Problem 51 - Calculus Single Variable

Question

Show that $$ \int \frac{1}{|x|} d x=\ln (|x|)+C $$ is not a valid integration formula.

Step-by-Step Solution

Answer

The function \( \frac{1}{|x|} \) is not continuous at \( x = 0 \), so the integral \( \ln (|x|) + C \) is incorrect as a universal result.

Step 1: Understand the Function

We need to integrate the function \( \frac{1}{|x|} \). This function is defined as \( \frac{1}{x} \) for \( x > 0 \) and \( -\frac{1}{x} \) for \( x < 0 \).

Step 2: Integrate Positive x

For \( x > 0 \), we integrate \( \frac{1}{x} \), which results in \( \ln|x| + C \). This part of the integral evaluates correctly, covering half of the domain.

Step 3: Integrate Negative x

For \( x < 0 \), we are integrating \( -\frac{1}{x} \), leading to \(-\ln|-x|\), which still equals \( \ln|x| + C \). Again, integration is straightforward.

Step 4: Consider the Absolute Value

The function \( \frac{1}{|x|} \) introduces problems at \( x = 0 \) because the absolute value changes signs. The function is not defined continuously across zero, causing an improper integral since it would be discontinuous at zero.

Step 5: Evaluate Discontinuities

The integral must be split at the point where \( x = 0 \), indicating a discontinuity. Any valid integration over an interval should not include a point of discontinuity inside without proper limits.

Step 6: Conclusion

Thus, the integration over all real numbers including 0 is not defined because \( \frac{1}{|x|} \) is undefined for \( x = 0 \). The expression \( \ln(|x|) + C \) cannot accommodate this discontinuity, showing it is incorrect over the real number line.

Key Concepts

Improper Integral

When it comes to calculus, understanding improper integrals can be a bit challenging at first. Unlike regular integrals, improper integrals involve functions that are not defined at certain points within the interval of integration or when the interval itself is infinite. This happens when:

• The integrand becomes infinite at one or more points in the range.
• The interval of integration stretches to infinity.

In the context of the given exercise, we are looking at the function \( \frac{1}{|x|} \). This function is problematic because it becomes undefined at \( x = 0 \) due to division by zero, making it an improper integral. Therefore, when integrating \( \frac{1}{|x|} \), it's crucial to consider these undefined points. We use limits to approach these points rather than evaluating them directly. Ignoring these particular points, like \( x = 0 \) in this case, can lead to incorrect integration results. This is why saying that \( \int \frac{1}{|x|} dx = \ln(|x|) + C \) over the entire real line is not valid.

Absolute Value Function

The absolute value function is another interesting concept frequently encountered in calculus. It is represented mathematically as \( |x| \), and is a function that returns the magnitude of a number, regardless of its sign. Here's how it behaves:

• For \( x > 0 \), \( |x| = x \).
• For \( x < 0 \), \( |x| = -x \).
• For \( x = 0 \), \( |x| = 0 \).

In our exercise, we are dealing with the absolute value function in the context of \( \frac{1}{|x|} \). Because the absolute value changes the sign depending on whether \( x \) is positive or negative, it splits the domain into two parts: one for positive \( x \) and one for negative \( x \). This splitting is crucial because the integrand changes significantly over these ranges. For instance, \( \frac{1}{|x|} = \frac{1}{x} \) when \( x > 0 \), and \( \frac{1}{|x|} = -\frac{1}{x} \) when \( x < 0 \), affecting how the integral is computed across different regions.

Discontinuity in Integration

Discontinuities occur in integrals when there are breaks or undefined points in the function or interval. The point \( x = 0 \) in the function \( \frac{1}{|x|} \) serves as a classic example of a discontinuity. Here's why:

• The function is not defined at \( x = 0 \) because it involves division by zero.
• The absolute value inherently alters the behavior of the integrand as you cross \( x = 0 \).

To properly integrate across a discontinuity, we typically need to split the integral at the point of discontinuity and consider the limit of the integral as we approach the point from either side. This makes sure that we account for the behavior of the function on either side of the discontinuity. In our exercise, failing to handle the discontinuity at \( x = 0 \) with limits is why \( \ln(|x|) + C \) is insufficient as a solution. Properly managing such discontinuities is crucial for getting correct solutions in calculus integration.