Integral Of An Absolute Value Of X

Evaluating the Definite Integral of the Absolute Value of x

The definite integral of the absolute value of x, ∫|x| dx, might seem daunting at first glance, but it's surprisingly straightforward once you understand the nature of the absolute value function. This article will guide you through the process, explaining the concept and providing a step-by-step solution, including consideration of different integration limits. Understanding this integral is crucial for various applications in calculus and beyond, such as finding areas under curves and solving differential equations.

Understanding the Absolute Value Function

The absolute value of x, denoted as |x|, is defined as:

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

This means the function reflects the negative portion of the x-axis onto the positive x-axis. Graphically, it's a V-shaped curve with its vertex at the origin (0,0). This piecewise definition is key to integrating |x|.

Solving the Definite Integral ∫|x| dx

To solve the definite integral ∫|x| dx from a to b, we need to consider the different cases based on the values of 'a' and 'b'.

Case 1: Both 'a' and 'b' are non-negative (a ≥ 0 and b ≥ 0)

If both limits are non-negative, the absolute value function simplifies to f(x) = x. The integral becomes a simple power rule integration:

∫_a^b |x| dx = ∫_a^b x dx = [x²/2]_a^b = (b²/2) - (a²/2)

Case 2: Both 'a' and 'b' are non-positive (a ≤ 0 and b ≤ 0)

If both limits are non-positive, the absolute value function becomes f(x) = -x. The integral becomes:

∫_a^b |x| dx = ∫_a^b -x dx = [-x²/2]_a^b = (-b²/2) - (-a²/2) = (a²/2) - (b²/2)

Case 3: 'a' is non-positive and 'b' is non-negative (a ≤ 0 and b ≥ 0)

This is the most common and challenging scenario. We need to split the integral into two parts at x = 0:

∫_a^b |x| dx = ∫_a⁰ -x dx + ∫₀^b x dx

Applying the power rule:

= [-x²/2]_a⁰ + [x²/2]₀^b = (0 - (-a²/2)) + (b²/2 - 0) = a²/2 + b²/2

Example: Evaluating ∫_-2³ |x| dx

Using Case 3, we split the integral:

∫_-2³ |x| dx = ∫_-2⁰ -x dx + ∫₀³ x dx = [-x²/2]_-2⁰ + [x²/2]₀³ = (0 - (-4/2)) + (9/2 - 0) = 2 + 4.5 = 6.5

Conclusion: Mastering the Integral of |x|

The integral of the absolute value of x is solved by carefully considering the piecewise definition of the absolute value function. By splitting the integral at x=0 when necessary and applying the fundamental theorem of calculus, we can accurately determine the definite integral for any given limits 'a' and 'b'. Remember to always account for the sign changes based on the interval of integration for an accurate result. This fundamental concept is essential for numerous applications in advanced calculus and related fields.