The value of $$\sum_{n=1}^{100} \int_{n-1}^{n} e^{x - [x]} dx$$, where $$[x]$$ is the greatest integer $$\leq x$$, is:

We need to evaluate $$\sum_{n=1}^{100} \int_{n-1}^{n} e^{x - [x]} \, dx$$, where $$[x]$$ denotes the greatest integer function.

On the interval $$[n-1, n)$$, the greatest integer function gives $$[x] = n - 1$$. Therefore $$x - [x] = x - (n-1)$$, and the integrand becomes $$e^{x-(n-1)}$$.

Each integral evaluates to $$\int_{n-1}^{n} e^{x-(n-1)} \, dx = \left[e^{x-(n-1)}\right]_{n-1}^{n} = e^{n-(n-1)} - e^{(n-1)-(n-1)} = e^1 - e^0 = e - 1$$.

Since each of the 100 integrals equals $$e - 1$$, the total sum is $$100(e - 1)$$.