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$\int\limits_{ - 1}^1 {\left[ {x + \left[ {x + \left[ x \right]} \right]} \right]} dx$ where $[.]$ denotes the greatest integer function, is equal toA. $- 2$B. $- 1$C. $- 3$D. $- 4$

Last updated date: 15th Aug 2024
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Hint: Whenever we have this type of question, one thing we need to remember is that the greatest integer function $\left[ x \right]$ breaks at integers which are usually discontinuous. So now we need to break the given limits as $- 1 \to 0$ and $0 \to 1$ then do the integration to arrive at the required answer.

Here in this type of problems, finding the integration of the greatest integer function. Greatest integer function is also known as step function or floor function. One thing we need to remember is that the greatest integer function $\left[ x \right]$ breaks at integers which are usually discontinuous. Hence we can divide the given limit $- 1 \to 1$ into two separate limit as $- 1 \to 0$ and $0 \to 1$. therefore we can write the given function which is $\int\limits_{ - 1}^1 {\left[ {x + \left[ {x + \left[ x \right]} \right]} \right]} dx$ as below.
$\int\limits_{ - 1}^1 {\left[ {x + \left[ {x + \left[ x \right]} \right]} \right]} dx = \int\limits_{ - 1}^0 {\left[ {x + \left[ {x + \left[ x \right]} \right]} \right]} dx + \int\limits_0^1 {\left[ {x + \left[ {x + \left[ x \right]} \right]} \right]} dx$
Whenever we have limit $- 1 \to 0$ the value inside the box or braces becomes $- 1$ and in case of limit $0 \to 1$ we get the value inside the box or braces as $0$, which can be written as below.
$\Rightarrow \int\limits_{ - 1}^1 {\left[ {x + \left[ {x + \left[ x \right]} \right]} \right]} dx = \int\limits_{ - 1}^0 {\left[ {x + \left[ {x + \left[ { - 1} \right]} \right]} \right]} dx + \int\limits_0^1 {\left[ {x + \left[ {x + \left[ 0 \right]} \right]} \right]} dx$
$\Rightarrow \int\limits_{ - 1}^1 {\left[ {x + \left[ {x + \left[ x \right]} \right]} \right]} dx = \int\limits_{ - 1}^0 {\left[ { - 1 + \left[ { - 1 + \left[ { - 1} \right]} \right]} \right]} dx + \int\limits_0^1 {\left[ {0 + \left[ {0 + \left[ 0 \right]} \right]} \right]} dx$
$\Rightarrow \int\limits_{ - 1}^1 {\left[ {x + \left[ {x + \left[ x \right]} \right]} \right]} dx = \int\limits_{ - 1}^0 { - 3} dx + \int\limits_0^1 0 dx$
Now integrate the function, we get
$\Rightarrow \int\limits_{ - 1}^1 {\left[ {x + \left[ {x + \left[ x \right]} \right]} \right]} dx = \left[ { - 3x} \right]_{ - 1}^0 + 0$
Now apply the limit, and simplify the expression. We get
$\Rightarrow \int\limits_{ - 1}^1 {\left[ {x + \left[ {x + \left[ x \right]} \right]} \right]} dx = - 3(0 - ( - 1)) + 0$
$\Rightarrow \int\limits_{ - 1}^1 {\left[ {x + \left[ {x + \left[ x \right]} \right]} \right]} dx = - 3(1) + 0 = - 3$

Hence the integration of greatest integer function $\int\limits_{ - 1}^1 {\left[ {x + \left[ {x + \left[ x \right]} \right]} \right]} dx$ is $- 3$. Therefore, the option C is the correct answer.

Note:
Whenever we have this type of problems, first we need to know the concept of greatest integer function, integration and simplifying the limits. And when integrating the function you should be clear with the integration concepts then only you can get the correct answer, also when simplifying the limits be careful.