Problem 76 - University Calculus: Early Transcendentals

Question

Unboundedregion Integrate \(f(x, y)=1 /\left[\left(x^{2}-x\right)(y-1)^{2 / 3}\right]\) over the infinite rectangle \(2 \leq x<\infty, 0 \leq y \leq 2\)

Step-by-Step Solution

Answer

The integral diverges over the given region.

Step 1: Set Up the Integral

We want to integrate the function \(f(x, y) = \frac{1}{\left[(x^2 - x)(y-1)^{2/3}\right]}\) over the region \(2 \leq x < \infty, 0 \leq y \leq 2\). This requires setting up a double integral: \[\int_{2}^{\infty} \int_{0}^{2} \frac{1}{\left[(x^2 - x)(y-1)^{2/3}\right]} \, dy \, dx\].

Step 2: Simplify the Integrand

Notice that the integrand \(\frac{1}{(y-1)^{2/3}}\) becomes problematic when \(y=1\). Observe whether the limits of integration \(y\) causes the expression to diverge. The function \((x^2 - x)\) for the limits given ensures the numerator is zero when \( x = 0\) or \(x = 1\). However, our lower limit is \( x = 2 \).

Step 3: Evaluate the Inner Integral with respect to y

Since the integrand is separable, consider the integral with respect to \(y\): \[\int_{0}^{2} \frac{1}{(y-1)^{2/3}} \, dy\]. Evaluate this integral by considering: \(\int \frac{1}{(y-1)^{2/3}} \, dy = 3(y-1)^{1/3}+C\). Let's find definite integral from 0 to 2. The result becomes: \[3 \times (1)^{1/3} - 3 \times (-1)^{1/3}=3\times 1 + 3 \times (-1) = 3(1-(-1)) = 6\].

Step 4: Evaluate the Outer Integral with respect to x

Now, evaluate the outer integral: \(\int_{2}^{\infty} \frac{1}{x(x-1)} \, dx\). This can be handled by partial fraction decomposition, rewriting as: \[\frac{1}{x(x-1)} = \frac{A}{x} + \frac{B}{x-1}\]. Solving for A and B, we get \(A = 1\) and \(B = -1\), so the integral becomes: \[\int_{2}^{\infty} \left(\frac{1}{x} - \frac{1}{x-1}\right) \, dx\].

Step 5: Compute the Integral Result

Integrating each term separately: \(\int \frac{1}{x} \, dx = \ln|x| + C\) and \(-\int \frac{1}{x-1} \, dx = -\ln|x-1| + C\). Therefore, the definite integral from 2 to \(\infty\) is the following calculation: \(\lim_{b \to \infty} [\ln b - \ln(b-1) + C - (\ln 2 - \ln 1)] = \infty\). Since \(\ln b - \ln(b-1) \to \ln 1 = 0\). The integral diverges.

Key Concepts

Improper Integrals

In calculus, improper integrals are used when an integral has at least one infinite limit of integration or a point where the integrand becomes unbounded. These integrals extend the concept of covering areas to more complex situations.
For the given exercise, the integral is improper because the upper limit of integration for the variable \(x\) is infinity. This means the area we are attempting to measure extends infinitely in one direction.
To obtain a solution, we need to find the limit of a definite integral as one or both of the limits of integration approach infinity or a point of discontinuity. This often involves evaluating the integral up to a finite boundary \(b\) and then considering \(\lim_{b \to \infty}\). Improper integrals can converge to a finite number or diverge to infinity. Evaluating convergence requires careful application of limits during integration.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify complex rational expressions, which appear during integral calculations. It involves breaking down a complicated fraction into simpler fractions that are easier to integrate. In the given problem, we use partial fraction decomposition to solve the outer integral with respect to \(x\), which is \[\int_{2}^{\infty} \frac{1}{x(x-1)} \, dx\].
By expressing \(\frac{1}{x(x-1)}\) as \(\frac{A}{x} + \frac{B}{x-1}\), we identify constants \(A\) and \(B\) by equating and simplifying: \(A = 1\) and \(B = -1\). Now, the problem breaks into two simpler integrals, \(\int \frac{1}{x} \, dx\) and \(-\int \frac{1}{x-1} \, dx\), both of which are straightforward to solve.
This method is very useful when dealing with integrals involving rational polynomials.

Integration Techniques

There are various integration techniques that help solve calculus problems. Choosing the right one depends on the problem structure and function type. For the double integral in this exercise, two main techniques were used: partial fraction decomposition and substitution.
1. **Partial Fraction Decomposition**: As explained earlier, this is used for dealing with rational expressions.
2. **Substitution**: Here, we derived the integral of \(\frac{1}{(y-1)^{2/3}}\) by evaluating the antiderivative as \(3(y-1)^{1/3}+C\). The substitution can often simplify integration by transforming the integrand into a more manageable form.
Utilizing these techniques can ease complicated integrals and bring clarity in solving challenging calculus problems.

Convergence and Divergence of Integrals

A key aspect of improper integrals is determining whether they converge to a finite value or diverge to infinity or does not exist.
For the evaluated double integral, the integral with respect to \(y\) was a proper integral, converging to a value. However, the outer integral with respect to \(x\) posed a challenge. By applying the limit for the definite integral \(\int_{2}^{\infty} \left(\frac{1}{x} - \frac{1}{x-1}\right) \, dx\), we found that \(\lim_{b \to \infty} [\ln b - \ln(b-1)] = \ln 1 = 0\). This showed divergence because the result does not approach a finite number.
When an integral diverges, it implies that the area under the curve is not bounded, signifying infinite extension. Distinguishing between convergence and divergence requires careful analysis and is central to understanding improper integrals.