Problem 42 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^{2}_{1} \frac{(x - 1)^3}{x^2} \,dx \)

Step-by-Step Solution

Answer

The integral evaluates to \(-2 + 3\ln 2\).

Step 1: Simplify the Integrand

First, simplify the expression \( \frac{(x - 1)^3}{x^2} \) by expanding the numerator. \((x - 1)^3 = x^3 - 3x^2 + 3x - 1\). Thus, the integrand becomes \( \frac{x^3 - 3x^2 + 3x - 1}{x^2} \). Divide each term in the numerator by \( x^2 \), resulting in the expression \( x - 3 + \frac{3}{x} - \frac{1}{x^2} \).

Step 2: Set Up the Integral with Simplified Integrand

Rewrite the integral using the simplified integrand: \[ \int_{1}^{2} \left( x - 3 + \frac{3}{x} - \frac{1}{x^2} \right) \, dx. \]

Step 3: Integrate Each Term Separately

Find the antiderivative of each term:- The antiderivative of \( x \) is \( \frac{x^2}{2} \).- The antiderivative of \( -3 \) is \( -3x \).- The antiderivative of \( \frac{3}{x} \) is \( 3\ln|x| \).- The antiderivative of \( -\frac{1}{x^2} \) is \( \frac{1}{x} \).Thus, the antiderivative of the integrand is: \( \frac{x^2}{2} - 3x + 3\ln|x| + \frac{1}{x} + C \).

Step 4: Evaluate the Definite Integral

Plug in the limits of integration into the antiderivative and subtract:1. Evaluate at the upper limit (2): \[ \left( \frac{2^2}{2} - 3(2) + 3\ln|2| + \frac{1}{2} \right) = \left( 2 - 6 + 3\ln 2 + 0.5 \right). \] Simplify to \( -3.5 + 3\ln 2 \).2. Evaluate at the lower limit (1): \[ \left( \frac{1^2}{2} - 3(1) + 3\ln|1| + \frac{1}{1} \right) = \left( 0.5 - 3 + 0 + 1 \right). \] Simplify to \( -1.5 \).3. The result is: \[ -3.5 + 3\ln 2 - (-1.5) = -2 + 3\ln 2. \]

Step 5: Simplify the Final Result

Simplify the result to find the final value of the definite integral: \(-2 + 3\ln 2\). This is the exact value of the integral after evaluation.

Key Concepts

Integrand Simplification

Simplifying the integrand is the first step in evaluating a definite integral. It involves transforming the original complex expression into simpler terms, making it easier to integrate. In our example, we began with the integrand \( \frac{(x - 1)^3}{x^2} \).

To simplify this, we expanded \((x - 1)^3\) using the distributive property, resulting in \(x^3 - 3x^2 + 3x - 1\).

We then divided each term in the numerator by \(x^2\):

• \(\frac{x^3}{x^2} = x\),
• \(-\frac{3x^2}{x^2} = -3\),
• \(\frac{3x}{x^2} = \frac{3}{x}\),
• \(-\frac{1}{x^2} = -\frac{1}{x^2}\).

The integrand becomes \(x - 3 + \frac{3}{x} - \frac{1}{x^2}\). This simplification is crucial as it breaks down the integral into more manageable parts.

Antiderivative

Finding the antiderivative is a critical step in solving integrals. The antiderivative, also known as the indefinite integral, is a reverse process of differentiation.

After simplification, each term of the integrand \( \left( x - 3 + \frac{3}{x} - \frac{1}{x^2} \right) \) is integrated separately:

• The antiderivative of \(x\) is \(\frac{x^2}{2}\). This follows from reversing the power rule in differentiation.
• The antiderivative of \(-3\) is \(-3x\). Constant factors remain in place as constants when integrated.
• The antiderivative of \(\frac{3}{x}\) is \(3 \ln |x|\). The rule for \(\frac{1}{x}\) is that it integrates to \(\ln|x|\).
• The antiderivative of \(-\frac{1}{x^2}\) is \(\frac{1}{x}\). This uses the rule that \(x^n\) integrates to \(\frac{x^{n+1}}{n+1}\) when \(neq -1\).

Together, these lead to the expression: \( \frac{x^2}{2} - 3x + 3\ln|x| + \frac{1}{x} + C \), where \(C\) represents the constant of integration.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation and integration, two central operations in calculus. It states that if a function is continuous on an interval, the integral of a function over this interval can be computed using its antiderivative.

When we have our antiderivative, \( \frac{x^2}{2} - 3x + 3\ln|x| + \frac{1}{x} \), we use the Fundamental Theorem to evaluate the definite integral from 1 to 2.

This involves:

• Calculating the antiderivative at the upper limit (2) and simplifying it to give \(-3.5 + 3\ln 2\).
• Evaluating at the lower limit (1) and simplifying to \(-1.5\).
• Subtracting the result at the lower limit from the result at the upper limit, leading to \(-3.5 + 3\ln 2 - (-1.5)\).

The result is \(-2 + 3\ln 2\). This theorem provides an efficient way to evaluate definite integrals and connect antiderivatives with accumulation of quantities over an interval.