Problem 16 - Applied Calculus

Question

Use integration by parts twice to evaluate the integral. $$ \int(\ln t)^{2} d t $$

Step-by-Step Solution

Answer

The integral is \( t(\ln t)^2 - 2t \ln t + 2t + C \).

Step 1: Identify the Components for Integration by Parts

The integration by parts formula states: \[ \int u \, dv = uv - \int v \, du \]For the integral \( \int (\ln t)^2 \, dt \), let \( u = (\ln t)^2 \) and \( dv = dt \).Calculate \( du \) and \( v \) using derivatives and integrals respectively. \( du = 2 \ln t \, \frac{1}{t} \, dt = \frac{2 \ln t}{t} \, dt \) and \( v = t \).

Step 2: Apply Integration by Parts - First Time

Apply the integration by parts formula:\[ \int (\ln t)^2 \, dt = (\ln t)^2 t - \int t \frac{2 \ln t}{t} \, dt = t( \ln t)^2 - 2 \int \ln t \, dt \].Here, the problem simplifies to evaluating \( 2 \int \ln t \, dt \) next.

Step 3: Choose Components for Second Integration by Parts

To compute \( \int \ln t \, dt \), choose \( u = \ln t \) and \( dv = dt \).Calculate \( du \) and \( v \) again, yielding \( du = \frac{1}{t} \, dt \) and \( v = t \).

Step 4: Apply Integration by Parts - Second Time

Using integration by parts formula for \( \int \ln t \, dt \):\[ \int \ln t \, dt = t \ln t - \int t \times \frac{1}{t} \, dt = t\ln t - \int 1 \, dt = t\ln t - t + C \].Substitute this result back into the expression found in Step 2: \( t(\ln t)^2 - 2(t\ln t - t + C) \).

Step 5: Simplify the Expression

Simplify the expression obtained:\[ t(\ln t)^2 - 2(t \ln t - t + C) = t(\ln t)^2 - 2t \ln t + 2t - 2C \].Since \(-2C\) is a constant term, adjust the constant of integration to \(C'\) if needed.

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus. They provide the area under the curve of a function within specified limits. Unlike indefinite integrals, which represent a family of functions, definite integrals have numerical bounds that yield a specific number. This concept is essential for applications involving total quantities, such as distances, areas, and volumes.

The integral of a function from point 'a' to point 'b' is represented as \( \int_{a}^{b} f(x) \, dx \). To solve such integrals, especially when dealing with complex functions like logarithmic ones, we use various techniques including substitution, partial fractions, and integration by parts.

When integrating a logarithmic function like \((\ln t)^{2}\), we're not dealing with a regular definite integral since no bounds are provided. However, understanding how definite integrals work helps set the stage for these more complex calculations.

Logarithmic Functions

Logarithmic functions are crucial in mathematics due to their unique properties. Represented as \(\ln(x)\), the natural logarithm is the inverse operation of exponentiation with base \(e\). This function grows slowly compared to polynomial and exponential counterparts, which has important implications in calculus.

When dealing with integrals involving logarithmic functions, the challenge often lies in manipulating these functions using rules such as the logarithm derivative \(\frac{d}{dx} \ln(x) = \frac{1}{x}\). These derivatives facilitate the integration process, especially under the integration by parts technique.

In our case, integrating \((\ln t)^{2}\) required a clever choice of \(u\) and \(dv\) in the process of integration by parts to reduce the problem into simpler components.

Calculus Techniques

Calculus techniques enable the solving of complex mathematical problems like integration and differentiation. Integration by parts is one such technique, particularly used when the integrand is the product of two functions. It is represented as:

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

In our example, integration by parts was applied twice to evaluate \(\int (\ln t)^{2} \, dt\).

Steps involved:

• Selecting \(u\) and \(dv\) correctly for simplifying calculations.
• Deriving \(du\) and \(v\) through differentiation and integration.
• Applying the formula methodically to break down the integral step by step.

A deep understanding of these steps and calculus rules is vital in mastering integration techniques, allowing for the solution of a variety of integral problems.

Mathematical Solutions

Mathematical solutions involve consistent methods and logical derivations to solve problems effectively. In this exercise, sequential application of integration by parts provided the means to handle the complexity of the original problem.

Key Highlights:

• Breaking down the integral using by parts into simpler components.
• Substituting back the evaluation of simpler integrals.
• Simplifying expressions to reach the final answer.

The solution's clarity is enhanced by the step-by-step manner, demonstrating how complex expressions can be tackled iteratively. A well-thought-out approach not only solves the problem but also strengthens the intuition for handling different calculus problems. Applying such mathematical solutions builds a solid foundation for solving future integration challenges.