Problem 22 - Calculus and its applications

Question

Evaluate using integration by parts or substitution. (Assume \(u>0\) in \(\ln\) u. Check by differentiating. $$ \int x^{2} \ln (5 x) d x $$

Step-by-Step Solution

Answer

The integral is \( \frac{x^3}{3} \ln(5x) - \frac{x^3}{9} + C \).

Step 1: Choosing the Technique

Since we're asked to evaluate \( \int x^2 \ln(5x) \, dx \), we'll use integration by parts. Integration by parts is helpful when you have a product of functions, one of which is easily differentiable and the other easily integrable.

Step 2: Identify Parts for Integration by Parts

In the integration by parts formula, \( \int u \, dv = uv - \int v \, du \). We will choose \( u = \ln(5x) \) and \( dv = x^2 \, dx \) because \( \ln(5x) \) is nicely differentiable and \( x^2 \, dx \) is easily integrable.

Step 3: Differentiate and Integrate Parts

Differentiate \( u = \ln(5x) \) to get \( du = \frac{1}{x} \, dx \). Integrate \( dv = x^2 \, dx \) to find \( v = \frac{x^3}{3} \).

Step 4: Apply the Integration by Parts Formula

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula:\[ \int x^2 \ln(5x) \, dx = \frac{x^3}{3} \ln(5x) - \int \left( \frac{x^3}{3} \right) \frac{1}{x} \, dx \] Simplify the integrand of the remaining integral: \( \int \frac{x^3}{3} \frac{1}{x} \, dx = \int \frac{x^2}{3} \, dx \).

Step 5: Solve the Remaining Integral

Evaluate the remaining integral \( \int \frac{x^2}{3} \, dx \). This becomes \( \frac{1}{3} \int x^2 \, dx = \frac{1}{3} \cdot \frac{x^3}{3} = \frac{x^3}{9} \).

Step 6: Write the Final Solution

Combine all parts of the integration:\[ \int x^2 \ln(5x) \, dx = \frac{x^3}{3} \ln(5x) - \frac{x^3}{9} + C \]where \( C \) is the constant of integration.

Step 7: Verify by Differentiating the Result

Differentiate the result: Find the derivative of \( \frac{x^3}{3} \ln(5x) - \frac{x^3}{9} + C \): Using the product rule on \( \frac{x^3}{3} \ln(5x) \), \( \frac{d}{dx} \left( \frac{x^3}{3} \right) = x^2 \) and \( \frac{d}{dx} \ln(5x) = \frac{1}{x} \).This confirms correctness as it gives the original integrand \( x^2 \ln(5x) \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus used to determine how a function changes as its input changes. Essentially, it provides us with the rate of change or the slope of the function at any given point. This is particularly useful when you want to understand steepness or the trends of curves.

In mathematical terms, the derivative of a function gives us the function that describes its slope.

• The notation for a derivative is generally represented by \( f'(x) \) or \( \frac{df}{dx} \), which means the derivative of \( f \) with respect to \( x \).
• To differentiate a simple function, you increase each exponent by one, then multiply by the original exponent. For example, if \( f(x) = x^2 \), then its derivative, \( f'(x) = 2x \).
• Exponential and logarithmic functions, like \( \ln(x) \), need special rules like the chain rule, where the derivative of \( \ln(5x) \) becomes \( \frac{1}{x} \).

Applying these rules, we executed differentiation in our integration by parts process, making sure that each component of our integration analysis aligns with the original function's behavior.

Integration

Integration is essentially the reverse process of differentiation. While differentiation is about finding the rate of change, integration is about finding the accumulation of quantities. Integration is useful in various applications such as calculating areas under curves, displacement from velocity, and other cumulative sums.

One of the key integral operations is evaluating integrals by breaking them into simpler parts via methods like integration by parts:

• Integration by parts is used primarily when you encounter products of functions, especially where one part is easier to integrate and the other easier to differentiate.
• The formula \( \int u \, dv = uv - \int v \, du \) guides us to choose which function to differentiate \( (u) \) and which to integrate \( (dv) \).
• For example, in our problem, \( u = \ln(5x) \) and \( dv = x^2 \, dx \) was selected because \( \ln(5x) \) is more manageable when differentiated, while \( x^2 \, dx \) is easily integrable.

Following these structured steps allows solving the integrals that, at first glance, seem rather intricate.

Definite and Indefinite Integrals

Integrals come in two main types: definite and indefinite. Understanding their differences and applications is important in solving calculus problems.

• **Indefinite Integrals**: These represent a family of functions and include the constant of integration, \( C \). When solving indefinite integrals, you are essentially uncovering all possible antiderivatives of a function. For instance, \( \int x^2 \, dx = \frac{x^3}{3} + C \).
• **Definite Integrals**: These calculate the exact value of the area under the curve between two points, typically with no constant- Just a number. They are denoted by the integral sign with limits of integration, for example, \( \int_{a}^{b} f(x) \, dx \).
• In the context of our exercise, we tackled an indefinite integral. The solution included "\( C \)" because we were interested in the general form of the antiderivative that could explain the integral's behavior over its domain.

Grasping these concepts allows one to effectively navigate between finding an accumulation function and evaluating actual numeric values from functions.