Problem 12 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int x^{2} \ln x d x $$

Step-by-Step Solution

Answer

The integral of \( x^2 \ln x \) is \( \frac{x^3}{3} \ln x - \frac{x^3}{9} + C \).

Step 1: Identify Functions for Integration by Parts

For integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). We need to choose \( u \) and \( dv \). Let \( u = \ln x \) since its derivative \( du/dx = 1/x \) simplifies, and let \( dv = x^2 \, dx \), making \( v \) the antiderivative.

Step 2: Differentiate and Integrate

Differentiate \( u \) to get \( du = \frac{1}{x} \, dx \). Integrate \( dv \) to obtain \( v = \frac{x^3}{3} \). This gives us the necessary components for integration by parts.

Step 3: Apply the Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \( \int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx \). Simplify the equation to: \( \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \int \frac{x^2}{3} \, dx \).

Step 4: Evaluate the Remaining Integral

Evaluate \( \int \frac{x^2}{3} \, dx \). The antiderivative of \( \frac{x^2}{3} \) is \( \frac{x^3}{9} \). Thus, the integral becomes \( \frac{x^3}{3} \ln x - \frac{x^3}{9} + C \), where \( C \) is the constant of integration.

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of calculus focused on integrals and antiderivatives. It is concerned with the accumulation of quantities and the area under curves. In integration by parts, one tool of integral calculus, we break down complex integrals into simpler parts to solve them effectively.

The main goal of integral calculus is finding an antiderivative or solving an integral. In the given exercise, employing integration by parts facilitates this process. Integration by parts is based on the product rule for differentiation and is particularly useful when integrating the product of two functions, such as in \(\int x^{2} \ln x \, dx\). By carefully selecting our functions for \(u\) and \(dv\), we create a simpler problem to tackle.

This technique transforms the original integral into an expression that includes both a product of functions and another, hopefully simpler, integral. The formula for integration by parts is:

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

Understanding how to apply this formula is crucial when dealing with integral calculus problems that involve multiplicative combinations.

Antiderivative

An antiderivative is a function whose derivative is the original function. In simple terms, it is the reverse process of differentiation. Finding the antiderivative is a vital skill in solving integral calculus problems, as it allows us to evaluate definite and indefinite integrals.

The exercise we are analyzing requires us to find the antiderivative of the function \(x^2 \ln x\). To achieve this, we use integration by parts, which strategically selects parts of the function to differentiate and integrate. In this case:

• Choose \(u = \ln x\) to simplify differentiation.
• Select \(dv = x^2 \, dx\), so \(v\) becomes the antiderivative of \(dv\).

After differentiating and integrating, substitute back into the integration by parts formula to isolate and evaluate the remaining integral. This process gives us the antiderivative for the expression within the problem.

Antiderivatives play a critical role in calculus as they allow us to solve for unknowns in integral expressions, leading us to evaluate the total accumulation or area represented by the original function.

Logarithmic Integration

Logarithmic integration involves integrating functions that include logarithms, such as \(\ln x\). These types of integrals often appear challenging due to the properties of logarithmic functions and their derivatives. However, using appropriate techniques like integration by parts can simplify the process significantly.

In the exercise, \(x^2 \ln x\) features a logarithmic function within the integral. We achieve logarithmic integration by first choosing \(u = \ln x\) because its derivative \(du = \frac{1}{x} \, dx\) simplifies nicely. The logarithms' properties make them a suitable choice for \(u\) when using integration by parts since their derivatives lead to simple to work with expressions.

By integrating the non-logarithmic part, \(dv = x^2 \, dx\), we get \(v = \frac{x^3}{3}\), which is then used within the formula for integration by parts:

• \(\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx\)

Applying this strategic approach helps to solve the given integral and demonstrates the effectiveness of combining logarithmic functions with other polynomial expressions in integration by parts.