Problem 67 - Calculus Early Transcendentals

Question

Use integration by parts to evaluate the following integrals. $$\int_{0}^{1} x \ln x d x$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral \(\int_{0}^{1} x \ln x d x\) using integration by parts. Answer: \(-\frac{1}{4}\)

Step 1: Identify u and dv

We will use integration by parts with \(u = \ln x\) and \(dv = x dx\). We choose \(\ln x\) as our \(u\) function because it will simplify when we take the derivative. On the other hand, when we integrate \(x\), which is \(dv\), it will remain relatively simple as well.

Step 2: Compute du and v

Now, we will calculate du and v by differentiating u and integrating dv: $$du = \frac{1}{x}dx \quad \text{and} \quad v = \int x dx = \frac{1}{2}x^2$$

Step 3: Apply integration by parts formula

We will now apply the integration by parts formula: \(\int u dv = uv - \int v du\). Substituting our functions into the formula, we get: $$\int_{0}^{1} x \ln x d x = \left[\frac{1}{2}x^2\ln x\right]_{0}^{1} - \int_{0}^{1} \frac{1}{2}x^2 \frac{1}{x} dx$$

Step 4: Simplify and solve the integral

Simplify the integral and solve it: $$\int_{0}^{1} x \ln x d x = \left[\frac{1}{2}x^2\ln x\right]_{0}^{1} - \int_{0}^{1} \frac{1}{2}x dx = \left[\frac{1}{2}x^2\ln x\right]_{0}^{1} - \left[\frac{1}{4}x^2\right]_{0}^{1}$$

Step 5: Evaluate the definite integral

Evaluate the definite integral using the computed bounds: $$\left[\frac{1}{2}x^2\ln x - \frac{1}{4}x^2\right]_{0}^{1} = \left(\frac{1}{2}(1)(0) - \frac{1}{4}(1)\right) - \left(\frac{1}{2}(0)(-\infty) - \frac{1}{4}(0)\right) = -\frac{1}{4}$$ So, the value of the definite integral is: $$\int_{0}^{1} x \ln x d x = -\frac{1}{4}$$

Key Concepts

Definite Integral

The concept of a definite integral is central to understanding areas under curves and accumulations. When evaluating a definite integral, we perform integration across a specific interval, often denoted as \( [a, b] \). This means we find the total area under the curve of a function from \( x = a \) to \( x = b \), where \( a \) and \( b \) are the limits of integration.

• Definite integrals result in a number, unlike indefinite integrals, which result in a function.
• A definite integral is usually solved by finding the antiderivative first and then applying it to the upper and lower limits.
• The Fundamental Theorem of Calculus connects the process of differentiation and integration, and helps evaluate definite integrals efficiently.

In the provided exercise, the integral \( \int_{0}^{1} x \ln x \, dx \) is solved with bounds 0 and 1, showing the application of a definite integral. Substituting the limits into the antiderivative lets you find the area under the curve from 0 to 1, providing a concise final value.

Logarithmic Functions

Logarithmic functions are inverse operations to exponential functions, characterized by the notation \( \ln x \) when using natural logs with base \( e \). Understanding these functions is crucial, especially when dealing with integration problems, as they have properties that can simplify calculations.

• The derivative of \( \ln x \) is \( \frac{1}{x} \), making it a natural choice for the \( u \) term in integration by parts, as seen in the exercise.
• Logarithmic functions can transform multiplicative relationships into additive ones, which simplifies integration in these situations.
• In contexts involving integration, \( \ln x \) often appears with the need to apply integration by parts, especially when a polynomial function combines with the logarithmic function.

This understanding of logarithmic functions is crucial, as it influenced how \( u = \ln x \) was chosen in the integration by parts step. The logarithm's properties directly impacted the simplification and solution of the integral.

Differentiation and Integration

Differentiation and integration are fundamental calculus processes. They are often considered opposites: differentiation finds the rate of change, and integration aggregates these changes over an interval. These tools are indispensable in mathematical analysis and applied mathematics.

• Differentiation is used to compute \( du \), or the derivative of \( u \), in the integration by parts formula. For \( u = \ln x \), the derivative is \( \frac{1}{x} \).
• Integration helps calculate \( v \), the antiderivative of \( dv \). In the example, this involves integrating \( x \) to get \( \frac{1}{2}x^2 \).
• The integration by parts formula, \( \int u \, dv = uv - \int v \, du \), illustrates a blend of these concepts, allowing transformation of complex integrals into more manageable forms.

In our specific problem, the mastery of these operations allowed the solution to proceed effectively through each step of integration by parts, leading to the successful evaluation of the definite integral.