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

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

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

\int_{- 1}^{1} (x + ∣ x - 1 ∣) d x

2

Solution steps

\int_{- 1}^{1} x + ∣ x - 1 ∣ d x

Eliminate Absolutes

= \int_{- 1}^{1} x - x + 1 d x

Simplify

= \int_{- 1}^{1} 1 d x

Integral of a constant:

\int a d x = a x

= [1 \cdot x]_{- 1}^{1}

Simplify

= [x]_{- 1}^{1}

Compute the boundaries:

2

= 2

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