Evaluate `int_(-1)^(3) [x]dx` ,where [.] denotes the greatest integer function.

Answer

A, C

Explanation

`S="Area of" Delta OAC+ "Area of" DeltaBCD`
`=(1.x)/(2)+(1-x)(y-1)/(2),0 lt x lt 1`
`therefore S=(x)/(2)-(x-1)(y-1)/(2)` (1)
Now,`Delta`'s CBD and OCA are similar. Therfore,
`(y-1)/(1)=(1-x)/(x)`
`therefore S=(x)/(2)-((x-1){(1//x)-1})/(2)`
`=(x)/(2)+(x-1)^(2)/(2x)`
`=(x^2+(x-1)^1)/(2x)=((2x^2-2x+1))/(2x)`
`=x+(1)/(2x)-1=(sqrtx-(1)/(sqrt(2x)))^(2)-1+sqrt(2)`
Therefore, A is minimum if `sqrt(x)=1//sqrt(2x)`. therefore , `x=1//sqrt(2)`
which lies in `(2//3,1)` and `A_("min")=sqrt(2)` which lies in `(1//3,1//2)`.