Key Concepts and Formulas

• Greatest Integer Function: $[x]$ denotes the greatest integer less than or equal to $x$.
• Fractional Part Function: $\{x\} = x - [x]$. This function has a period of 1 and its range is $[0, 1)$.
• Definite Integral Property: For a continuous function $f(x)$ and an integer $n$, $\int_{a}^{a+n} f(x) dx$ can be evaluated by understanding the behavior of $f(x)$ over the interval. If $f(x)$ is periodic, this becomes simpler.
• Properties of Exponents: $e^a e^b = e^{a+b}$ and $e^0 = 1$.

Step-by-Step Solution

Step 1: Understand the Integrand and the Summation The problem asks for the value of the sum $S = \sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}dx} }$. The integrand is $e^{x - [x]}$. We know that $x - [x]$ is the fractional part of $x$, denoted as $\{x\}$. So, the integrand is $e^{\{x\}}$. The summation involves integrating this function over consecutive unit intervals: $[0, 1), [1, 2), [2, 3), \dots, [99, 100)$.

Step 2: Analyze the Integrand over a Unit Interval Consider a single integral term in the summation: $I_n = \int\limits_{n - 1}^n {{e^{x - [x]}}dx}$. For any $x$ in the interval $[n-1, n)$, the greatest integer $[x]$ is equal to $n-1$. For example, if $n=1$, the interval is $[0, 1)$, and for $x \in [0, 1)$, $[x] = 0$. If $n=2$, the interval is $[1, 2)$, and for $x \in [1, 2)$, $[x] = 1$. In general, for $x \in [n-1, n)$, we have $[x] = n-1$. Thus, within the interval of integration $[n-1, n)$, the term $x - [x]$ simplifies to $x - (n-1)$.

Step 3: Evaluate a Single Integral Term Substitute $[x] = n-1$ into the integral: $I_n = \int\limits_{n - 1}^n {{e^{x - (n - 1)}}dx}$ Let $u = x - (n-1)$. Then, $du = dx$. When $x = n-1$, $u = (n-1) - (n-1) = 0$. When $x = n$, $u = n - (n-1) = 1$. The integral becomes: $I_n = \int\limits_{0}^1 {{e^u}du}$ The antiderivative of $e^u$ is $e^u$. Evaluating the definite integral: $I_n = \left[ {{e^u}} \right]_{0}^1 = e^1 - e^0 = e - 1$ So, each integral term in the summation evaluates to $e-1$.

Step 4: Calculate the Total Sum The problem asks for the sum of these integral terms from $n=1$ to $n=100$: $S = \sum\limits_{n = 1}^{100} {I_n}$ Since $I_n = e-1$ for every value of $n$, we are adding the constant value $(e-1)$ one hundred times: $S = \sum\limits_{n = 1}^{100} {(e - 1)}$ $S = (e - 1) + (e - 1) + \dots + (e - 1) \quad \text{(100 times)}$ $S = 100 \times (e - 1)$

Common Mistakes & Tips

• Incorrectly applying the greatest integer function: Remember that $[x]$ is constant over intervals of the form $[k, k+1)$.
• Forgetting the limits of integration when changing variables: When substituting $u = x - (n-1)$, ensure the limits of integration are correctly transformed from $x$ to $u$.
• Algebraic errors with exponents: Be careful when simplifying terms like $e^{n - (n-1)}$.

Summary

The problem involves summing integrals of the fractional part of $x$, i.e., $e^{\{x\}}$, over consecutive unit intervals. For each interval $[n-1, n)$, the term $x-[x]$ is equal to $x-(n-1)$. Evaluating the integral $\int_{n-1}^n e^{x-(n-1)}dx$ yields $e-1$. Since this result is constant for all $n$ from 1 to 100, the total sum is simply 100 times $(e-1)$.

The final answer is \boxed{100(e - 1)}.