1.

Let `[x]` denote the greatest integer less than or equal to `x`. Then, `int_(1)^(-1)[x]dx=?`A. `-1`B. `0`C. `(1)/(2)`D. `2`

Answer» `{-1 le x lt 0implies[x]=-1}` and `{0 le x lt 1implies[x]=0}`
`:. I=int_(-1)^(0)[x]dx+int_(0)^(1)[x]dx=int_(-1)^(0)(-1)dx+int_(0)^(1)0*dx`
`=[-x]_(-1)^(0)=0-[-(-1)]=-1`.


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