Question

Let $f(x) = x - [x] $, for every real number x, where [x] is the integral part of x Then, $ \int^1_{-1} \, f ( x ) \, dx$ is

• A. 1
• B. 2
• C. 0
• D. $- \frac{ 1 }{2 }$

Answer 1

The Correct Option is A

The correct option is(A): 1.

\(\int^1_{-1} \ f ( x ) \ dx \ = \int^1_{-1} ( x - [x] ) \ dx = \int^1_{-1} \ x \ dx - \int^1_{-1} [x ] \ dx\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0 - \int^1_{-1} [ x ] \ dx [ \because \ x \ is \ an \ odd \ number ]\) 
But $[ x ] = \bigg \{ \begin{array}
\ -1 \\
0, \\
1, \\
\end {array} \begin{array}
\ \ \ if \\
\ \ if \\
\ \ if \\
\end {array} \begin{array}
\ \ \ -1 \le x < 0 \\
\ \ \le x < 1 \\
\ \ \ x = 1 \\
\end {array}$ 
\(\therefore \int^1_{-1} [ x ] \ dx = \int^0_{-1} [ x ] \ dx + \int^1_0 [ x ] \ dx\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int^0_{-1} ( - 1) dx + \int^1 _0 \ 0 \ dx\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - [ x ]^0_{-1} + 0 = {-1}\)
\(\therefore \int^1_{-1} f ( x ) \ dx = 1\)