The area of the region bounded by $y = [x]$ and the ordinates $x = 1, x = 2$ in sq. units is

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Solution

The correct option is A 1
$y = [x] = 1$ for $1 \leq x < 2$
$\int_{1}^{2} [x] d x = \int_{1}^{2} 1 \cdot d x = x {∣ ∣}_{1}^{2} = 2 - 1 = 1 s q . u n i t .$

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