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Step-by-Step Solution
Step 1: Understand the Problem
We want to evaluate the definite integral
$ I = \int_0^{10} \dfrac{[x]\, e^{[x]}}{e^{x-1}} \, dx ,$
where $[x]$ denotes the greatest integer less than or equal to $x$.
Step 2: Rewrite the Integrand
Notice that
$ \dfrac{[x]\, e^{[x]}}{e^{x-1}} = [x]\, e^{[x] + 1 - x}. $
Thus, the integral becomes
$ I = \int_0^{10} [x]\, e^{[x] + 1 - x} \, dx. $
Step 3: Break the Integral into Integer Sub-Intervals
The greatest integer function $[x]$ is constant on each interval $[k, k+1)$ for integer $k$. Therefore, we split the integral from $0$ to $10$ into sums of integrals over these intervals:
$ I = \int_0^1 0 \cdot e^{0 + 1 - x} \, dx \;+\; \int_1^2 1 \cdot e^{1 + 1 - x} \, dx \;+\; \int_2^3 2 \cdot e^{2 + 1 - x} \, dx \;+\; \dots + \int_9^{10} 9 \cdot e^{9 + 1 - x} \, dx. $
Note that $[x] = 0$ for $x$ from $0$ to $1$, so the integrand is $0$ in the first interval, which makes
$ \int_0^1 0 \, dx = 0. $
Step 4: Evaluate a Typical Sub-Integral
Consider the general term for $k \ge 1$:
$ I_k = \int_k^{k+1} k \cdot e^{k + 1 - x} \, dx, $
where $k = 1, 2, \dots, 9$.
We factor out the constant $k$:
$ I_k = k \int_k^{k+1} e^{k + 1 - x} \, dx. $
Use the substitution $u = (k+1) - x$, so $du = -dx$. When $x = k$, $u = (k+1) - k = 1$, and when $x = k+1$, $u = 0$. Thus,
$ I_k = k \int_{u=1}^{u=0} e^u \, (-du) = k \int_0^1 e^u \, du. $
We evaluate $ \int_0^1 e^u \, du $:
$ \int_0^1 e^u \, du = e^1 - e^0 = e - 1. $
Hence,
$ I_k = k \, (e - 1). $
Step 5: Sum Over All Relevant k
We sum $I_k$ for $k$ from 1 to 9:
$ I = \sum_{k=1}^{9} I_k = \sum_{k=1}^{9} k \, (e - 1) = (e - 1) \sum_{k=1}^{9} k. $
The sum of the integers from 1 to 9 is $ \frac{9 \cdot 10}{2} = 45 $. Therefore,
$ I = (e - 1) \times 45 = 45 \, (e - 1). $
Step 6: Conclusion
Thus, the value of the integral is
$ \boxed{45 \, (e - 1)}. $
