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Popular >

integral from 0 to 2 of 1-|1-x|

Solution

\int_{0}^{2} 1 - ∣ 1 - x ∣ d x

Solution

1

Solution steps

\int_{0}^{2} 1 - ∣ 1 - x ∣ d x

Eliminate Absolutes

= \int_{0}^{1} 1 - (1 - x) d x + \int_{1}^{2} 1 - (- 1 + x) d x

\int_{0}^{1} 1 - (1 - x) d x = \frac{1}{2}

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

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

Simplify

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

= 1

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\int_{0}^{l n (14)} tanh (x) d x

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\int_{- 1}^{3} 5 x^{2} - 4 d x

Frequently Asked Questions (FAQ)

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