What is the integral from 0 to 1-|x-1| of 1 ?

The integral from 0 to 1-|x-1| of 1 is -|x-1|+1

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integral from 0 to 1-|x-1| of 1

\int_{0}^{1 - ∣ x - 1 ∣} 1 d y

- ∣ x - 1 ∣ + 1

Solution steps

\int_{0}^{1 - ∣ x - 1 ∣} 1 d y

Integral of a constant:

\int a d x = a x

= [1 \cdot y]_{0}^{1 - ∣ x - 1 ∣}

Simplify

= [y]_{0}^{1 - ∣ x - 1 ∣}

Compute the boundaries:

- ∣ x - 1 ∣ + 1

= - ∣ x - 1 ∣ + 1

\frac{\partial}{\partial x} (- cos^{2} (x))

\frac{d}{d x} (\frac{x ^{2}}{x - 7})

\int_{0}^{1} 2 π (x + 11) (x^{2}) d x

\int \frac{arctan ( x )}{x ^{2}} d x

What is the integral from 0 to 1-|x-1| of 1 ?
The integral from 0 to 1-|x-1| of 1 is -|x-1|+1