Question

\( \int_{-1}^{1} (1 - |x|) \, dx \) is equal to :

• A. \( 2 \int_{0}^{1} (1 + x) \, dx \)
• B. \( 2 \int_{-1}^{0} (1 + x) \, dx \)
• C. \( 0 \)
• D. \( 2 \int_{-1}^{0} (1 - x) \, dx \)

Hint:

Before integrating symmetric limits, always check if the function is even or odd. For even functions, doubling the integral over half the interval often simplifies the math, especially when absolute values are involved.

Answer 1

The Correct Option is B

Concept:
• Parity of Functions: A function \( f(x) \) is even if \( f(-x) = f(x) \).
• Integration Property: For an even function, \( \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx = 2 \int_{-a}^{0} f(x) \, dx \).
• Definition of Absolute Value: \( |x| = x \) if \( x \geq 0 \) and \( |x| = -x \) if \( x < 0 \).

Step 1: Determine the parity of the integrand
Let \( f(x) = 1 - |x| \). Substitute \( -x \) for \( x \): \[ f(-x) = 1 - |-x| = 1 - |x| = f(x) \] Since \( f(x) = f(-x) \), the function is an even function.

Step 2: Apply the property of even functions
Using the definite integral property for even functions: \[ \int_{-1}^{1} (1 - |x|) \, dx = 2 \int_{-1}^{0} (1 - |x|) \, dx \]

Step 3: Simplify the expression within the chosen sub-interval
In the interval \( [-1, 0] \), \( x \) is negative, so \( |x| = -x \). The integrand becomes: \[ 1 - |x| = 1 - (-x) = 1 + x \] Therefore: \[ 2 \int_{-1}^{0} (1 - |x|) \, dx = 2 \int_{-1}^{0} (1 + x) \, dx \]