Problem 42 - Calculus Single Variable

Question

Simplify the integrand before integrating by parts. $$ \int x \ln \left(x^{5}\right) d x $$

Step-by-Step Solution

Answer

The evaluated integral of the expression is: \( \frac{5x^2}{2} \ln(x) - \frac{5x^2}{4} + C \).

Step 1: Simplify the Expression Inside the Integral

Start by simplifying the expression inside the integral: \( \int x \ln(x^5) \, dx \). We know that \( \ln(x^5) = 5 \ln(x) \) because of the logarithmic identity \( \ln(a^b) = b \ln(a) \). Substitute this back into the integral: \( \int x \cdot 5 \ln(x) \, dx = 5 \int x \ln(x) \, dx \).

Step 2: Identify Functions for Integration by Parts

Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \). Choose \( u = \ln(x) \) and \( dv = x \, dx \). Therefore, \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \) after integrating \( dv \).

Step 3: Apply the Integration by Parts Formula

Now apply the integration by parts formula: \( \int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \). Simplify the remaining integral: \( \int \frac{x^2}{2x} \, dx = \int \frac{x}{2} \, dx \).

Step 4: Solve the Remaining Integral

Calculate the integral \( \int \frac{x}{2} \, dx \). This is \( \frac{1}{2} \int x \, dx = \frac{1}{2} \cdot \frac{x^2}{2} + C = \frac{x^2}{4} + C \).

Step 5: Combine Results

Substitute back the result from Step 4 into the expression from Step 3: \( \frac{x^2}{2} \ln(x) - \left( \frac{x^2}{4} + C \right) = \frac{x^2}{2} \ln(x) - \frac{x^2}{4} - C \).

Step 6: Evaluate the Final Answer

Utilize the factor of 5 from Step 1: \( 5 \cdot \left( \frac{x^2}{2} \ln(x) - \frac{x^2}{4} \right) = \frac{5x^2}{2} \ln(x) - \frac{5x^2}{4} + C' \). This is the simplified and integrated result of the original expression.

Key Concepts

Logarithmic Identity

When simplifying expressions that involve logarithms, knowing some key properties can be extremely helpful. In this case, we use the logarithmic identity:

• \( \ln(a^b) = b \ln(a) \)

This identity tells us that when you have a logarithm of a number raised to a power, you can pull the exponent in front of the logarithm.
In the given exercise, the expression inside the integral is \( \ln(x^5) \). Using the identity, this simplifies to \( 5 \ln(x) \).
This simplification is crucial as it reduces the complexity of the integrand, making it easier to manage in subsequent steps of the problem-solving process. Every time you simplify an expression, you’re setting yourself up for easier calculations later on!

Integrals in Calculus

Understanding integrals is a core part of calculus, and they're used to find areas under curves, among other applications. An integral of a function can be indefinite or definite:

• **Indefinite Integral**: Represents a family of functions and includes a constant of integration \( C \).
• **Definite Integral**: Calculates a specific area and has numerical limits of integration, resulting in a number, not a function.

In this exercise, we work with an indefinite integral: \( \int x \ln(x^5) \, dx \).
The task is to find a function whose derivative gives us the original integrand. Integrals can often be complex, requiring techniques like integration by parts to solve. This technique uses the formula: \[ \int u \, dv = uv - \int v \, du \]Carefully choosing \( u \) and \( dv \) simplifies further problem-solving, and it's essential for tackling complex integrals like the one in the exercise.

Definite Integrals

Though the current problem deals with an indefinite integral, understanding definite integrals can give insights into practical applications. A definite integral computes the accumulated value of a function over an interval \([a, b]\), providing a precise total area under the curve between these bounds.
The formula for a definite integral is given as:\[ \int_a^b f(x) \, dx \]This results in a number, representing the net area. To solve it, you typically:

• Find the indefinite integral, adding a constant of integration \( C \).
• Evaluate this antiderivative at the upper and lower bounds \( b \) and \( a \).
• Subtract the two results to find the area.

Understanding definite integrals expands the scope of integrals beyond theoretical calculus and into practical applications, such as computing the total change in a quantity or the accumulated value over time.