Answer
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Hint: -To solve this question use this property $\left( {x = \left[ x \right] + \left\{ x \right\}} \right)$ that is any number is equal to sum of its integral part and fractional part. And now just integrate simple terms and put the limit to get an answer.
Complete step-by-step answer:
We have
$f\left( x \right) = x - \left[ x \right]$
Now using this property $\left( {x = \left[ x \right] + \left\{ x \right\}} \right)$ we get,
$
f\left( x \right) = \left[ x \right] + \left\{ x \right\} - \left[ x \right] \\
\therefore f\left( x \right) = \left\{ x \right\} \\
$
Here f(x) is the fractional part of ‘x’ so to solve further we have to understand the fractional part of ‘x’ and we should be familiar with the properties of the fractional part of x.
Domain of fractional part is real number and its range is $\left[ {0,1} \right)$
Fractional part is discontinuous at all integers.
Now we have given in the question
$\int\limits_{ - 1}^1 {f\left( x \right)dx} $$ = \int\limits_{ - 1}^1 {\left\{ x \right\}dx} $
Now using properties of integration $\left( {\int\limits_{ - a}^a {f\left( x \right)dx} =
2\int\limits_0^a {f\left( x \right)dx} } \right)$
( this property can be used when function is periodic with period ‘a’, here it use be used
because period of fractional part is ‘1’ and we will use this property here)
Using this property we get,
= $2\int\limits_0^1 {\left\{ x \right\}dx} $
( here limit is from 0 to 1 and between 0 to 1 fractional part of x is nothing but x)
So from property of above line we get,
= $2\int\limits_0^1 {xdx} = 2{\left[ {\dfrac{{{x^2}}}{2}} \right]_0}^1 = 2\left[ {\dfrac{1}{2} - 0}
\right] = 1$
Hence option B is the correct option.
Note: - Whenever we get this type of question the key concept of solving is we have all knowledge about fractional part of x and also remember properties of integration so that we can solve the question easily. Property written in hints is very important so it should be remembered.
Complete step-by-step answer:
We have
$f\left( x \right) = x - \left[ x \right]$
Now using this property $\left( {x = \left[ x \right] + \left\{ x \right\}} \right)$ we get,
$
f\left( x \right) = \left[ x \right] + \left\{ x \right\} - \left[ x \right] \\
\therefore f\left( x \right) = \left\{ x \right\} \\
$
Here f(x) is the fractional part of ‘x’ so to solve further we have to understand the fractional part of ‘x’ and we should be familiar with the properties of the fractional part of x.
Domain of fractional part is real number and its range is $\left[ {0,1} \right)$
Fractional part is discontinuous at all integers.
Now we have given in the question
$\int\limits_{ - 1}^1 {f\left( x \right)dx} $$ = \int\limits_{ - 1}^1 {\left\{ x \right\}dx} $
Now using properties of integration $\left( {\int\limits_{ - a}^a {f\left( x \right)dx} =
2\int\limits_0^a {f\left( x \right)dx} } \right)$
( this property can be used when function is periodic with period ‘a’, here it use be used
because period of fractional part is ‘1’ and we will use this property here)
Using this property we get,
= $2\int\limits_0^1 {\left\{ x \right\}dx} $
( here limit is from 0 to 1 and between 0 to 1 fractional part of x is nothing but x)
So from property of above line we get,
= $2\int\limits_0^1 {xdx} = 2{\left[ {\dfrac{{{x^2}}}{2}} \right]_0}^1 = 2\left[ {\dfrac{1}{2} - 0}
\right] = 1$
Hence option B is the correct option.
Note: - Whenever we get this type of question the key concept of solving is we have all knowledge about fractional part of x and also remember properties of integration so that we can solve the question easily. Property written in hints is very important so it should be remembered.
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