Key Concepts and Formulas

• Greatest Integer Function: $[x]$ is the largest integer less than or equal to $x$. It is a step function, constant on intervals $[n, n+1)$.
• Definite Integral Decomposition: An integral over an interval $[a, b]$ can be split into a sum of integrals over subintervals $[a, c]$ and $[c, b]$. This is crucial when the integrand's form changes, as with the greatest integer function.
• Integral of $e^{ax+b}$: The indefinite integral of $e^{ax+b}$ is $\frac{1}{a}e^{ax+b} + C$.

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Step-by-Step Solution

Step 1: Simplify the Integrand

The given integral is $I = \int_0^{10} {{{[x]{e^{[x]}}} \over {{e^{x - 1}}}}dx}$. We can simplify the exponential part of the integrand using the property $\frac{a^m}{a^n} = a^{m-n}$. $I = \int_0^{10} {[x]{e^{[x] - (x - 1)}}dx}$ $I = \int_0^{10} {[x]{e^{[x] - x + 1}}dx}$ This simplification makes the integrand easier to work with in subsequent steps.

Step 2: Decompose the Integral Based on the Greatest Integer Function

The greatest integer function $[x]$ changes its value at integer points. Since the integral is from $0$ to $10$, $[x]$ will take integer values $0, 1, 2, \dots, 9$ over the intervals $[0,1), [1,2), [2,3), \dots, [9,10)$ respectively. We must decompose the integral into a sum of integrals over these unit intervals. $I = \sum_{n=0}^{9} \int_n^{n+1} {[x]{e^{[x] - x + 1}}dx}$ Within each interval $[n, n+1)$, we know that $[x] = n$. Substituting this into the integral: $I = \sum_{n=0}^{9} \int_n^{n+1} {n{e^{n - x + 1}}dx}$

Step 3: Evaluate the Integral for Each Interval

Now, we evaluate the sum of integrals. For the first interval, $n=0$: $\int_0^1 {0 \cdot {e^{0 - x + 1}}dx} = \int_0^1 {0 \, dx} = 0$ For $n \ge 1$, the integral is: $\int_n^{n+1} {n{e^{n + 1 - x}}dx}$ We can pull the constant $n$ out of the integral: $n \int_n^{n+1} {{e^{n + 1 - x}}dx}$ To evaluate this integral, let $u = n+1-x$. Then $du = -dx$. When $x=n$, $u = n+1-n = 1$. When $x=n+1$, $u = n+1-(n+1) = 0$. The integral becomes: $n \int_1^0 {{e^u (-du)}}$ $n \int_0^1 {{e^u du}}$ $n \left[ {e^u} \right]_0^1$ $n (e^1 - e^0)$ $n (e - 1)$

Step 4: Sum the Results from All Intervals

Now we sum the results for $n=1$ to $n=9$ (since the integral for $n=0$ is 0). $I = 0 + \sum_{n=1}^{9} n(e - 1)$ We can factor out $(e-1)$ from the sum: $I = (e - 1) \sum_{n=1}^{9} n$ The sum of the first $k$ natural numbers is given by the formula $\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$. Here, $k=9$. $\sum_{n=1}^{9} n = \frac{9(9+1)}{2} = \frac{9 \times 10}{2} = 45$ Substituting this sum back into the expression for $I$: $I = (e - 1) \times 45$ $I = 45(e - 1)$

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Common Mistakes & Tips

• Incorrectly handling the $n=0$ term: The term for $[x]=0$ over $[0,1)$ results in an integrand of $0 \cdot e^{0-x+1} = 0$, so its integral is $0$. Do not assume the first term will be non-zero.
• Mistakes in substitution or integration: Be careful with the sign change when performing substitution for the exponential integral. Ensure the limits of integration are correctly transformed.
• Forgetting to sum all terms: The integral is a sum over all relevant intervals. Ensure all non-zero contributions from each interval are included.

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Summary

The integral $I$ was first simplified by combining exponential terms. The presence of the greatest integer function $[x]$ necessitated decomposing the integral into a sum of integrals over unit intervals $[n, n+1)$. In each interval, $[x]$ was replaced by the constant integer $n$. After evaluating the integral for each interval, a pattern emerged where the contribution from interval $[n, n+1)$ was $n(e-1)$ for $n \ge 1$. Summing these contributions from $n=1$ to $9$ and adding the zero contribution from $n=0$, we obtained the final result $45(e-1)$.

The final answer is $\boxed{45(e-1)}$.