Question

If \[\int\limits_0^{25} {{e^{x - \left[ x \right]}}dx = k(e - 1)} \] , then find the value of k.
A.12
B.25
C.23
D.24

Answer 1

Hints First write x as the sum of integer part and the fractional part then obtain the value of box x and substitute in the given function. Then integrate the given function to obtain the required value.

Formula used
\[\int {{e^x} = } {e^x}\] , where x is any value.
If f(x) is a periodic function with period n then, \[\int\limits_0^{n.a} {f(x)dx = n} \int\limits_0^a {f(x)dx} \] .

Complete step by step solution
The given integral is \[\int\limits_0^{25} {{e^{x - \left[ x \right]}}dx} \].
Now, we know that any number is the sum of its integer part and fractional part, therefore,
\[x = \left[ x \right] + \left\{ x \right\}\], where \[\left[ x \right]\] is the integer part and \[\left\{ x \right\}\] is the fractional part.
So, \[x - \left[ x \right] = \left\{ x \right\}\]
Substitute \[x - \left[ x \right] = \left\{ x \right\}\]in the given integral we have\[\int\limits_0^{25} {{e^{\left\{ x \right\}}}dx} \] .
Now, \[\left\{ x \right\}\]is a periodic function with period 1.
So, \[{e^{\left\{ x \right\}}}\]has period 1.
Therefore, \[\int\limits_0^{25} {{e^{\left\{ x \right\}}}dx} = \int\limits_0^{1.25} {{e^{\left\{ x \right\}}}dx} \]
 =\[25\int\limits_0^1 {{e^{\left\{ x \right\}}}dx} \]
=\[25\left[ {{e^{\left\{ x \right\}}}} \right]_0^1\]
 =\[25(e - 1)\]
Hence, k=25.
The correct option is B.

Note Sometimes students directly integrate the given function using the concept that \[x - \left[ x \right] = \left\{ x \right\}\] but here we need to use the periodic function part to obtain the correct answer as 25.