The value of $\int_{- 2}^{2} min {x - [x], - x - [- x]} d x$ is equal to (where $[.]$ denotes the greatest integer function.)

\frac{1}{2}

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1

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\frac{3}{2}

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2

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Solution

The correct option is D $1$
From the graph of $f (x) = m i n (x - [x], - x - [x])$
$∴ \int_{- 2}^{2} f (x) d x = 4 \int_{0}^{1} f (x) d x = 4 \times \frac{1}{2} \times 1 \times \frac{1}{2} = 1$

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The value of ∫2−2min{x−[x],−x−[−x]}dx is equal to (where [.] denotes the greatest integer function.)

The correct option is D 1From the graph of f(x)=min(x−[x],−x−[x])∴∫2−2f(x)dx=4∫10f(x)dx=4×12×1×12=1

The value of $\int_{- 2}^{2} min {x - [x], - x - [- x]} d x$ is equal to (where $[.]$ denotes the greatest integer function.)

The correct option is D $1$
From the graph of $f (x) = m i n (x - [x], - x - [x])$
$∴ \int_{- 2}^{2} f (x) d x = 4 \int_{0}^{1} f (x) d x = 4 \times \frac{1}{2} \times 1 \times \frac{1}{2} = 1$