What is the integral of |1-x| ?

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

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integral of |1-x|

\int ∣ 1 - x ∣ d x

- \frac{( 1 - x ) ∣ 1 - x ∣}{2} + C

Solution steps

\int ∣ 1 - x ∣ d x

Apply u-substitution

: \int - ∣ u ∣ d u

= \int - ∣ u ∣ d u

Take the constant out:

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

= - \int ∣ u ∣ d u

Use the common integral:

\int ∣ u ∣ d u = \frac{u ∣ u ∣}{2}

= - \frac{u ∣ u ∣}{2}

Substitute back

u = 1 - x

= - \frac{( 1 - x ) ∣ 1 - x ∣}{2}

Add a constant to the solution

= - \frac{( 1 - x ) ∣ 1 - x ∣}{2} + C

\frac{d}{d x} (\frac{\frac{d}{d x} ( 3 )}{x})

\frac{d}{d x} (\frac{1}{3 + 2 x})

y^{^{'}} = - x y^{3}

x \to 0 - lim (g (x))

\int x^{3} 2^{x} d x