Problem 11 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify Parts for Integration by Parts

In integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). For the integral \( \int x \ln x \, dx \), let \( u = \ln x \) and \( dv = x \, dx \). This is because \( \ln x \) becomes simpler when differentiated rather than integrated.

Step 2: Differentiate and Integrate Chosen Parts

Differentiate \( u = \ln x \) to get \( du = \frac{1}{x} \, dx \). Integrate \( dv = x \, dx \) to obtain \( v = \frac{x^2}{2} \).

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \[ \int x \ln x \, dx = \ln x \cdot \frac{x^2}{2} \Big| - \int \left( \frac{x^2}{2} \cdot \frac{1}{x} \right) \, dx \] Simplify this expression to start solving the integral.

Step 4: Simplify and Solve Remaining Integral

Simplify the remaining integral: \[ = \frac{x^2}{2} \ln x - \int \frac{x^2}{2x} \, dx \] This further simplifies to \[ = \frac{x^2}{2} \ln x - \int \frac{x}{2} \, dx \] Calculate the integral: \[ = \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx \] \[ = \frac{x^2}{2} \ln x - \frac{1}{2} \cdot \frac{x^2}{2} + C \] \[ = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \]

Step 5: Final Solution

The integral \( \int x \ln x \, dx \) evaluates to \( \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \).

Key Concepts

Integral Calculus

Integral calculus is a fundamental part of mathematics that deals with integrals, which are essentially the reverse operation of differentiation. Two primary types of integrals exist: definite and indefinite.

• **Definite Integrals** involve the calculation of the area under a curve, bounded by two limits.
• **Indefinite Integrals** represent a broader set of functions whose derivatives give the original function.

In indefinite integrals, integration by parts is a powerful technique used for integrating products of functions. It transforms the product into a simpler integral that's more straightforward to calculate. Choosing the right parts for differentiation and integration is crucial for simplifying calculations, especially in cases where standard methods might complicate the process.

Differentiation

Differentiation is the process of finding the derivative of a function, providing the rate at which one quantity changes with respect to another. It's the backbone for functions' analysis in calculus due to its ability to reveal all manner of vital information about the graph.

In the context of integration by parts, we use differentiation to simplify one part of the product in the integral. For example, in the integral \( \int x \ln x \, dx \), choosing \( u = \ln x \) makes sense because \( \ln x \), when differentiated, becomes \( \frac{1}{x} \). This differentiation transforms a complex function into a simpler one, facilitating a simpler integration process.

Integration Techniques

Integration techniques are varied, with integration by parts being one of the more elegant methods for dealing with certain products of functions. The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). Here’s how this technique helps:

• **Choose \( u \) and \( dv \) judiciously**: Selecting \( u = \ln x \) and \( dv = x \, dx \) simplifies the problem because differentiating \( \ln x \) and integrating \( x \) are straightforward.
• **Transform and Simplify**: Once you differentiate \( u \) and integrate \( dv \), substitute back into the integration by parts formula. Simplifying the resultant expression often leads to an integral that is easier to solve.
• **Work towards simpler integrals**: The goal is to transform complex integrals into simpler forms that can be handled with basic integration techniques.

Integration by parts is particularly useful when standard integration techniques alone are insufficient. Understanding how to manipulate functions this way broadens the toolkit for solving integral calculus problems.