Question

The value of \int\limits_{-1}^{2}{\left| \left[ x \right]-\left\\{ x \right\\} \right|}dx where $\left[ x \right]$ the greatest integer less than or equal to x and \left\\{ x \right\\} is the fractional part of x is:
(a). ${7}/{2}\;$
(b). ${5}/{2}\;$
(c). ${1}/{2}\;$
(d). ${3}/{2}\;$

Answer 1

In this question, we will break the integration sign in three intervals such that we can write the values of functions used and then integrate them to find required value.

Complete step-by-step solution -
For any real number x, $\left[ x \right]$, that is greatest integer function, represents that integer number which is less than x, but greater than or equal to all other integers which are less than x and $\left[ x \right]$is equal to x if x is integer.
Example: $\left[ 7.5 \right]=7$ , $7<7.5$ but $7\ge 7,6,5\ldots$
Also, {x} is a fractional part function, that is for any real number x, digits after decimal with zero before decimal will be{x}.
Example: \left\\{ 7.5 \right\\}=0.5
Here, when we add [x] and {x}, we actually add the value of x with decimal points with its decimal points, so we will get x.
\begin{aligned} & \left[ 7.5 \right]+\left\\{ 7.5 \right\\}=7+0.5 \\\ & =7.5 \\\ \end{aligned}
\begin{aligned} & \left[ x \right]+\left\\{ x \right\\}=x \\\ & \Rightarrow \left\\{ x \right\\}=x-\left[ x \right]\ldots \ldots (i) \\\ \end{aligned}
Now, we are to find value of,
\int\limits_{-1}^{2}{\left| \left[ x \right]-\left\\{ x \right\\} \right|}dx
Using equation (i), we get,
\begin{aligned} & \int\limits_{-1}^{2}{\left| \left[ x \right]-\left\\{ x \right\\} \right|}dx=\int\limits_{-1}^{2}{\left| \left[ x \right]-\left( x-\left[ x \right] \right) \right|}dx \\\ & =\int\limits_{-1}^{2}{\left| \left[ x \right]-x-\left[ x \right] \right|}dx \\\ & =\int\limits_{-1}^{2}{\left| 2\left[ x \right]-x \right|}dx \\\ \end{aligned}
Breaking integer sign in three parts, we get,
\int\limits_{-1}^{2}{\left| \left[ x \right]-\left\\{ x \right\\} \right|}dx=\int\limits_{-1}^{0}{\left| 2\left[ x \right]-x \right|}dx+\int\limits_{0}^{1}{\left| 2\left[ x \right]-x \right|}dx+\int\limits_{1}^{2}{\left| 2\left[ x \right]-x \right|}dx\ldots \ldots (ii)
Now, value of $\left[ x \right]$can be given as,

& -1,\,-1\le x<0 \\\ & 0,\,0\le x<1 \\\ & 1,\,1\le x<2 \\\ \end{aligned} \right.$$ Using this value of $\left[ x \right]$in equation (ii), we get, $$\begin{aligned} & \int\limits_{-1}^{2}{\left| \left[ x \right]-\left\\{ x \right\\} \right|}dx=\int\limits_{-1}^{0}{\left| 2\left[ -1 \right]-x \right|}dx+\int\limits_{0}^{1}{\left| 2\left[ 0 \right]-x \right|}dx+\int\limits_{1}^{2}{\left| 2\left[ 1 \right]-x \right|}dx \\\ & =\int\limits_{-1}^{0}{\left| -2-x \right|}dx+\int\limits_{0}^{1}{\left| 0-x \right|}dx+\int\limits_{1}^{2}{\left| 2-x \right|}dx\ldots \ldots (iii) \\\ \end{aligned}$$ Here, in range -1 to 0, x is negative and less than 2 so, $-2-x$ is also negative. $\left| -2-x \right|=-\left[ -2-x \right]=2+x$ . In range 0 to 1 of x, x is positive, So, -x is negative. Therefore $\left| -x \right|=-\left( -x \right)=x$ . In the range of 1 to 2, x is positive, but less than 2. So, $2-x$ is positive. Therefore $\left| 2-x \right|=2-x$ . Using these values in equation (iii), we get, $$\begin{aligned} & \int\limits_{-1}^{2}{\left| \left[ x \right]-\left\\{ x \right\\} \right|}dx=\int\limits_{-1}^{0}{\left( 2+x \right)}dx+\int\limits_{0}^{1}{x}dx+\int\limits_{1}^{2}{\left( 2-x \right)}dx \\\ & =\left[ 2x+\dfrac{{{x}^{2}}}{2} \right]_{-1}^{0}+\left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{1}+\left[ 2x-\dfrac{{{x}^{2}}}{2} \right]_{1}^{2} \\\ & =\left( 2\times 0+\dfrac{{{0}^{2}}}{2} \right)-\left( 2\left( -1 \right)+\dfrac{{{\left( -1 \right)}^{2}}}{2} \right)+\left( \dfrac{{{1}^{2}}}{2} \right)-\left( \dfrac{{{0}^{2}}}{2} \right)+\left( 2\left( 2 \right)-\dfrac{{{2}^{2}}}{2} \right)-\left( 2\times 1-\dfrac{{{\left( 1 \right)}^{2}}}{2} \right) \\\ \end{aligned}$$ Opening brackets, we get, $\begin{aligned} & \int\limits_{-1}^{2}{\left| \left[ x \right]-\left\\{ x \right\\} \right|}dx=0-\left( -2+\dfrac{1}{2} \right)+\dfrac{1}{2}-0+\left( 4-\dfrac{4}{2} \right)-\left( 2-\dfrac{1}{2} \right) \\\ & =2-\dfrac{1}{2}+\dfrac{1}{2}+4-\dfrac{4}{2}-2+\dfrac{1}{2} \\\ & =4-\dfrac{3}{2} \\\ \end{aligned}$ Taking LCM, we get, $\int\limits_{-1}^{2}{\left| \left[ x \right]-\left\\{ x \right\\} \right|}dx=\dfrac{8-3}{2}=\dfrac{5}{2}$ Hence, the value of the given integral is $\dfrac{5}{2}$ . **Note:** In this type of questions, where $\left\\{ x \right\\}$is used, try to cancel it or write it in the form $x-\left[ x \right]$ , otherwise integrating it will be difficult.