Problem 12 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int x^{2} \ln x d x $$

Step-by-Step Solution

Answer

\( \int x^2 \ln x \, dx = \frac{x^3}{9} (3 \ln x - 1) + C \), where \( C \) is the constant of integration.

Step 1: Identify Parts for Integration by Parts

To solve the integral \( \int x^{2} \ln x \, dx \) using integration by parts, we need to choose parts for \( u \) and \( dv \). Integration by parts follows the formula \( \int u \, dv = uv - \int v \, du \). A common strategy is to let \( u \) be a part that simplifies when differentiated. Here, choose \( u = \ln x \) and thus \( dv = x^{2} \, dx \).

Step 2: Differentiate and Integrate

With \( u = \ln x \), differentiate it to get \( du = \frac{1}{x} \, dx \). Now, integrate \( dv = x^{2} \, dx \) to find \( v \). The integral of \( x^{2} \) is \( \frac{x^{3}}{3} \), so \( v = \frac{x^{3}}{3} \).

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \[\int x^{2} \ln x \, dx = \left[ \ln x \cdot \frac{x^{3}}{3} \right] - \int \frac{x^{3}}{3} \cdot \frac{1}{x} \, dx\]Simplifying the integral on the right gives: \[= \frac{x^{3}}{3} \ln x - \frac{1}{3} \int x^{2} \, dx\]

Step 4: Evaluate the Resulting Integral

Calculate \( \int x^{2} \, dx \). This integral is straightforward: \[\int x^{2} \, dx = \frac{x^{3}}{3}\]Substitute this back into the expression: \[\int x^{2} \ln x \, dx = \frac{x^{3}}{3} \ln x - \frac{1}{3} \cdot \frac{x^{3}}{3} + C\]After simplifying, this results in:\[= \frac{x^{3}}{3} \ln x - \frac{x^{3}}{9} + C\]

Step 5: Simplify the Expression

Combine the terms to express the solution neatly:\[\int x^{2} \ln x \, dx = \frac{x^{3}}{9} (3 \ln x - 1) + C\]where \( C \) is the constant of integration.

Key Concepts

Integral Calculus

Integral calculus is a branch of calculus focused on finding the antiderivative, also known as the integral, of functions. It is the reverse process of differentiation and is crucial for solving problems involving areas, volumes, central points, and many other practical applications. The process involves finding a function whose derivative matches the original function.

• **Definite Integrals** provide a number representing the area under the curve of a function over a specified interval.
• **Indefinite Integrals**, like our problem involving \( \int x^{2} \ln x \, dx \), result in a family of functions plus a constant \( C \).

For our exercise, we used **integration by parts**, a method that simplifies complex integrals by transforming them into simpler parts. This technique is based on the product rule for differentiation, but turned around for integration. By choosing appropriate parts for \( u \) and \( dv \), we make the integration more manageable.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions and are generally written as \( \ln x \) for the natural logarithm, using the base \( e \). These functions have useful properties that make them valuable in calculus and other fields of mathematics. They grow very slowly compared to exponential functions, making them ideal for solving amplifying functions without bursting computational limits.

• **Key Property**: The derivative of \( \ln x \) is \( \frac{1}{x} \).
• In integration problems, the natural logarithm often simplifies the differentiation process, as seen in our problem where we chose \( u = \ln x \), simplifying to \( du = \frac{1}{x} \ dx \).

By understanding these properties, we can effectively apply logarithmic functions in integration processes, especially when they simplify expressions significantly, as demonstrated in integration by parts.

Polynomial Integration

Polynomial integration is one of the basic types of integration, dealing with powers of \( x \). The general rule is to increase the power by one and divide by the new power, a process known as **power rule for integrals**.

• The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), assuming \( n eq -1 \).
• In our exercise, integrating \( x^2 \) gives us \( \frac{x^3}{3} \), a straightforward application of this rule.

Combining polynomial integration with techniques such as integration by parts helps tackle more complicated integrals, like those involving products of polynomial terms and other functions. Mastery of polynomial integration opens up various mathematical and real-world applications, from engineering to physics, where such calculations frequently occur.