left( \frac{e^2 - 1}{4} \right) = \frac{e^2}{2} - \frac{e^2}{4} + \frac{1}{4} = \frac{e^2}{4} + \frac{1}{4} \].

Step 8: Simplifying the Result

Combine the terms: \( \frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4} \). This is the final result of the definite integral.

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus. They assign a numerical value to the function over a specific interval. For example, in the exercise, we need to find the integral \[ \int_{1}^{e} x \ln(x) \, dx \] which represents the area under the curve from \( x = 1 \) to \( x = e \). Integrating by parts is a common method for evaluating integrals that involve products of functions, particularly when straightforward integration is complex.

When you compute a definite integral, the first step is to find its antiderivative. After determining the antiderivative, you evaluate it at the upper and lower limits and subtract the results to find the exact area or accumulated value over that range. This process ties into the Fundamental Theorem of Calculus, which connects differentiation and integration.

• The limits of integration in our exercise are \( 1 \) and \( e \).
• The definite integral results in a specific numerical value rather than an expression with a variable.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is approximately equal to 2.718. It is a continuous, monotonically increasing function often appearing in calculus, especially within integrals and derivatives.

In our exercise, \( \ln(x) \) is chosen as the function \( u \) when using the integration by parts formula. This is because its derivative, \( \frac{1}{x} \), is simpler and more manageable than differentiating \( x \).

Understanding properties of \( \ln(x) \) such as \( \ln(e) = 1 \) and \( \ln(1) = 0 \) is crucial. These properties simplify boundaries in definite integrals and allow us to efficiently solve problems involving natural logarithms.

• \( \ln(e) = 1 \)
• \( \ln(1) = 0 \)
• Key to integration strategies, where choosing \( u = \ln(x) \) simplifies calculations.

Antiderivatives

An antiderivative, or indefinite integral, is a function whose derivative is the original function. The goal of finding an antiderivative is to reverse the process of differentiation. This is especially important when solving definite integrals, as they rely on evaluating antiderivatives at specific limits.

In this exercise, the function \( x \) is paired with \( \ln(x) \) to apply integration by parts. After choosing \( u = \ln(x) \) and \( dv = x \, dx \), you calculate \( v \) as an antiderivative of \( dv \). Specifically, \( v = \frac{x^2}{2} \).

The next step involves using the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \]. In our case, substituting our expressions leads to simpler integrals that can be evaluated at the boundaries.

Key points when dealing with antiderivatives include:

• The antiderivative of \( x \) is \( \frac{x^2}{2} \).
• You always subtract the evaluated bounds to find the definite integral's value.
• Integration by parts requires strategic selection for \( u \) and \( dv \) to simplify integration tasks.