Problem 37 - College Algebra and Calculus: An Applied Approach

Question

$$ \text { Evaluate the definite integral. } $$ $$ \int_{0}^{1} \frac{x^{3}}{x^{2}-2} d x $$

Step-by-Step Solution

Answer

The value of the definite integral is \( \frac{1}{2} + \ln 2 \).

Step 1: Polynomial Long Division

Rewrite the given integral by dividing the numerator by the denominator. The integrand \( \frac{x^{3}}{x^{2}-2} \) is a rational function, and one can get a simpler integrand by dividing \(x^{3}\) by \(x^{2}-2\). This results in the expression \(x - 2\cdot\frac{2x}{x^{2}-2}\). Therefore, the given integral is equivalent to \(\int_{0}^{1} (x - 2\cdot\frac{2x}{x^{2}-2}) dx\).

Step 2: Split the Integral into Two

Writing the integral as a sum of two integrals, which are \(\int_{0}^{1} x dx - 2 \int_{0}^{1} \frac{2x}{x^{2}-2} dx\). This will be easier to integrate since each integral now only contains a single fraction or a polynomial.

Step 3: Evaluate Each Integral

To find the antiderivative of the first integral (\(\int x dx\)), use the power rule for integration, which gives the result as \(\frac{1}{2} x^{2}\). To evaluate the second integral \(2\int_{0}^{1} \frac{2x}{x^{2}-2}dx\), use a u-substitution where \(u = x^{2}-2\) and \(du = 2x dx\). The antiderivative is \(-\ln|x^{2}-2|\). Evaluate these at the limits of integration 0 to 1.

Step 4: Calculate the Definite Integral

Substitute 1 and then 0 into the antiderivative and subtract to find the definite integral. The final answer is \(\frac{1}{2} - \ln|1-2| - 0 + \ln|0-2| = \frac{1}{2} - \ln 1 + \ln 2 = \frac{1}{2} + \ln 2\).

Key Concepts

Polynomial Long Division

Polynomial long division is a technique used to divide one polynomial, which is the dividend, by another polynomial, the divisor. Similar to the long division process for numbers, this method involves subtracting multiples of the divisor from the dividend to find a quotient and a remainder. The process continues until the remainder has a degree less than the divisor. In the context of evaluating integrals of rational functions, polynomial long division is often used when the degree of the numerator is equal to or greater than the degree of the denominator. Dividing allows us to represent the integrand in a more manageable form, typically a polynomial plus a proper fraction.

For instance, consider the integrand \( \frac{x^{3}}{x^{2}-2} \). Here, the numerator has a higher degree than the denominator. By dividing \(x^{3}\) by \(x^{2}-2\), we simplify the rational function, which helps in further integration steps. The quotient and the remainder together provide a clear path for integration, especially when the integral is broken down into simpler parts that are easier to evaluate.

Rational Functions

Rational functions are ratios of two polynomials. They can take various forms, but a common issue arises when attempting to integrate them, especially when the degree of the polynomial in the numerator is greater than or equal to that of the denominator. In the original exercise, the integrand \( \frac{x^{3}}{x^{2}-2} \) is an example of a rational function. To integrate such a function, one may need to employ different techniques—such as polynomial long division, as previously mentioned, or partial fraction decomposition if applicable—to convert it into a more integrable form.

Rational functions are important in calculus due to their relevance in modeling various phenomena and their frequent appearance in various mathematical contexts. Understanding how to manipulate and integrate these functions can shed light on many practical problems.

Integration by Substitution

Integration by substitution, commonly known as u-substitution, is a method used to find the antiderivative of a function by changing the variable of integration to simplify the integral. This technique is essentially the reverse process of the chain rule for derivatives. When facing a complex integral, identifying a part of the integrand that can be substituted with a new variable \(u\), hence simplifying the integral into a form that's easier to handle, is a crucial step.

In the exercise, the integral \(2\int_{0}^{1} \frac{2x}{x^{2}-2}dx\) is simplified through substitution by setting \(u = x^{2}-2\) and accordingly \(du = 2x dx\), making the integration straightforward. After substitution, using the antiderivative and applying the limits of integration become much more feasible. Integration by substitution requires a careful choice of \(u\) and keeping track of the differential \(du\), but it efficiently streamlines the integration process for appropriate functions.

Power Rule for Integration

The power rule for integration is one of the most fundamental rules in calculus for finding antiderivatives of monomials. It states that the integral of \(x^n\) with respect to \(x\), where \(n\) is any real number different from -1, is \(\frac{x^{n+1}}{n+1}\) plus a constant of integration. This rule is based on reversing the process of differentiation. When applying the power rule, it's important to remember to increase the exponent by one and then divide by the new exponent.

For the simple case of the integral \(\int_{0}^{1} x dx\) in the given exercise, the power rule applies directly because we have a monomial with exponent 1. The application of this rule results in the antiderivative \(\frac{1}{2} x^{2}\). The power rule simplifies the task of finding antiderivatives for polynomials and is a staple in the toolkit for evaluating definite integrals.