Question

The value of the integral $\int {\left[ x \right]dx} $ is
A) $\dfrac{{{{[x]}^2}}}{2}\forall x > 0$
B) $\dfrac{{[{x^2}]}}{2}\forall x > 0$
C) $5x - 15$ for $x \in [0,t],t \in (5,6)$
D) Cannot be evaluated

Answer 1

Hint: Greatest integer function is a discontinuous function, so to determine the integral of a discontinuous function, the function is separated into intervals of continuous parts and then summing up all the integrals.

Complete step by step answer:
We are required to evaluate the integral $\int {\left[ x \right]dx} $. We will proceed by considering each option separately.
Consider (A), that is $\dfrac{{{{[x]}^2}}}{2}\forall x > 0$.
To prove that the given option is not correct, we require a counter example, let us say for $x = 3.7$, so we need to determine the integral of the given function and check it with the given option if the answer is correct or not.
$\int\limits_0^{3.7} {[x]dx} $
Since it is a discontinuous function, so break it into different continuous parts, since for any integral of $[x]$ between $n$and $n + 1$,$\int\limits_n^{n + 1} {[x]dx = } \int\limits_n^{n + 1} {ndx} $, so we get
$\int\limits_0^{3.7} {[x]dx} = \int\limits_0^1 {0dx} + \int\limits_1^2 {1dx} + \int\limits_2^3 {2dx + \int\limits_3^{3.7} {3dx} } $
$\int\limits_0^{3.7} {[x]dx} = [0]_0^1 + [1]_1^2 + [2]_2^3 + [3]_3^{3.7}$
$\int\limits_0^{3.7} {[x]dx} = 0(1 - 0) + 1(2 - 1) + 2(3 - 2) + 3(3.7 - 3)$
$\int\limits_0^{3.7} {[x]dx} = 0 + 1 + 2 + 3(0.7) = 5.1$
Now, substitute the value of $x = 3.7$ in the value of integral given in option (A) $\dfrac{{{{[x]}^2}}}{2}$
$\dfrac{{{{[x]}^2}}}{2} = \dfrac{{{{[3.7]}^2}}}{2} = \dfrac{{{3^2}}}{2} = \dfrac{9}{2} = 4.5$
Since the value is not the same, so option (A) is not correct.
Now, consider option (B) that is $\dfrac{{[{x^2}]}}{2}\forall x > 0$
To prove that the given option is not correct, we require a counter example, let us say for $x = 3.7$, so we need to determine the integral of the given function and check it with the given option if the answer is correct or not.
As, we have already evaluated that $\int\limits_0^{3.7} {[x]dx} = 5.1$, we only need to substitute the value of $x = 3.7$ in the value of integral given in option (B) $\dfrac{{[{x^2}]}}{2}$
$\dfrac{{[{x^2}]}}{2} = \dfrac{{[{{3.7}^2}]}}{2} = \dfrac{{[13.69]}}{2} = \dfrac{{13}}{2} = 6.5$
Since the value is not the same, option (B) is not correct.
Consider (C), that is $5x - 15$ for $x \in [0,t],t \in (5,6)$,
Consider any arbitrary $t \in (5,6)$. Now we must determine the integral $\int\limits_0^t {[x]dx} $.
Break the integral into continuous parts,
$\int\limits_0^t {[x]dx} = \int\limits_0^1 {0dx} + \int\limits_1^2 {1dx} + \int\limits_2^3 {2dx} + \int\limits_3^4 {3dx} + \int\limits_4^5 {4dx} + \int\limits_5^t {5dx} $
Now, evaluate the other integral
$\int\limits_0^t {[x]dx} = [0]_0^1 + [1]_1^2 + [2]_2^3 + [3]_3^4 + [4]_4^5 + [5]_5^t$
$\int\limits_0^t {[x]dx} = 0(1 - 0) + 1(2 - 1) + 2(3 - 2) + 3(4 - 3) + 4(5 - 4) + 5(t - 5)$
$\int\limits_0^t {[x]dx} = 0 + 1 + 2 + 3 + 4 + 5t - 25$
$\int\limits_0^t {[x]dx} = 5t - 15$
The value of the integral is $5x - 15$ for $x \in [0,t],t \in (5,6)$.
So, option (C) is the correct.

Note:The greatest integer function is also known as the step function or floor function. The greatest integer function goes up to the nearest integer that is less than or equal to the provided number. The biggest integer function has R as its domain and $\mathbb{Z}$ as its range.