Problem 18 - Calculus and its applications

Question

Evaluate using integration by parts or substitution. (Assume \(u>0\) in \(\ln\) u. Check by differentiating. $$ \int(x-2) \ln x d x $$

Step-by-Step Solution

Answer

Using integration by parts and simplifying the expression gives the antiderivative as: \[ \left( \ln x \right) \left( \frac{x^2}{2} - 2x \right) - \left( \frac{x^2}{4} - 2x \right) + C \].

Step 1: Identify the Method

Integration by parts is suitable here because the integrand is a product of two functions. Identify the parts to use in the formula.

Step 2: Assign Functions

Let \( u = \ln x \) and \( dv = (x-2) \, dx \). Differentiate and integrate these parts: \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} - 2x \).

Step 3: Apply Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substituting, we have: \( \int (x-2) \ln x \, dx = \left( \ln x \right) \left( \frac{x^2}{2} - 2x \right) - \int \left( \frac{x^2}{2} - 2x \right) \frac{1}{x} \, dx \).

Step 4: Simplify and Integrate

Simplify the integral \( \int \left( \frac{x}{2} - 2 \right) \, dx \), which breaks down to \( \frac{1}{2} \int x \, dx - 2 \int 1 \times dx \).

Step 5: Perform Integration

Calculate the integrals: \( \frac{1}{2} \int x \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4} \) and \( - 2 \int dx = -2x \).

Step 6: Compile the Solution

Putting it all together, the antiderivative is \( \left( \ln x \right) \left( \frac{x^2}{2} - 2x \right) - \left( \frac{x^2}{4} - 2x \right) + C \).

Step 7: Verify by Differentiation

Differentiate the result to ensure it returns to the original integrand \((x-2) \ln x\). Confirm each derivative component matches, verifying the solution.

Key Concepts

Understanding Antiderivatives

An antiderivative, also known as the indefinite integral, is a function that reverses the process of differentiation. When you find an antiderivative of a function, you're essentially looking for a new function whose derivative equals the original function. This concept is crucial in calculus, as it allows us to determine the accumulation of quantities. In your exercise, the antiderivative is determined using the integration by parts technique, which is designed to tackle integrals of products of functions.
In this context, finding the antiderivative of \( (x-2) \ln x \) involves establishing which parts of the function to differentiate and which to integrate, leading to the final integral that represents the original function plus a constant \( C \) to account for any constant differences across antiderivatives.

Mastering Integration Techniques

Integration techniques are a set of methods used to find antiderivatives of complex functions. These techniques include substitution, integration by parts, partial fractions, and more. Integration by parts is particularly powerful when dealing with products of functions, as it transforms one integral into another potentially easier one to evaluate.

• To apply integration by parts, identify two parts of the integrand: one to differentiate and another to integrate.
• Use the formula \( \int u \, dv = uv - \int v \, du \).

This formula stems from the product rule for differentiation and is suitable for our example \( \int(x-2) \ln x \, dx \). By carefully selecting \( u \) and \( dv \), where \( u = \ln x \) and \( dv = (x-2) \, dx \), we transform the task into a simpler integration, computed step by step.

Verification by Differentiation

After finding the antiderivative, it's crucial to verify the result by differentiation to ensure accuracy. Differentiation is the reverse operation of integration, and confirming the differentiation leads back to the original integrand reassures us that the antiderivative is correct.
In your problem, once you calculate the antiderivative of \( \int(x-2) \ln x \, dx \), apply the derivative rules:

• Differentiating \( \left( \ln x \right) \left( \frac{x^2}{2} - 2x \right) - \left( \frac{x^2}{4} - 2x \right) + C \) should give you back \( (x-2) \ln x \).
• Confirm each term's derivative to match the components of the original function.

This step acts like a check and balance process to certify your integration technique, reinforcing the solution's accuracy.