Question

Let $I=\int\limits_{0}^{25}{{{e}^{x-\left[ x \right]}}dx}$ and I be equal to \[I\text{ }=\text{ }\left( n+1 \right)\cdot \left( e\text{ }+\text{ }k \right)\]. Find n + k?

Answer 1

Hint: First of all, we must evaluate the integral correctly by employing the correct definition of $x-\left[ x \right]$ and integrate the function by using integration formulas and its properties ,then we would further proceed for obtaining a relationship between different variables involved in our problem.

Complete step-by-step answer:
Now let us see a different way of expressing x in terms of integral part and fractional part.
$x=\left[ x \right]+\left\{ x \right\}$ where the first counterpart is known as the integral part of x and the other counterpart is called the fractional part of x.
So, $x-\left[ x \right]=\left\{ x \right\}$. One most basic feature of such expression is related to the range of fractional parts which is $0<\left\{ x \right\}<1$. This statement also signifies that the period of fractional function is 1 unit such that it repeats its cycle after every one step in its cycle.
So, our integral can be modified as $I=\int\limits_{0}^{25}{{{e}^{\left\{ x \right\}}}dx}$.
In definite integral if the time period of a function is n such that $f(x+nT)=f(x)$. then its integral can be easily evaluated as:
${{I}_{n}}=\int\limits_{0}^{nT}{f(x)dx}$
${{I}_{n}}=n\int\limits_{0}^{T}{f\left( x \right)dx}$
So, our integral would be:
$\begin{align}
  & I=\int\limits_{0}^{25\times 1}{{{e}^{\left\{ x \right\}}}dx} \\
 & I=25\times \int\limits_{0}^{1}{{{e}^{\left\{ x \right\}}}dx} \\
\end{align}$
When $0<\times<1$ then the fractional part of x becomes x without any restrictions.
So, $\left( x \right)={{e}^{\left\{ x \right\}}}$ can be converted as $\left( x \right)={{e}^{x}}$.
$I=25\times \int\limits_{0}^{1}{{{e}^{x}}dx}$
So, the integral can be solved by using the property $\int\limits_{\alpha }^{\beta }{{{e}^{x}}dx}=\left[ {{e}^{x}} \right]_{\alpha }^{\beta }$
$\begin{align}
  & I=25\left[ {{e}^{x}} \right]_{0}^{1} \\
 & I=25\left( e-1 \right) \\
\end{align}$
Therefore, the integral could be expressed as:
$I=\left( 24+1 \right)\cdot \left( e+\left( -1 \right) \right)$
From the above given expression in the question \[I\text{ }=\text{ }\left( n+1 \right)\cdot \left( e\text{ }+\text{ }k \right)\]
On comparing both of the above expression we get:
n = 24 and k = -1
So, to evaluate the expression the final expression (n + k) = (24 – 1)
Hence, the obtained solution is 23.

Note: We must correctly define the fractional part of a given variable and then evaluate the integral carefully. Without expanding the value of x, we are bound to invite errors in solving problems.Integral must be evaluated by using proper limits to the function.Students should remember important formulas of integration and its properties.