Problem 11 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int x \ln x d x $$

Step-by-Step Solution

Answer

\( \int x \ln x \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \).

Step 1: Choose Functions for Integration by Parts

For integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). Here, we choose \( u = \ln x \) and \( dv = x \, dx \). This implies \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \).

Step 2: Apply the Integration by Parts Formula

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

Step 3: Simplify the Remaining Integral

Simplify the integral: \( \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \int \frac{x}{2} \, dx \). This further simplifies to \( \int \frac{1}{2} x \, dx \).

Step 4: Compute the Simplified Integral

Compute the integral \( \int \frac{1}{2} x \, dx \). The integral of \( \frac{1}{2} x \) with respect to \( x \) is \( \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4} \).

Step 5: Complete the Solution

Substitute back the result from Step 4 into Step 2: \[ \left( \ln x \right) \left( \frac{x^2}{2} \right) - \frac{x^2}{4} + C. \] Simplifying gives us \( \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \).

Key Concepts

Integral Calculus

Integral calculus is a crucial part of mathematical analysis. It allows us to find quantities like areas, volumes, and accumulation of other quantities by integrating functions. In this process, the integral acts as an "accumulator," summing up infinitesimally small factors across a range of values.

In the exercise above, we used a method called integration by parts to solve \( \int x \ln x \, dx \). This is particularly useful when dealing with integrals of products of functions, like a polynomial and a logarithmic function, as seen here. Integration by parts is derived from the product rule of differentiation. The formula is:

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

We choose one function to differentiate (\( u \)) and another to integrate (\( dv \)) to simplify the integral. This method transforms the original problem into simpler, more manageable pieces by shifting differentiation and integration back and forth between chosen functions.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions. They are often encountered in problems of integral calculus, especially when using methods like integration by parts.

The natural logarithm, \( \ln x \), is the logarithm to the base \( e \) (where \( e \) is approximately 2.71828). This function has specific properties that make it convenient for differentiation and integration. For example, the derivative of \( \ln x \) is \( \frac{1}{x} \), a simple expression that plays well into integration formulas.

• \( \ln(ab) = \ln a + \ln b \)
• \( \ln(a^b) = b \ln a \)

These properties and the derivative make \( \ln x \) a common choice for \( u \) in integration by parts, as its differentiation simplifies the integral process. In the given problem, picking \( u = \ln x \) and \( dv = x \, dx \) helped transition the integral into a more simple form.

Antiderivatives

Antiderivatives are the reverse process of taking derivatives, commonly known as finding the original function that was differentiated to get the given function. This is central to integral calculus and essential in solving problems involving integration.

To solve \( \int x \ln x \, dx \), after applying integration by parts, we derived an expression that required us to find the antiderivative of \( \frac{1}{2}x \), which is a straightforward task:

• \( \int \frac{1}{2}x \, dx = \frac{1}{2} \times \frac{x^2}{2} = \frac{x^2}{4} \)

The antiderivative process can be thought of as finding the area under a curve, represented by the original function. In this exercise, computing the antiderivative enabled us to determine the evaluation of the integral fully and reach our final answer: \( \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \), with \( C \) representing the constant of integration.