Problem 25 - Calculus Single Variable

Question

Determine whether the given improper integral is convergent or divergent. If it converges, then evaluate it. \(\int_{-2}^{4}(x+1)^{-2 / 3} d x\)

Step-by-Step Solution

Answer

The integral is divergent due to a singularity at \(x = -1\).

Step 1: Analyze the Integrand

The function to integrate is \((x+1)^{-2/3}\). Notice the exponent is negative, which implies the function will have an asymptote where the base \((x+1)\) equals zero. This happens at \(x = -1\). Therefore, the function has an improper behavior at \(x = -1\), which is within our interval of integration \([-2, 4]\).

Step 2: Decompose the Integral for Analysis

Split the integral at the point of improper behavior \(x = -1\). This forms two separate integrals: \[ \int_{-2}^{-1}(x+1)^{-2/3} dx \] and \[ \int_{-1}^{4}(x+1)^{-2/3} dx \]. Both integrals need to be analyzed separately to determine their convergence.

Step 3: Analyze Convergence of Each Part

Consider the first integral \(\int_{-2}^{-1}(x+1)^{-2/3} dx\). As \(x \to -1^{-}\), \(x+1\) approaches zero from the negative side, causing the function \((x+1)^{-2/3}\) to become undefined. Similarly, consider the second integral \(\int_{-1}^{4}(x+1)^{-2/3} dx\). As \(x \to -1^{+}\), the function again becomes undefined.

Step 4: Conclude on Convergence

Both integrals have a point \(x = -1\) where they are undefined and where \((x+1)^{-2/3}\) diverges to infinity. Since there is no way to integrate across these points without the integral becoming infinite, neither part converges. Hence, the original integral \(\int_{-2}^{4}(x+1)^{-2/3} dx\) is divergent.

Key Concepts

Convergence

Convergence refers to the behavior of an improper integral as it approaches a specific value. An improper integral converges if its total area under the curve approaches a finite number. For example, when evaluating the improper integral \( \int_{-2}^{-1}(x+1)^{-2/3} dx \), convergence would mean that the area between \(-2\) and \(-1\) is finite as \(x\) approaches \(-1\).
In simpler terms:

• If the area under the curve becomes constant or reaches a fixed point, the integral converges.
• A convergent integral means we can calculate a finite result.

In the original exercise, we could not manage to make this happen, which is why the integral is found to be divergent, meaning the area becomes infinitely large instead.

Divergence

The concept of divergence in integrals means that the integral does not have a finite value or result. Essentially, the area under the curve continues to grow indefinitely. In our case, the integral \( \int_{-2}^{4}(x+1)^{-2/3} dx \) is divergent.
The divergence arises here because:

• The function \((x+1)^{-2/3}\) has undefined behavior at \(x = -1\).
• Both parts of the splitted integral (\( \int_{-2}^{-1}(x+1)^{-2/3} dx \) and \( \int_{-1}^{4}(x+1)^{-2/3} dx \)) become infinite at this point.

Therefore, the integral cannot be evaluated to give any finite answer, which results in divergence.

Asymptote

An asymptote in mathematics is a line that a curve gradually approaches but never actually reaches. In our integral problem, the expression \((x+1)^{-2/3}\) has a vertical asymptote at \(x = -1\).
Understanding asymptotes is crucial because:

• They indicate points where the function becomes infinitely large or small, as \(x\) approaches the asymptote value.
• In the integral \( \int_{-2}^{4}(x+1)^{-2/3} dx \), as the curve nears \(x = -1\), it shoots to infinity or negative infinity.

This behavior causes problems in calculations, as it leads to divergence rather than convergence.

Interval of Integration

When working with integrals, the interval of integration is the range over which we calculate the integral. It defines the starting and ending values for \(x\). For \( \int_{-2}^{4}(x+1)^{-2/3} dx \), the interval of integration is from \(-2\) to \(4\).
The interval is significant because:

• It includes \(-1\), where an asymptote exists, causing the function to become undefined.
• By splitting the interval at the asymptote \(x = -1\) into \([-2, -1]\) and \([-1, 4]\), each section can be analyzed for convergence or divergence.

In our situation, neither part of this interval can produce a finite result due to the asymptotic behavior at \(x = -1\), resulting in an overall divergent integral.