Question

The integral \( \int_{-1}^{1} (1 - |x|) \, dx \) 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:

For definite integrals with symmetric limits \( [-a, a] \), always evaluate if the function is even or odd. If even, remember that \( \int_{-a}^{a} f(x)dx = 2\int_{-a}^{0} f(x)dx \), and then replace \( |x| \) with \( -x \) since \( x \) is negative in that domain.

Answer 1

The Correct Option is B

Concept: This problem involves using the properties of definite integrals, specifically the behavior of even functions over a symmetric interval \( [-a, a] \). Recall that if a function \( f(x) \) satisfies \( f(-x) = f(x) \), it is called an even function, and its integral over a symmetric domain satisfies: \[ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx = 2 \int_{-a}^{0} f(x) \, dx \]

Step 1: Determine the parity (even/odd property) of the integrand.
Let our integrand function be defined as: \[ f(x) = 1 - |x| \] Let us substitute \( -x \) in place of \( x \): \[ f(-x) = 1 - |-x| \] Since the absolute value function removes negative signs, \( |-x| = |x| \). Therefore: \[ f(-x) = 1 - |x| = f(x) \] Since \( f(-x) = f(x) \), the function \( f(x) \) is strictly an even function.

Step 2: Split and evaluate using interval properties.
For an even function, the total area under the curve from \( -1 \) to \( 1 \) is twice the area of either the positive half or the negative half: \[ \int_{-1}^{1} (1 - |x|) \, dx = 2 \int_{0}^{1} (1 - |x|) \, dx = 2 \int_{-1}^{0} (1 - |x|) \, dx \]

Step 3: Analyze the definition of absolute value in the intervals.
Let's check the behavior of \( |x| \) within the interval of integration for the options given: - In the interval \( [-1, 0] \), \( x \le 0 \), which implies by definition that \( |x| = -x \). Substituting this into our integral expression over \( [-1, 0] \): \[ 2 \int_{-1}^{0} (1 - (-x)) \, dx = 2 \int_{-1}^{0} (1 + x) \, dx \] This precisely matches option (B).