Problem 56 - Calculus: Early Transcendentals

Question

What is wrong with the equation? \( \displaystyle \int^2_{-1} \frac{4}{x^3} \, dx = -\frac{2}{x^2} \Bigg]^2_{-1} = -\frac{3}{2} \)

Step-by-Step Solution

Answer

The function is undefined at \(x=0\), making the integral improper and cannot be evaluated as it crosses a discontinuity.

Step 1: Identify the Function and Integral Bounds

The given equation involves integrating the function \( f(x) = \frac{4}{x^3} \) from \( x = -1 \) to \( x = 2 \). Check for any points where the function is undefined within this interval.

Step 2: Check for Discontinuities

The function \( f(x) = \frac{4}{x^3} \) is undefined at \( x = 0 \), as it would involve division by zero. Since \( x = 0 \) lies within the interval \([-1, 2]\), the integral diverges.

Step 3: Analyze the Final Calculation

The solution provided attempts to evaluate the indefinite integral and compute the definite integral as \( \int \frac{4}{x^3} \, dx = -\frac{2}{x^2} \), leading to the evaluation \(-\frac{2}{2^2} + \frac{2}{(-1)^2}\), which simplifies to \(-\frac{3}{2}\). However, this computation is invalid due to the discontinuity.

Step 4: Conclusion on the Error

Since the function is discontinuous within the given bounds, the definite integral cannot be evaluated directly over \([-1, 2]\), as it's not defined at every point of the interval.

Key Concepts

Discontinuities in Functions

In calculus, a discontinuity in a function occurs when the function is not continuous at a point within its domain. Simply put, there is a break or gap in the function.
For example, in the function \( f(x) = \frac{4}{x^3} \), this discontinuity appears at \( x = 0 \), where the function is undefined because dividing by zero is impossible.
This forms a vertical asymptote at \( x = 0 \), creating a significant challenge when dealing with definite integrals.

• Discontinuities break the interval into separate parts.
• In integrals, especially _improper integrals_, these discontinuities need careful handling, often requiring limits to solve them.
• Functions across intervals containing discontinuities cannot be integrated directly without considering these breaks.

Definite Integrals

Definite integrals calculate the net area under a curve between two points on the x-axis.
They are fundamental to understanding accumulated change and can provide the total value between bounds \( a \) and \( b \).
For example, the expression \( \int_{a}^{b} f(x) \, dx \) represents the area between the curve \( f(x) \) and the x-axis from \( x=a \) to \( x=b \).

• This operation relies on the function being _continuous_ over the interval \([a, b]\).
• Discontinuities can prevent using straightforward calculation methods.
• The solution can involve breaking the integral at points of discontinuity and evaluating limits.

Integration Techniques

Various integration techniques allow us to solve complex integrals with varying methods.
In the process of solving definite integrals with discontinuities, choosing the correct technique is vital.
Some common techniques include:

• Substitution: Often used when the integral matches a derivative's present form.
• Integration by Parts: Useful when the product of two functions is present.
• Partial Fractions: Effective for rational functions, decomposing them into simpler fractions.
• Improper Integrals and Limits: When dealing with vertical asymptotes or discontinuities, breaking the integral at these points and calculating limits can help.

Integrals over discontinuous intervals may require dividing at the point of discontinuity and assessing the limit behavior near these points.
This allows for a proper evaluation of intervals that aren't initially clear.