Problem 40 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^3_1 \frac{y^3 - 2y^2 - y}{y^2} \,dy \)

Step-by-Step Solution

Answer

The value of the integral is \(4 - \ln 3\).

Step 1: Simplify the integrand

First, simplify the expression inside the integral. The integrand is \( \frac{y^3 - 2y^2 - y}{y^2} \). We can simplify this by dividing each term by \( y^2 \):\[ \frac{y^3}{y^2} - \frac{2y^2}{y^2} - \frac{y}{y^2} = y - 2 - \frac{1}{y} \]

Step 2: Set up the integral with the simplified expression

Now that we have simplified the integrand to \( y - 2 - \frac{1}{y} \), we set up the integral with respect to \( y \):\[ \int_1^3 (y - 2 - \frac{1}{y}) \, dy \]

Step 3: Integrate term by term

We integrate each term separately. The integral becomes:\[ \int (y - 2 - \frac{1}{y}) \, dy = \int y \, dy - 2 \int \, dy - \int \frac{1}{y} \, dy \]Calculating each term:1. \( \int y \, dy = \frac{y^2}{2} + C \)2. \( -2 \int \, dy = -2y + C \)3. \( -\int \frac{1}{y} \, dy = -\ln|y| + C \)

Step 4: Evaluate the definite integral

Now evaluate the integral from 1 to 3:\[ \Bigg[ \frac{y^2}{2} - 2y - \ln|y| \Bigg]_1^3 \]Substitute 3:\[ \left( \frac{3^2}{2} - 2(3) - \ln|3| \right) = \left( \frac{9}{2} - 6 - \ln 3 \right) \]Substitute 1:\[ \left( \frac{1^2}{2} - 2(1) - \ln|1| \right) = \left( \frac{1}{2} - 2 - 0 \right) \]Subtract:\[ \left( \frac{9}{2} - 6 - \ln 3 \right) - \left( \frac{1}{2} - 2 \right) \]

Step 5: Simplify the result

Subtract the lower bound from the upper bound:\( \frac{9}{2} - 6 - \ln 3 - \frac{1}{2} + 2 = \)Simplify:\( \frac{8}{2} - 4 - \ln 3 = 4 - \ln 3 \)Thus, the value of the definite integral is \(4 - \ln 3\).

Key Concepts

Simplification

In calculus, the simplification process is crucial in making complex integrals easier to solve. It involves breaking down expressions into simpler forms. Simplifying an integrand, like our example, means rewriting it in a form that facilitates easier integration.
Here we have the integrand \( \frac{y^3 - 2y^2 - y}{y^2} \). By dividing each term in the numerator by the denominator \( y^2 \), the expression becomes \( y - 2 - \frac{1}{y} \).

• \( \frac{y^3}{y^2} = y \)
• \( \frac{2y^2}{y^2} = 2 \)
• \( \frac{y}{y^2} = \frac{1}{y} \)

This simplified form makes subsequent integration straightforward, enabling us to easily apply integration techniques to each term separately.

Definite Integral

A definite integral calculates the net area under a curve over a specific interval. Unlike an indefinite integral, it results in a numerical value, not a function with a constant of integration. In our case, the integral \( \int_1^3 (y - 2 - \frac{1}{y}) \, dy \) is evaluated between the limits 1 and 3.
The key features of definite integrals include:

• Bounds: The upper and lower limits \([1, 3]\) indicate where to evaluate the integral.
• Area Interpretation: It gives the total change—or net area—of the function \( y - 2 - \frac{1}{y} \) between these bounds.
• Fundamental Theorem of Calculus: This theorem relates the evaluation of definite integrals to antiderivatives.

Therefore, when computing a definite integral, we determine the antiderivative at each bound and find their difference.

Integration Techniques

Integration techniques refer to methods for finding antiderivatives, which are essential in solving integrals. Here, we'll explore how to apply basic integration techniques, specifically to simple polynomials and logarithmic functions.
For the integral \( \int (y - 2 - \frac{1}{y}) \, dy \), each term's integration follows:

• Polynomial Integration: The integral of \( y \) is \( \frac{y^2}{2} \), using the power rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \).
• Constant Function: \(-2 \int \, dy\) integrates to \(-2y\), simple since the integral of a constant is multiplied by the variable.
  
• Logarithmic Integration: \(-\int \frac{1}{y} \, dy\) results in \(-\ln|y|\), leveraging the fact that the derivative of \( \ln y \) is \( \frac{1}{y} \).

By applying these techniques, integrals of complex expressions become substantially manageable.

Step by Step Solution

A step by step solution ensures clarity and understanding in problem-solving. Let's retrace our steps for solving the given integral \( \int_1^3 (y - 2 - \frac{1}{y}) \, dy \):

1. Simplification: Convert \( \frac{y^3 - 2y^2 - y}{y^2} \) to \( y - 2 - \frac{1}{y} \).
2. Setup the Integral: Write it in the form \( \int_1^3 (y - 2 - \frac{1}{y}) \, dy \).
3. Integrate: Calculate each term:
   • \( \int y \, dy = \frac{y^2}{2} \)
   • \(-2 \int \, dy = -2y \)
   • \(-\int \frac{1}{y} \, dy = -\ln|y| \)
4. Evaluate the Integral: Use the bounds from 1 to 3:
   • Substitute \( y = 3 \): \( \frac{9}{2} - 6 - \ln 3 \)
   • Substitute \( y = 1 \): \( \frac{1}{2} - 2 - 0 \)
5. Subtract Bounds: \( \left( \frac{9}{2} - 6 - \ln 3 \right) - \left( \frac{1}{2} - 2 \right) \)
6. Simplify Result: Final result \( 4 - \ln 3 \)

Using this structured approach, students can logically follow each calculation step, building a comprehensive understanding.