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

The integral from-1 to 1 of (x+|x|)^2 is 4/3

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

\int_{- 1}^{1} (x + ∣ x ∣)^{2} d x

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

\int_{- 1}^{1} (x + ∣ x ∣)^{2} d x

Eliminate Absolutes

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

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

\int_{0}^{1} (x + x)^{2} d x = \frac{4}{3}

= 0 + \frac{4}{3}

Simplify

= \frac{4}{3}

What is the integral from-1 to 1 of (x+|x|)^2 ?
The integral from-1 to 1 of (x+|x|)^2 is 4/3