What is the integral of |t-1| ?

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

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integral of |t-1|

\int ∣ t - 1 ∣ d t

\frac{( t - 1 ) ∣ t - 1 ∣}{2} + C

Solution steps

\int ∣ t - 1 ∣ d t

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 = t - 1

= \frac{( t - 1 ) ∣ t - 1 ∣}{2}

Add a constant to the solution

= \frac{( t - 1 ) ∣ t - 1 ∣}{2} + C

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