Question

\[\int_0^{1.5} {{{\left[ x \right]}^2}dx} \], where $\left[ . \right]$denotes the greatest integer function, equals.

Answer 1

Hint:The greatest integer function gives different values in different situations, so we break it into parts to solve and then break limits as per the break parts of the greatest integer function.

Complete step-by-step answer:
Given: we have given the\[\int_0^{1.5} {{{\left[ x \right]}^2}dx} \]where $\left[ . \right]$is the greatest function and we have to find the value of \[\int_0^{1.5} {{{\left[ x \right]}^2}dx} \].
First, we will calculate the upper limit and the lower limit of the ${x^2}$.
Limit of $x$ is 0 to $1.5$.
So, limit of ${x^2}$is ${\left( 0 \right)^2}$to ${\left( {1.5} \right)^2}$.
That is 0 to $2.25$.
Now, we will break our limits in different interval because the greatest integer function behave differently in different interval.
So, we break it like.
\[I = \int_0^{1.5} {{{\left[ x \right]}^2}dx} \]
The limit can be break at that point where the function is discontinuous, the function is discontinuous at 1 and $\sqrt 2 $.
So we can break it into the limits 0 to 1 and 1 to $\sqrt 2 $ because the value at $\sqrt 2 $is 2, it is discontinued at $\sqrt 2 $, and $\sqrt 2 $ to $1.5$.
So,
\[I = \int_0^1 {{{\left[ x \right]}^2}dx} + \int_1^{\sqrt 2 } {{{\left[ x \right]}^2}dx} + \int_{\sqrt 2 }^{1.5} {{{\left[ x \right]}^2}dx} \]
Now, calculate the values of greater integer function.
G.I.F ( Greater integer function ) values from 0 to 1 is 0.
G.I.f value from 1 to $\sqrt 2 $is 1.
G.I.F value from $\sqrt 2 $to $1.5$ is 2.
\[I = \int_0^1 {{0^2}dx} + \int_1^{\sqrt 2 } {{1^2}dx} + \int_{\sqrt 2 }^{1.5} {{2^2}dx} \]
Now, solve this integral by using the standard formulas of integration that is $\int\limits_a^b {ndx} = \left[ {nx} \right]_a^b$,
$ = 0 + \mathop {\left| x \right|}\nolimits_1^{\sqrt 2 } + 2\mathop {\left| x \right|}\nolimits_{\sqrt 2 }^{1.5} $
Now, use second Fundamental theorem of integral calculus - definition of definite integral to simplify further.
$
   = \sqrt 2 - 1 + 2\left( {1.5 - \sqrt 2 } \right) \\
   = \sqrt 2 - 1 + 3 - 2\sqrt 2 \\
   = 2 - \sqrt 2 \\
$
So, $\int_0^{1.5} {{{\left[ x \right]}^2}dx} = 2 - \sqrt 2 $is the required answer.

Note:
Break the upper limit and lower limit by keeping in mind about ${x^2}$not about $x$, due to this we may lead to the incorrect answer.