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Evaluate: ∫01|x| dx

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Question

Evaluate: `int_0^1 |x|  "d"x`

Sum
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Solution

`int_0^1 |x|  "d"x - int_0^1 x  "d"x`   ......[|x| = x, x ≥ 0]

= `[x^2/2]_0^(-1)`

= `1/2(1^2 - 0)`

= `1/2`

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Methods of Evaluation and Properties of Definite Integral
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Chapter 2.4: Definite Integration - Very Short Answers

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