- 1 1 ( | x - 2 | + | x | ) ? d x =

Question

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

Smart Achievers · Accepted Answer

(2)

Ans. 5

Explanation :

$I = \int_{- 1}^{1} (| x - 2 | + | x |) d x$

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

$= - {[\frac{x^{2}}{2} - 2 x]}_{- 1}^{1} - {[\frac{x^{2}}{2}]}_{- 1}^{0} + {[\frac{x^{2}}{2}]}_{0}^{1}$

$= 1 + \frac{3}{2} + \frac{5}{2} + \frac{1}{2} + \frac{1}{2}$

$= \frac{10}{2} = 5$

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