Problem 84 - University Calculus: Early Transcendentals

Question

Improper double integral Evaluate the improper integral $$\int_{0}^{1} \int_{0}^{3} \frac{x^{2}}{(y-1)^{2 / 3}} d y d x$$

Step-by-Step Solution

Answer

The improper double integral diverges due to a singularity, resulting in an infinite value.

Step 1: Re-evaluate the Limits for Integration

The given integral \[\int_{0}^{1} \int_{0}^{3} \frac{x^{2}}{(y-1)^{2 / 3}} d y d x\]involves an improper integral due to the factor \((y-1)^{-2/3}\), which becomes singular at \(y = 1\). This requires us to consider a limit as \(y\) approaches 1 from either side.

Step 2: Split the Integral

Split the integral at the point of singularity:\[\int_{0}^{1} \int_{0}^{3} \frac{x^{2}}{(y-1)^{2 / 3}} d y d x = \int_{0}^{1} \left( \int_{0}^{1} \frac{x^{2}}{(y-1)^{2 / 3}} d y + \int_{1}^{3} \frac{x^{2}}{(y-1)^{2 / 3}} d y \right) d x\]

Step 3: Evaluate First Integral and Address Singularity

Evaluate \(\int_{0}^{1} \frac{x^{2}}{(y-1)^{2 / 3}} d y\) using a limit.Consider the limit as \(y\) approaches 1 from the left:\[\lim_{a \to 1^-} \int_{0}^{a} \frac{x^{2}}{(y-1)^{2/3}} dy\]Since the function approaches infinity as \(y \to 1^-\), this integral is improper diverging to infinity.

Step 4: Evaluate Second Integral

Next, evaluate \(\int_{1}^{3} \frac{x^{2}}{(y-1)^{2 / 3}} d y\):Use the substitution \(u = y - 1\), \(du = dy\), and change the limits accordingly. The new limits are from 0 to 2.The integral becomes:\[\int_{0}^{2} \frac{x^{2}}{u^{2/3}} du\]Now evaluate:\[x^{2} \int_{0}^{2} u^{-2/3} du = x^{2} \cdot \left[\frac{u^{1/3}}{1/3}\right]_{0}^{2}\]This simplifies to \(3x^{2} \left[2^{1/3}\right]\).

Step 5: Evaluate and Sum Contributions

Since \(\int_{0}^{1} \frac{x^{2}}{(y-1)^{2 / 3}} d y\) diverges, the first term in the summation from Step 2 does not contribute a finite number to the solution overall, as one part of the integral results in infinity, which affects the outcome of this improper integral.

Step 6: Concluding Result

Due to the integral in Step 3 diverging, the entire improper double integral results in an undefined (infinite) value upon evaluation. Typically, specific boundary behaviors might allow redefining the integral in a conditional convergence, but this integral straightforwardly diverges.

Key Concepts

Divergence of Integrals

Improper integrals often involve expressions where functions approach infinity at a point within or at the boundaries of the region of integration. This is exactly the case for the integral \( \int_{0}^{1} \int_{0}^{3} \frac{x^{2}}{(y-1)^{2/3}} \, dy \, dx \), where we encounter a divergence at \( y = 1 \). When trying to evaluate such improper integrals, it becomes crucial to investigate the behavior of the integral near the singularity point.

In this scenario, the factor \((y-1)^{-2/3}\) causes the integral to diverge when \( y \) approaches 1. This is because as \( y \to 1 \), the denominator approaches zero, leading the function to infinity. The divergence of integrals indicates that the area under the curve between certain limits becomes infinite, thus not yielding a finite numerical result. In practice, an improper integral like this can be split at the point of singularity and re-evaluated through limits on either side to better understand its behavior.

Double Integrals

Double integrals are used to compute areas and volumes for functions of two variables over a given region. The original exercise features a double integral of the form \( \int_{0}^{1} \int_{0}^{3} \frac{x^{2}}{(y-1)^{2/3}} \, dy \, dx \). Here, the outer integral is with respect to \( x \) and goes from 0 to 1, while the inner integral is with respect to \( y \), ranging from 0 to 3.

Double integrals are solved by integrating one variable at a time. In this scenario, we started from the inside, tackling the \( y \)-integral first, because it directly attributed to the singular behavior of the function. However, it's also common to swap the order of integration when it's more convenient for evaluation. The main takeaway with double integrals is they provide the tools to evaluate multi-dimensional space, illustrating why understanding singularity and behavior of integration over problematic regions is vital.

Integration Limits

Understanding integration limits is essential in resolving integrals, particularly those that might be improper like in the given task. Integration limits define the range over which the function is evaluated. In the case of \( \int_{0}^{1} \int_{0}^{3} \frac{x^{2}}{(y-1)^{2/3}} \, dy \, dx \), we see how crucial it is to approach singular points carefully.

Initially, the integration bounds are simply 0 to 3 for \( y \), and 0 to 1 for \( x \). However, due to the improper nature of the integral involving \( (y-1)^{-2/3} \), the region near \( y = 1 \) needs special attention. Splitting the integral at \( y = 1 \) and analyzing the limits can prevent overlooking issues tied to the function’s behavior. Properly handling these limits and understanding their implications allow us to address whether an integral converges to a finite value or diverges, highlighting their importance when dealing with complex integrative problems.