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

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

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

\int_{- 1}^{x} 1 - ∣ x ∣ d x

x + 1 - \frac{x ∣ x ∣ + 1}{2}

Solution steps

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

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

\int_{- 1}^{x} ∣ x ∣ d x = \frac{x ∣ x ∣ + 1}{2}

= x + 1 - \frac{x ∣ x ∣ + 1}{2}

\int_{- 3}^{- 1} \frac{x ^{5} - 2 x}{x ^{3}} d x

\int_{0}^{2} x (\frac{3}{4} x (2 - x)) d x

\int_{y}^{y + 1} (x + y) d x

\int_{0}^{3} (x^{3}) d x

\int_{0}^{11} 2000 - 12 x^{2} d x

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