Question

\(\displaystyle \int_{-a}^{a}|x|\,dx=\)

• A. \(a\)
• B. \(2a\)
• C. \(0\)
• D. \(a^2\)

Hint:

Since \(|x|\) is an even function, use \(\int_{-a}^{a}|x|dx=2\int_0^a xdx\).

Answer 1

The Correct Option is D

We need to evaluate: \[ \int_{-a}^{a}|x|\,dx. \] The function \(|x|\) is an even function because \[ |-x|=|x|. \] For an even function \(f(x)\), \[ \int_{-a}^{a}f(x)\,dx=2\int_0^a f(x)\,dx. \] Therefore, \[ \int_{-a}^{a}|x|\,dx=2\int_0^a |x|\,dx. \] For \(x\geq 0\), \[ |x|=x. \] So, \[ 2\int_0^a |x|\,dx=2\int_0^a x\,dx. \] Now integrate: \[ 2\int_0^a x\,dx=2\left[\frac{x^2}{2}\right]_0^a. \] \[ =\left[x^2\right]_0^a. \] \[ =a^2-0^2. \] \[ =a^2. \] Hence, \[ \int_{-a}^{a}|x|\,dx=a^2. \]