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

\int_{0}^{20} x^{2} - 17 x + 70 d x

\int_{- 1}^{1} 6 t 4 t^{2} + 1 d t

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

\int_{e}^{e^{5}} \frac{1}{x ln ( x )} d x

\int_{0}^{12} 462 e^{0.019 x} d x

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