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Let f(x) = x-[x], for every real number ...

Let `f(x) = x-[x]`, for every real number x, where [x] is integral part of x. Then `int_(-1) ^1 f(x) dx` is

A

1

B

2

C

0

D

`-(1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Let `int_(-1)^(1)f(x)dx=int_(-1)^(1)(x-[x])dx=int_(-1)^(1) xdx-int_(-1)^(1)[x]dx`
`=0-int_(-1)^(1)[x]dx[:'x "is an odd function"]`
But `[x]={{:(-1_(,),if,1lexlt0),(0_(,),if,0lexlt1),(1_(,),if,x=1):}`
`:. int_(-1)^(1)[x]dx = int_(-1)^(0)[x]dx+int_(0)^(1)[x]dx`
`=int_(-1)^(0)(-1)dx+int_(0)^(1)0dx`
`=-[x]_(-1)^(0)+0=-1,:. int_(-1)^(1)f(x)dx =1`
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