Problem 18 - Thomas Calculus

Question

In Exercises \(17-20\) , express the integrands as a sum of partial fractions and evaluate the integrals. $$ \int_{-1}^{0} \frac{x^{3} d x}{x^{2}-2 x+1} $$

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{5}{2}\).

Step 1: Recognize the Denominator

First, observe that the denominator is a perfect square trinomial. It can be rewritten as \((x-1)^2\). This expression indicates that the denominator \(x^2 - 2x + 1\) can be factored.

Step 2: Rewrite the Integral

Rewrite the integral using the factored form of the denominator: \(\int_{-1}^{0} \frac{x^{3}}{(x-1)^{2}} \, dx\). Now our task is to decompose the integrand into partial fractions, if possible.

Step 3: Set Up Partial Fraction Decomposition

Here we consider \(\frac{x^3}{(x-1)^2}\) as \(\frac{A}{(x-1)} + \frac{B}{(x-1)^2}\). However, since \(x^3\) is a higher degree than \((x-1)^2\), direct partial fraction decomposition is not appropriate. We need polynomial long division.

Step 4: Perform Polynomial Long Division

Divide \(x^3\) by \((x-1)^2\). This division gives a polynomial part and a remainder. The result is \(x + 2 + \frac{3}{(x-1)^2}\).

Step 5: Integrate the Result

Now, integrate each part: \(\int_{-1}^{0} (x + 2 + \frac{3}{(x-1)^2}) \, dx\). This breaks into three separate integrals: \(\int x \, dx\), \(\int 2 \, dx\), and \(\int \frac{3}{(x-1)^2} \, dx\).

Step 6: Evaluate Each Integral

1. \(\int x \, dx = \frac{x^2}{2} \bigg|_{-1}^{0} = 0 - \frac{1}{2} = -\frac{1}{2}\).2. \(\int 2 \, dx = 2x \bigg|_{-1}^{0} = 0 - (-2) = 2\).3. \(\int \frac{3}{(x-1)^2} \, dx = -\frac{3}{x-1} \bigg|_{-1}^{0} = 0 + \frac{3}{2}\).

Step 7: Combine Results for Final Answer

Combine the results of the integrals: \(-\frac{1}{2} + 2 + \frac{3}{2} = 1 + \frac{3}{2} = \frac{5}{2}\). This is the final value of the integral over the given limits.

Key Concepts

Polynomial Long Division

Polynomial long division is a technique similar to arithmetic long division. It helps break down complex algebraic expressions into simpler parts. Here, we divide a polynomial by another, generally of lower degree.

• First, identify the terms: the dividend (the expression being divided) and the divisor (the expression by which we divide).
• In our example, divide the polynomial numerator, \(x^3\), by the factored polynomial in the denominator, \( (x-1)^2 \).
• Align the terms by degree and divide the leading term of the dividend by the leading term of the divisor.
• Multiply the entire divisor by this result and subtract it from the dividend.
• Repeat the process with the new polynomial obtained after subtraction until the remainder has a lower degree than the divisor.

The result of this process is a simpler expression plus a remainder, as seen with the expression \( x + 2 + \frac{3}{(x-1)^2} \). This method notably eases complex integral evaluation.

Integral Evaluation

Integral evaluation is about calculating the integral's value, which represents the area under a curve within specific limits. In calculus, it's a fundamental technique used to find solutions to various problems.

• Break down complex expressions into more manageable pieces.
• Once polynomial long division is done, integrate each part of the simpler polynomial separately.
• For example, the integrals involved are \(\int x \, dx\), \(\int 2 \, dx\), and \(\int \frac{3}{(x-1)^2} \, dx\).
• Evalute each from the limits \(-1\) to \(0\) to find the portion of the total area under the curve for each term.

Each part requires a different approach. Some integrals can be found using basic antiderivative rules, and others may need more careful consideration of limits and function properties. The sum of these separate areas gives the final integral value, \(\frac{5}{2}\) in our problem.

Perfect Square Trinomial

A perfect square trinomial is a special polynomial form that can be easily factored into a squared binomial. Recognizing these trinomials significantly simplifies the task of integration by allowing easier manipulation of the expression.

• Such trinomials take the form \(a^2 \pm 2ab + b^2\), reducing to \((a\pm b)^2\).
• In our exercise, \(x^2 - 2x + 1\) is identified as a perfect square trinomial.
• It appears in factored form as \((x-1)^2\), simplifying the denominator of our integral.

Recognizing and utilizing this form early on fosters easier manipulation for division and integration, turning what initially appears complex into a straightforward process. Being able to spot these forms efficiently is a key skill in calculus.