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Chapter 8: Problem 18

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

Short Answer

Expert verified
The integral evaluates to \(-2\).

Step by step solution

01

Identify the Denominator as a Perfect Square

First, notice that the denominator \(x^2 - 2x + 1\) is a perfect square. It can be rewritten as \((x-1)^2\).
02

Express the Integrand in Partial Fractions

The integrand \(\frac{x^3}{(x-1)^2}\) already has \((x-1)^2\) as the denominator, and \(x^3\) is the numerator. Since the denominator is a power of a linear factor, a partial fraction decomposition can be written as \(\frac{x^3}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}\).
03

Determine the Values of A and B

To find \(A\) and \(B\), multiply through by \((x-1)^2\) to clear the denominators: \(x^3 = A(x-1) + B\). Expand and combine like terms: \(x^3 = Ax - A + B\). Equate coefficients to find \(A = 1\) and \(B = 1\) (assuming you equate with \(x^3 - x = 0\)).
04

Rewrite the Integrand with Determined Coefficients

Substitute \(A\) and \(B\) back into the partial fraction expression: \(\frac{x^3}{(x-1)^2} = \frac{x}{x-1} + \frac{1}{(x-1)^2}\).
05

Integrate Each Term Separately

Now integrate each term separately: \(\int_{-1}^{0} \frac{x}{x-1} \, dx \) and \(\int_{-1}^{0} \frac{1}{(x-1)^2} \, dx \). Start with the first term; use the substitution \(u = x-1, \; du = dx\) for both integrals.
06

Integrate the First Term

The integral \(\int \frac{x}{x-1} \, dx\) transforms into \(\int \frac{x}{u} \, du\), which can further be evaluated by substitution \(u = x - 1\), resulting in \(\int \left(\frac{x}{x-1} + 1\right) \, dx - \int 1 dx\). Simplifies to \(\int \left(1 + \frac{1}{(x-1)}\right)\,dx\).
07

Integrate the Second Term

The integral \(\int \frac{1}{(x-1)^2} \, dx\) uses the power rule for integration, resulting in \(-\frac{1}{x-1}\).
08

Evaluate the Definite Integrals

Evaluate each integral between the bounds \(-1\) and \(0\). Substitute the bounds into the integrated expressions and simplify. Sum the definite integrals of both terms.
09

Simplify the Evaluation

Perform the calculations: the definite integral of each term with the bounds \(-1\) to \(0\). Sum \([-2 - (0) - (-2) + 0]\) to yield a final value of \(-2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Fraction Decomposition
Partial fraction decomposition is a technique used in calculus to simplify complex rational expressions, making them easier to integrate. It involves breaking down a single rational fraction into a sum of simpler fractions. This method is particularly useful when dealing with integrals of rational functions.
  • Identify the type of decomposition needed: Look at the factors in the denominator to determine how to split up the fraction.
  • Set up the partial fraction decomposition: Express the original fraction as a sum of simpler fractions, each with an undetermined constant.
  • Determine coefficients: Multiply through by the common denominator to clear the fractions, then use algebraic techniques to find the constants.
For example, given the rational function \( \frac{x^3}{(x-1)^2} \), we write it as:\[\frac{x^3}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}\]By equating coefficients after multiplying through by \((x-1)^2\), we find the values of \(A\) and \(B\). This step allows us to integrate each simpler fraction individually.
Definite Integral
A definite integral is a fundamental concept in calculus, representing the area under a curve over a specific interval. Unlike indefinite integrals, definite integrals result in a numerical value that quantifies this area.
  • Evaluate definite integrals by computing the antiderivative of the function within the given bounds.
  • The notation \( \int_{a}^{b} f(x) \, dx \) denotes integration from \(a\) to \(b\).
For example, consider the integral:\[\int_{-1}^{0} \frac{x^3}{(x-1)^2} \, dx\]After decomposing and integrating the simpler fractions, you would substitute the upper and lower bounds of the interval into the antiderivative and subtract. The computation concludes with a specific value, representing the total area under the function's curve between \(-1\) and \(0\). It's crucial to work carefully through this process to ensure accuracy.
Power Rule Integration
Power rule integration is one of the most straightforward techniques for finding antiderivatives. It applies to terms of the form \( x^n \), where \(n\) is a real number.
  • For a function \( x^n \), the antiderivative is \( \frac{x^{n+1}}{n+1} + C \), where \(n eq -1\) and \(C\) is the constant of integration for indefinite integrals.
  • The power rule allows for the quick evaluation of integrals of polynomial terms.
In our example, the term \( \frac{1}{(x-1)^2} \) required integration using a modified approach. Applying the power rule to \( (x-1)^{-2} \), we integrate to find:\[\int \frac{1}{(x-1)^2} \, dx = -\frac{1}{x-1} \text{ over the specified interval}\]Remember, when utilizing the power rule for definite integrals, always respect the given limits, substituting them into the anti-differentiated function to find the exact area under the curve.

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