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Popular >

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

Solution

\int_{0}^{2} 1 - ∣ x - 1 ∣ d x

Solution

1

Solution steps

\int_{0}^{2} 1 - ∣ x - 1 ∣ d x

Eliminate Absolutes

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

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

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

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

Simplify

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

= 1

Graph

Popular Examples

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\int_{0}^{\frac{π}{2}} cos (x) \cdot sin (x) d x

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\int_{\frac{π}{6}}^{\frac{π}{3}} tan (x) sec^{2} (x) d x

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\int_{0}^{2} cos (x - 2) d x

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\int_{0}^{π} - 3 sin (t) cos (t) d t

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\int_{- 2}^{2} 4 x - x^{2} d x

Frequently Asked Questions (FAQ)

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