Consider the integral $$I = \int_0^{10} \frac{[x]e^{[x]}}{e^{x-1}}dx$$ where $$[x]$$ denotes the greatest integer less than or equal to $$x$$. Then the value of $$I$$ is equal to:

We need to evaluate $$I = \int_0^{10} \frac{[x]e^{[x]}}{e^{x-1}} dx = \int_0^{10} [x] \cdot e^{[x] - x + 1} \, dx$$.

For $$x \in [n, n+1)$$ where $$n$$ is a non-negative integer, $$[x] = n$$. The integral becomes:

$$I = \sum_{n=0}^{9} \int_n^{n+1} n \cdot e^{n - x + 1} \, dx$$

The $$n = 0$$ term is zero. For $$n \geq 1$$:

$$\int_n^{n+1} n \cdot e^{n-x+1} \, dx = n \cdot e^{n+1} \left[-e^{-x}\right]_n^{n+1} = n \cdot e^{n+1} \left(e^{-n} - e^{-(n+1)}\right) = n \cdot e^{n+1} \cdot e^{-n}\left(1 - e^{-1}\right) = n \cdot e \cdot \frac{e-1}{e} = n(e-1)$$

Therefore: $$I = (e-1) \sum_{n=1}^{9} n = (e-1) \cdot \frac{9 \cdot 10}{2} = 45(e-1)$$.