What is the integral from-1 to 1 of (|x|-1/2)^2 ?

The integral from-1 to 1 of (|x|-1/2)^2 is 1/6

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integral from-1 to 1 of (|x|-1/2)^2

\int_{- 1}^{1} (∣ x ∣ - \frac{1}{2})^{2} d x

\frac{1}{6}

Decimal

0.16666 \dots

Solution steps

\int_{- 1}^{1} (∣ x ∣ - \frac{1}{2})^{2} d x

Eliminate Absolutes

= \int_{- 1}^{0} (- x - \frac{1}{2})^{2} d x + \int_{0}^{1} (x - \frac{1}{2})^{2} d x

\int_{- 1}^{0} (- x - \frac{1}{2})^{2} d x = \frac{1}{12}

\int_{0}^{1} (x - \frac{1}{2})^{2} d x = \frac{1}{12}

= \frac{1}{12} + \frac{1}{12}

Simplify

\frac{1}{12} + \frac{1}{12} : \frac{1}{6}

= \frac{1}{6}

What is the integral from-1 to 1 of (|x|-1/2)^2 ?
The integral from-1 to 1 of (|x|-1/2)^2 is 1/6