What is the integral of-|x-1| ?

The integral of-|x-1| is -((x-1)|x-1|)/2+C

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integral of-|x-1|

\int - ∣ x - 1 ∣ d x

- \frac{( x - 1 ) ∣ x - 1 ∣}{2} + C

Solution steps

\int - ∣ x - 1 ∣ d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int ∣ x - 1 ∣ d x

Apply u-substitution

: \int ∣ u ∣ d u

= - \int ∣ u ∣ d u

Use the common integral:

\int ∣ u ∣ d u = \frac{u ∣ u ∣}{2}

= - \frac{u ∣ u ∣}{2}

Substitute back

u = x - 1

= - \frac{( x - 1 ) ∣ x - 1 ∣}{2}

Add a constant to the solution

= - \frac{( x - 1 ) ∣ x - 1 ∣}{2} + C

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