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Popular Calculus >

integral of 1-|x-1|

Solution

\int 1 - ∣ x - 1 ∣ d x

Solution

x - \frac{( x - 1 ) ∣ x - 1 ∣}{2} + C

Solution steps

\int 1 - ∣ x - 1 ∣ d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d x - \int ∣ x - 1 ∣ d x

\int 1 d x = x

\int ∣ x - 1 ∣ d x = \frac{( x - 1 ) ∣ x - 1 ∣}{2}

= x - \frac{( x - 1 ) ∣ x - 1 ∣}{2}

Add a constant to the solution

= x - \frac{( x - 1 ) ∣ x - 1 ∣}{2} + C

Graph

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Frequently Asked Questions (FAQ)

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