If \[\int_0^a {{e^{x - \left[ x \right]}}} dx = 10e - 9\],then what is the value of a? (Where \[\left[ . \right]\]is the greatest integer function)
A.\[9 + \ln 2\]
B. \[10 + \ln 2\]
C. \[10\]
D. \[9\]

Hint: We will try to remove the greatest integer function by breaking x in appropriate intervals and then applying the integration for \[{e^{x - \alpha }}\] (where α is real no.) to integrate them individually, after solving the integrals individually we will sum up all of them and comparing the calculated value of integral with the given value of integral.

Formula used:
1. \[\left[ x \right] = n,\,\,\,n\, \le x < n + 1\] (Here n is a natural no.)
2. \[\int {{e^{x - \alpha }}} dx = {e^{x - \alpha }} + c\] (Here α is any real no.)
3. \[\int\limits_a^{b + c} {f\left( x \right)} dx = \int\limits_a^b {f\left( x \right)} dx + \int\limits_b^c {f\left( x \right)} dx\] (Here a,b and c are real numbers.)

Complete step by step solution:
Let \[a = I + f\]where \[I\]be the integer and \[f\]be a fraction.
Then
\[\int_0^a {{e^{x - \left[ x \right]}}} dx = \]\[\int_0^{I + f} {{e^{x - \left[ x \right]}}} dx\] --- (1.1)
By the property of integral –
\[\int\limits_0^{b + c} {f\left( x \right)} dx = \int\limits_0^b {f\left( x \right)} dx + \int\limits_b^c {f\left( x \right)} dx\]
Using the above integral formula in equation (1.1)
\[\begin{array}{l}\int_0^{I + f} {{e^{x - \left[ x \right]}}} dx = \int\limits_0^I {{e^{x - \left[ x \right]}}} dx + \int\limits_I^{I + f} {{e^{x - \left[ x \right]}}} dx\\\end{array}\]
Let \[{I_1} = \int\limits_0^I {{e^{x - \left[ x \right]}}} dx\] and \[{I_2} = \int\limits_I^{I + f} {{e^{x - \left[ x \right]}}} dx\]
Then
\[I = {I_1} + {I_2}\]
Solving for \[{I_1}\]
\[{I_1} = \int\limits_0^I {{e^{x - \left[ x \right]}}} dx\]
Again, using the above integral formula in \[{I_1}\] to break the integral into intervals
\[{I_1} = \int\limits_0^1 {{e^{x - \left[ x \right]}}dx} + \int\limits_1^2 {{e^{x - \left[ x \right]}}dx} + \int\limits_2^3 {{e^{x - \left[ x \right]}}dx} + \int\limits_3^4 {{e^{x - \left[ x \right]}}dx} + ..... + \int\limits_{I - 1}^I {{e^{x - \left[ x \right]}}} dx\]
We know the property of the greatest integer function
\[\left[ x \right] = n,\,\,\,n\, \le x < n + 1\]
\[{I_1} = \int\limits_0^1 {{e^{x - \left[ x \right]}}dx} + \int\limits_1^2 {{e^{x - \left[ x \right]}}dx} + \int\limits_2^3 {{e^{x - \left[ x \right]}}dx} + \int\limits_3^4 {{e^{x - \left[ x \right]}}dx} + ..... + \int\limits_{I - 1}^I {{e^{x - \left[ x \right]}}} dx\]
Applying the above formula for the greatest integer function in \[{I_1}\]
\[{I_1} = \int\limits_0^1 {{e^x}dx} + \int\limits_1^2 {{e^{x - 1}}dx} + \int\limits_2^3 {{e^{x - 2}}dx} + \int\limits_3^4 {{e^{x - 3}}dx} + ..... + \int\limits_{I - 1}^I {{e^{x - (I - 1)}}} dx\]
\[ \Rightarrow {I_1} = (e - 1) + (e - 1) + (e - 1) + ............... + (e - 1)\]
\[ \Rightarrow {I_1} = Ie - I\]
Again, solving for \[{I_2}\]
\[{I_2} = \int\limits_I^{I + f} {{e^{x - \left[ x \right]}}} dx\]
Using the formula
\[\left[ x \right] = n,\,\,\,n\, \le x < n + 1\]
Since \[I < x < I + f\] here so \[\left[ x \right] = I\] .
\[{I_2} = \int\limits_I^{I + f} {{e^{x - I}}} dx\]
\[ \Rightarrow {I_2} = {e^f} - 1\]
  \[I = {I_1} + {I_2}\]
 \[ \Rightarrow I = Ie - I + {e^f} - 1\]
\[ \Rightarrow I = Ie - (I + 1) + {e^f}\]
Comparing the evaluated value of integral from the given value
\[Ie - (1 + I) + {e^f}\]=\[10e - 9\]
 By observation \[I = 10\]
 \[{e^f} - (1 + I) = - 9\]
Putting I=10
\[ \Rightarrow {e^f} - 11 = - 9\]
Rearranging the terms
\[ \Rightarrow {e^f} = 2\]
Taking in both sides of the equations
\[\ln {e^f} = \ln 2\]
\[ \Rightarrow f\ln e = \ln 2\]
(Using \[\ln {a^b} = b\ln a\])
 \[ \Rightarrow f = \ln 2\]
(Since \[\ln e = 1\])
∴ \[a = 10 + \ln 2\]
Hence option B is correct.

Note:
1. It is advised to tackle such type of question always try to start the question while removing an integral part by breaking the values of x in consecutive integrals.
2. The above integral\[{I_1}\] can be solved also if we know the concept of periodicity of the greatest integer function. If \[f(x)\] is a periodic function and \[T\] be the period of function then –
\[\int\limits_0^{nT} {f(x)dx = n\int\limits_0^T {f(x)dx} } \]
Where \[n\] is an integer.
The greatest integer function is periodic with the period 1 so
\[{I_1} = \int\limits_0^I {{e^{x - \left[ x \right]}}} dx\]
\[ \Rightarrow {I_1} = I\int\limits_0^1 {{e^{x - \left[ x \right]}}} dx\]
\[ \Rightarrow {I_1} = I(e - 1)\]