Problem 4 - Applied Calculus

Question

Find the integrals. $$ \int y \ln y d y $$

Step-by-Step Solution

Answer

\( \int y \ln y \, dy = \frac{y^2 \ln y}{2} - \frac{y^2}{4} + C \)

Step 1: Identify the Integration Method

The integral \( \int y \ln y \, dy \) is suitable for integration by parts. This method is useful when the integral consists of a product of two functions.

Step 2: Set Up Integration by Parts Formula

Recall the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \]We need to choose \( u \) and \( dv \) such that the integration becomes simpler.

Step 3: Choose \( u \) and \( dv \)

Let \( u = \ln y \) and \( dv = y \, dy \). This choice simplifies the differentiation of \( u \), which is a logarithmic function, and the integration of a polynomial function.

Step 4: Differentiate \( u \) and Integrate \( dv \)

Differentiate \( u = \ln y \) to find \( du \): \[ du = \frac{1}{y} \, dy \]Integrate \( dv = y \, dy \) to find \( v \): \[ v = \frac{y^2}{2} \]

Step 5: Substitute into Integration by Parts

Substitute \( u \), \( dv \), \( du \), and \( v \) into the integration by parts formula:\[ \int y \ln y \, dy = \ln y \cdot \frac{y^2}{2} - \int \frac{y^2}{2} \cdot \frac{1}{y} \, dy \]

Step 6: Simplify the Integral

Simplify the integral:\[ = \frac{y^2}{2} \ln y - \frac{1}{2} \int y \, dy \]

Step 7: Solve the Remaining Integral

Perform the integral \( \int y \, dy \):\[ \int y \, dy = \frac{y^2}{2} \]

Step 8: Combine and Simplify

Combine all parts and simplify:\[ \frac{y^2}{2} \ln y - \frac{1}{2} \cdot \frac{y^2}{2} = \frac{y^2 \ln y}{2} - \frac{y^2}{4} + C \]Where \( C \) is the constant of integration.

Key Concepts

Calculus Problems

Calculus is a branch of mathematics that deals with the study of rates of change and the accumulation of quantities. One frequent task in calculus is to solve integrals. Here, we consider the integral \( \int y \ln y \, dy \), which involves a function multiplied by a logarithmic function.
Understanding how to solve such integrals often requires selecting the right technique. For the given problem, integration by parts is the appropriate choice. Identifying the right method is crucial as it simplifies the problem and leads you to the solution more efficiently.
This type of calculus problem tests your ability to choose the correct method and see how different integration formulas interact together. It challenges the student to think critically about function composition and mathematical operations.

Integration Techniques

In calculus, various techniques are used for integration, each suited for different types of integrals. Integration by parts is one such technique, particularly beneficial when dealing with a product of functions.
According to the integration by parts formula, \( \int u \, dv = uv - \int v \, du \), it is vital to appropriately choose the functions \( u \) and \( dv \). In our problem, setting \( u = \ln y \) and \( dv = y \, dy \) simplifies both differentiation and integration aspects.

• Differentiating \( u = \ln y \) yields \( du = \frac{1}{y} \, dy \).
• Integrating \( dv = y \, dy \) provides \( v = \frac{y^2}{2} \).

This systematic choice leads to a simpler secondary integral and directly applies the formula, making the integration process more straightforward and manageable.

Mathematical Solutions

Achieving the correct solution in an integration problem requires precise calculations and simplification. Here, we continue by substituting our chosen values into the integration by parts formula. The initial substitution gives \[ \int y \ln y \, dy = \ln y \cdot \frac{y^2}{2} - \int \frac{y^2}{2} \cdot \frac{1}{y} \, dy \].
After simplifying the inner integral to \( \int y \, dy \), solving further yields \( \frac{y^2}{2} \). The complete simplification of the expression results finally as \[ \frac{y^2 \ln y}{2} - \frac{y^2}{4} + C \].

• Combining like terms helps in simplifying the integral expression efficiently.
• Understanding each component's role is key to arriving at the final, corrected form of the integral.

The introduction of a constant \( C \) represents the indefinite nature of the integral. Mastering this solution process is not just about the mechanics but also developing a deeper understanding of mathematical relationships.