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

Answer: (A)

1

Answer 2

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}
$\therefore \int^1_{-1} [ x ] \ dx = \int^0_{-1} [ x ] \ dx + \int^1_0 [ x ] \ dx$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int^0_{-1} ( - 1) dx + \int^1 _0 \ 0 \ dx$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - [ x ]^0_{-1} + 0 = {-1}$
$\therefore \int^1_{-1} f ( x ) \ dx = 1$