Let $\left[ x \right]$denotes the greatest integer less than or equal to x, then the value of the integral $\int\limits_{ - 1}^1 {\left( {\left| x \right| - 2\left[ x \right]} \right)dx} $ is equal to
$
(a){\text{ 3}} \\
(b){\text{ 2}} \\
(c){\text{ - 2}} \\
(d){\text{ - 3}} \\
$
Last updated date: 16th Mar 2023
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Answer
303.9k+ views
Hint – In this question we have to evaluate the integral and the functions involved inside are modulus function and greatest integer function. Firstly break the limits that are instead of directly having from -1 to 1 have them from -1 to 1 the form 0 to 1. The greatest integer function is defined only for intervals with stride of 1 unit. Then evaluate the integral.
Complete step-by-step answer:
Given integral is
$I = \int\limits_{ - 1}^1 {\left( {\left| x \right| - 2\left[ x \right]} \right)dx} $, where $\left[ x \right]$ denotes greatest integer function.
Now as we know in the interval $\left( { - \infty ,0} \right)$ the value of the greatest integer function is -1.
And in the interval $\left( {0,\infty } \right)$ the value of the greatest integer function is zero (0).
So break the integration limits and put the value of greatest integer function.
$I = \int\limits_{ - 1}^0 {\left( {\left| x \right| - 2\left( { - 1} \right)} \right)dx} + \int\limits_0^1 {\left( {\left| x \right| - 2\left( 0 \right)} \right)dx} $
$ \Rightarrow I = \int\limits_{ - 1}^0 {\left( {\left| x \right| + 2} \right)dx} + \int\limits_0^1 {\left( {\left| x \right|} \right)dx} $
Now as we know that the value of $\left| x \right|$ is -x on L.H.S and +x on R.H.S
I.e. $\left\{ \begin{gathered}
\left| x \right| = - x,{\text{ }}x \in \left[ { - 1,0} \right) \\
\left| x \right| = x,{\text{ }}x \in \left( {0,1} \right] \\
\end{gathered} \right.$
So, substitute this value in above equation we have,
$ \Rightarrow I = \int\limits_{ - 1}^0 {\left( { - x + 2} \right)dx} + \int\limits_0^1 {x{\text{ }}dx} $
Now apply the integration we have,
$ \Rightarrow I = \left[ {\dfrac{{ - {x^2}}}{2} + 2x} \right]_{ - 1}^0 + \left[ {\dfrac{{{x^2}}}{2}} \right]_0^1$
Now apply the integration limit we have,
$ \Rightarrow I = \left[ {0 + 0 - \left( {\dfrac{{ - {{\left( { - 1} \right)}^2}}}{2} + 2\left( { - 1} \right)} \right)} \right] + \left[ {\dfrac{{{1^2}}}{2} - 0} \right]$
Now simplify the above equation we have,
$ \Rightarrow I = \dfrac{1}{2} + 2 + \dfrac{1}{2} = 1 + 2 = 3$
So this is the required value of the integral.
Hence option (a) is correct.
Note – Whenever we face type of questions the key concept is to have the understanding of modulus function and greatest integer function. The basic definition of modulus function is $\left\{ \begin{gathered}
\left| x \right| = - x,{\text{ }}x < 0 \\
\left| x \right| = x,{\text{ }}x \geqslant 0 \\
\end{gathered} \right.$. Its plot is a pair of straight lines passing through origin and changing the slopes at point (0, 0). A greatest integer function is denoted by less than or equal to x for $\left[ x \right]$, it basically rounds down a real number to the nearest integer. These concepts along with some basic integration properties will help you get on the right track to reach the answer.
Complete step-by-step answer:
Given integral is
$I = \int\limits_{ - 1}^1 {\left( {\left| x \right| - 2\left[ x \right]} \right)dx} $, where $\left[ x \right]$ denotes greatest integer function.
Now as we know in the interval $\left( { - \infty ,0} \right)$ the value of the greatest integer function is -1.
And in the interval $\left( {0,\infty } \right)$ the value of the greatest integer function is zero (0).
So break the integration limits and put the value of greatest integer function.
$I = \int\limits_{ - 1}^0 {\left( {\left| x \right| - 2\left( { - 1} \right)} \right)dx} + \int\limits_0^1 {\left( {\left| x \right| - 2\left( 0 \right)} \right)dx} $
$ \Rightarrow I = \int\limits_{ - 1}^0 {\left( {\left| x \right| + 2} \right)dx} + \int\limits_0^1 {\left( {\left| x \right|} \right)dx} $
Now as we know that the value of $\left| x \right|$ is -x on L.H.S and +x on R.H.S
I.e. $\left\{ \begin{gathered}
\left| x \right| = - x,{\text{ }}x \in \left[ { - 1,0} \right) \\
\left| x \right| = x,{\text{ }}x \in \left( {0,1} \right] \\
\end{gathered} \right.$
So, substitute this value in above equation we have,
$ \Rightarrow I = \int\limits_{ - 1}^0 {\left( { - x + 2} \right)dx} + \int\limits_0^1 {x{\text{ }}dx} $
Now apply the integration we have,
$ \Rightarrow I = \left[ {\dfrac{{ - {x^2}}}{2} + 2x} \right]_{ - 1}^0 + \left[ {\dfrac{{{x^2}}}{2}} \right]_0^1$
Now apply the integration limit we have,
$ \Rightarrow I = \left[ {0 + 0 - \left( {\dfrac{{ - {{\left( { - 1} \right)}^2}}}{2} + 2\left( { - 1} \right)} \right)} \right] + \left[ {\dfrac{{{1^2}}}{2} - 0} \right]$
Now simplify the above equation we have,
$ \Rightarrow I = \dfrac{1}{2} + 2 + \dfrac{1}{2} = 1 + 2 = 3$
So this is the required value of the integral.
Hence option (a) is correct.
Note – Whenever we face type of questions the key concept is to have the understanding of modulus function and greatest integer function. The basic definition of modulus function is $\left\{ \begin{gathered}
\left| x \right| = - x,{\text{ }}x < 0 \\
\left| x \right| = x,{\text{ }}x \geqslant 0 \\
\end{gathered} \right.$. Its plot is a pair of straight lines passing through origin and changing the slopes at point (0, 0). A greatest integer function is denoted by less than or equal to x for $\left[ x \right]$, it basically rounds down a real number to the nearest integer. These concepts along with some basic integration properties will help you get on the right track to reach the answer.
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