Integral Of The Absolute Value Of X

Definite Integral of the Absolute Value of x: A Comprehensive Guide

The definite integral of the absolute value of x, often represented as ∫|x| dx, is a common calculus problem that requires understanding piecewise functions and their integration. This guide will walk you through the process, explaining the concept and providing step-by-step solutions. This article will cover both the indefinite and definite integrals, clarifying the nuances involved.

Understanding the Absolute Value Function

Before tackling the integral, let's refresh our understanding of the absolute value function, |x|. This function returns the non-negative value of x. Mathematically:

• |x| = x, if x ≥ 0
• |x| = -x, if x < 0

This piecewise definition is crucial for integrating |x|. We cannot directly integrate the absolute value symbol; we must break the function into its component parts.

Calculating the Indefinite Integral ∫|x| dx

Because of the piecewise nature of |x|, the indefinite integral will also be a piecewise function. We integrate separately for x ≥ 0 and x < 0:

• For x ≥ 0: ∫x dx = (1/2)x² + C (where C is the constant of integration)
• For x < 0: ∫-x dx = -(1/2)x² + C

Therefore, the indefinite integral of |x| is:

∫|x| dx = (1/2)x² + C, if x ≥ 0 -(1/2)x² + C, if x < 0

This can be more concisely represented as:

∫|x| dx = (1/2)x|x| + C

This form neatly encapsulates both cases within a single expression. Note that the constant of integration, C, is the same for both parts, although it can technically be different for each piece.

Calculating the Definite Integral ∫_a^b |x| dx

The definite integral requires considering the interval [a, b]. The approach depends on whether the interval contains 0:

Case 1: The interval [a, b] does not contain 0.

If both a and b are positive or both are negative, we can directly integrate the appropriate part of the piecewise function.

• Example: ∫₁³ |x| dx = ∫₁³ x dx = [(1/2)x²]₁³ = (1/2)(3)² - (1/2)(1)² = 4

• Example: ∫_-3^-1 |x| dx = ∫_-3^-1 -x dx = [-(1/2)x²]_-3^-1 = -(1/2)(-1)² + (1/2)(-3)² = 4

Case 2: The interval [a, b] contains 0.

If the interval includes 0, we must split the integral into two parts at x = 0:

• Example: ∫_-2² |x| dx = ∫_-2⁰ -x dx + ∫₀² x dx = [-(1/2)x²]_-2⁰ + [(1/2)x²]₀² = 2 + 2 = 4

Applications and Further Exploration

The integral of the absolute value of x finds applications in various fields, including:

• Physics: Calculating displacement from velocity when the velocity changes direction.
• Probability and Statistics: Working with probability density functions involving absolute values.
• Geometry: Determining areas bounded by functions involving absolute values.

Understanding the definite integral of the absolute value of x is fundamental to solving more complex integration problems involving piecewise functions and absolute value expressions. Practice with different intervals and boundaries will solidify your understanding and improve your problem-solving skills.