What is the integral of |x-1| ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of |x-1|

\int ∣ x - 1 ∣ d x

\frac{( x - 1 ) ∣ x - 1 ∣}{2} + C

Solution steps

\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

\int \frac{- 2}{x ^{2} + 16} d x

e^{\frac{y}{x}} \frac{( - y d x + x d y )}{x ^{2}} = 0

f (x) = \frac{8 x}{x ^{2} - 1}, (3, 3)

\frac{d}{d x} (3 x^{2} - 5 x + 4)

x \to 3 + lim (x^{2} - 9)