Question:
The value of the integral \( \int_{0}^{4} ||x - 2| - x| dx = \)
The value of the integral \( \int_{0}^{4} ||x - 2| - x| dx = \)
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For nested modulus functions, always work from the "inside out." Once the inner modulus is simplified for a specific range, the resulting expression is much easier to manage.
Updated On: Jun 3, 2026
- \( 2 \)
- \( 3 \)
- \( 6 \)
- \( 12 \)
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The Correct Option is C
Solution and Explanation
Concept:
To evaluate an integral involving absolute values, we must resolve the modulus signs by identifying where the expression inside the modulus changes sign. This requires splitting the interval of integration at the "critical points" where the internal expressions equal zero.
Step 1: Resolve the inner modulus \( |x - 2| \) based on the interval \( [0, 4] \).
The function is \( f(x) = ||x - 2| - x| \). The inner modulus \( |x - 2| \) behaves differently at \( x = 2 \):
• For \( 0 \leq x < 2 \), \( |x - 2| = -(x - 2) = 2 - x \).
• For \( 2 \leq x \leq 4 \), \( |x - 2| = x - 2 \).
Step 2: Simplify the function in sub-intervals.
Case 1: \( 0 \leq x < 2 \) \[ f(x) = |(2 - x) - x| = |2 - 2x| = 2|1 - x| \] This further splits at \( x = 1 \):
• For \( 0 \leq x < 1 \), \( f(x) = 2(1 - x) = 2 - 2x \).
• For \( 1 \leq x < 2 \), \( f(x) = 2(x - 1) = 2x - 2 \). Case 2: \( 2 \leq x \leq 4 \) \[ f(x) = |(x - 2) - x| = |-2| = 2 \]
Step 3: Evaluate the definite integral by splitting it.
\[ I = \int_{0}^{1} (2 - 2x) dx + \int_{1}^{2} (2x - 2) dx + \int_{2}^{4} 2 dx \] \[ I = \left[ 2x - x^2 \right]_{0}^{1} + \left[ x^2 - 2x \right]_{1}^{2} + \left[ 2x \right]_{2}^{4} \] \[ I = (2 - 1) + (4 - 4 - (1 - 2)) + (8 - 4) \] \[ I = 1 + (0 - (-1)) + 4 \] \[ I = 1 + 1 + 4 = 6 \]
Step 1: Resolve the inner modulus \( |x - 2| \) based on the interval \( [0, 4] \).
The function is \( f(x) = ||x - 2| - x| \). The inner modulus \( |x - 2| \) behaves differently at \( x = 2 \):
• For \( 0 \leq x < 2 \), \( |x - 2| = -(x - 2) = 2 - x \).
• For \( 2 \leq x \leq 4 \), \( |x - 2| = x - 2 \).
Step 2: Simplify the function in sub-intervals.
Case 1: \( 0 \leq x < 2 \) \[ f(x) = |(2 - x) - x| = |2 - 2x| = 2|1 - x| \] This further splits at \( x = 1 \):
• For \( 0 \leq x < 1 \), \( f(x) = 2(1 - x) = 2 - 2x \).
• For \( 1 \leq x < 2 \), \( f(x) = 2(x - 1) = 2x - 2 \). Case 2: \( 2 \leq x \leq 4 \) \[ f(x) = |(x - 2) - x| = |-2| = 2 \]
Step 3: Evaluate the definite integral by splitting it.
\[ I = \int_{0}^{1} (2 - 2x) dx + \int_{1}^{2} (2x - 2) dx + \int_{2}^{4} 2 dx \] \[ I = \left[ 2x - x^2 \right]_{0}^{1} + \left[ x^2 - 2x \right]_{1}^{2} + \left[ 2x \right]_{2}^{4} \] \[ I = (2 - 1) + (4 - 4 - (1 - 2)) + (8 - 4) \] \[ I = 1 + (0 - (-1)) + 4 \] \[ I = 1 + 1 + 4 = 6 \]
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