∫−11 max{x−[x],−x−[−x]} dx is equal to - Infinity Learn by Sri Chaitanya

A
3
B
$\frac{3}{2}$
C
$\frac{3}{4}$
D
2

Solution:

$\begin{array}{l} \int_{- 1}^{1} max ({x}, {- x}) dx \\ = \int_{- 1}^{- 0 .5} {- x} dx + \int_{- 0 .5}^{0} {x} dx + \int_{0}^{0 .5} {- x} dx + \int_{0 .5}^{1} {x} dx \\ = \int_{- 1}^{- 0 .5} (- x) dx + \int_{- 0 .5}^{0} {x - [x]} dx + \int_{0}^{0 .5} (- x - [- x]) dx \\ + \int_{0 .5}^{1} (x - [x]) dx \end{array}$

$\begin{array}{l} = {[- \frac{x^{2}}{2}]}_{- 1}^{- 0 .5} + {[\frac{x^{2}}{2} + x]}_{- 0 .5}^{0} + {[x - \frac{x^{2}}{2}]}_{0}^{0 .5} + {[\frac{x^{2}}{2}]}_{0 .5}^{1} \\ = - \frac{0 .25}{2} + \frac{1}{2} + (- \frac{0 .25}{2} + 0 .5) + 0 .5 - \frac{0 .25}{2} + \frac{1}{2} - \frac{0 .25}{2} \\ = 2 - \frac{1}{2} = \frac{3}{2} \end{array}$

$\int_{- 1}^{1} max {x - [x], - x - [- x]}$ dx is equal to

Solution:

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