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Popular Calculus >

integral from 0 to 2 of |t-1|

Solution

\int_{0}^{2} ∣ t - 1 ∣ d t

Solution

1

Solution steps

\int_{0}^{2} ∣ t - 1 ∣ d t

Eliminate Absolutes

= \int_{0}^{1} - t + 1 d t + \int_{1}^{2} t - 1 d t

\int_{0}^{1} - t + 1 d t = \frac{1}{2}

\int_{1}^{2} t - 1 d t = \frac{1}{2}

= \frac{1}{2} + \frac{1}{2}

Simplify

\frac{1}{2} + \frac{1}{2} : 1

= 1

Graph

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Frequently Asked Questions (FAQ)

What is the integral from 0 to 2 of |t-1| ?
The integral from 0 to 2 of |t-1| is 1