Question

The value of the integral `int_(-1)^2 [x]  dx` is 

Options

• `5/2`

• 3

• 1

• 0

Answer 1

0

Explanation:

`int_(-1)^2 [x]  dx`

`int_(-1)^0 [x] dx + int_0^1 [x] dx + int_1^2 [x] dx`

{Since [x] are discontinuous on integral values}

`int_(-1)^0 - 1 dx + int_0^1 0 dx + int_1^2 1 dx`

`- [x]_(-1)^0 + 0 + [x]_1^2`

`- (0 + 1) + 0 + [2 - 1]`

`- 1 + 0 + 1` = 0