Problem 14 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int x^{2} \ln x^{2} d x $$

Step-by-Step Solution

Answer

The integral is \( \frac{x^3}{3} \ln(x^2) - \frac{2x^3}{9} + C \).

Step 1: Identify the Functions for Integration by Parts

For integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). We need to identify \( u \) and \( dv \) from the integral \( \int x^2 \ln(x^2) \, dx \). Choose \( u = \ln(x^2) \) because it simplifies upon differentiation, and \( dv = x^2 \, dx \) because it's easily integrable.

Step 2: Differentiate and Integrate

Differentiate \( u = \ln(x^2) \), which gives \( du = \frac{2}{x} \, dx \). Integrate \( dv = x^2 \, dx \) to obtain \( v = \frac{x^3}{3} \).

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( v \), \( du \), and \( dv \) into the formula \( \int u \, dv = uv - \int v \, du \). We get: \[ \int x^2 \ln(x^2) \, dx = \left( \ln(x^2) \cdot \frac{x^3}{3} \right) - \int \left( \frac{x^3}{3} \cdot \frac{2}{x} \right) \, dx. \]

Step 4: Simplify and Integrate Remaining Integral

Simplify the integral: \( \int \frac{2x^2}{3} \, dx \). This is \( \frac{2}{3} \int x^2 \, dx \). Integrate to find \( \frac{2}{3} \cdot \frac{x^3}{3} = \frac{2x^3}{9} \).

Step 5: Write the Final Solution

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

Key Concepts

Calculus

Calculus is a branch of mathematics focused on studying how things change. It involves two primary operations: differentiation and integration. Differentiation is concerned with finding rates of change, while integration involves finding the total size or value accumulated over time. These two operations are essential for analyzing and understanding continuous change, helping us navigate a variety of fields ranging from physics to engineering.

In terms of application, calculus allows us to compute areas under curves, solve optimization problems, and model dynamic systems.

• Consider calculating the area under the curve of a function. This is where integration becomes a powerful tool.
• On the other hand, if you're looking to find the rate at which a car accelerates, you'd use differentiation.

Integration, including techniques like integration by parts, expands the scope of calculus, offering advanced tools for finding such areas precisely. Understanding calculus enriches problem-solving skills, making complex phenomena more tangible.

Definite Integrals

Definite integrals are used to calculate the exact area under a curve between two specific points on the x-axis. Unlike indefinite integrals, definite integrals have upper and lower limits of integration. These provide the specific interval over which you're calculating the area.

A definite integral is represented as: \[ \int_{a}^{b} f(x) \, dx \]Here, \( a \) and \( b \) are the limits of integration, and \( f(x) \) is the function being integrated. This computes the net area between \( f(x) \) and the x-axis, from \( x = a \) to \( x = b \).

Definite integrals are practical in various applications:

• Calculating the total distance traveled by an object.
• Finding the accumulated profit over a period of time.

By solving definite integrals, we can quantify these results in exact numerical terms, enabling us to understand the scope or magnitude of a quantity directly.

Differentiation

Differentiation is the mathematical process of finding the derivative of a function. The derivative represents the rate of change of a function with respect to a variable. In simple terms, differentiation tells us how a function is changing at any given point.

The derivative of a function \( f(x) \) can be denoted as \( f'(x) \) or \( \frac{df}{dx} \). This operation is incredibly useful when analyzing motions, such as velocity or acceleration. By understanding derivatives, we can predict how a function behaves: whether it is increasing, decreasing, or staying constant.

For instance, if \( f(x) \) represents the position of a car over time, then \( f'(x) \) gives the car's velocity at any point in time.

• You find where a function reaches its maximum or minimum value through differentiation.
• It's crucial in optimizing processes or figuring out the best possible conditions for certain outcomes.

Differentiation plays a key role, often providing the preliminary steps before moving into integration tasks, as seen in the integration by parts technique.

Indefinite Integrals

Indefinite integrals refer to the process of integrating a function without specific limits, resulting in a general form that includes a constant of integration, denoted as \( C \). This constant represents an infinite number of potential values the integral could take, as there are no bounds to restrict it.

The indefinite integral of a function \( f(x) \) is written as:\[ \int f(x) \, dx = F(x) + C \]where \( F(x) \) is the antiderivative or the original function whose derivative is \( f(x) \).
Indefinite integrals are essential because they serve as the foundation for finding definite integrals. By first solving the indefinite integral, you find the original function from which you can calculate areas or solve further calculus problems.

• They are essential in physics for solving motion problems where the full path or position is not immediately known.
• In economics, they help model overall cost functions or supply and demand dynamics over time.

In our exercise, the solution used integration by parts, a powerful technique aiding in solving more complex indefinite integrals involving products of functions.