Key Concepts and Formulas

• Greatest Integer Function (GIF): $[x]$ denotes the greatest integer less than or equal to $x$.
• Fractional Part Function: $\{x\} = x - [x]$. By definition, $0 \le \{x\} < 1$.
• Periodicity of $e^{\{x\}}$: The function $e^{\{x\}}$ is periodic with period 1, since $\{x+1\} = (x+1) - [x+1] = x+1 - ([x]+1) = x - [x] = \{x\}$.

Step-by-Step Solution

Step 1: Understand the integrand $e^{x - [x]}$ The integrand can be rewritten using the fractional part function: $e^{x - [x]} = e^{\{x\}}$. Since $0 \le \{x\} < 1$, the function $e^{\{x\}}$ is periodic with period 1.

Step 2: Decompose the integral based on the upper limit 'a' Let $a = n + \alpha$, where $n = [a]$ is the greatest integer less than or equal to $a$, and $\alpha = \{a\}$ is the fractional part of $a$, such that $0 \le \alpha < 1$. The integral can be split into two parts: an integral over full periods of length 1, and an integral over the remaining fractional part. $\int_0^a {{e^{x - [x]}}} dx = \int_0^n {{e^{x - [x]}}} dx + \int_n^a {{e^{x - [x]}}} dx$

Step 3: Evaluate the integral over full periods For the first part, $\int_0^n {{e^{x - [x]}}} dx$, since $e^{\{x\}}$ has a period of 1, we can use the property $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$ with $T=1$. $\int_0^n {{e^{x - [x]}}} dx = n \int_0^1 {{e^{x - [x]}}} dx$ In the interval $[0, 1)$, $[x] = 0$, so $x - [x] = x$. $\int_0^1 {{e^{x - [x]}}} dx = \int_0^1 {{e^x}} dx = [e^x]_0^1 = e^1 - e^0 = e - 1$ Therefore, the integral over full periods is: $\int_0^n {{e^{x - [x]}}} dx = n(e - 1)$

Step 4: Evaluate the integral over the remaining fractional part For the second part, $\int_n^a {{e^{x - [x]}}} dx$, let $x = n + u$. Then $dx = du$. When $x = n$, $u = 0$. When $x = a$, $u = a - n = \alpha$. Also, $[x] = [n+u] = n + [u]$. Since $0 \le u < \alpha < 1$, $[u] = 0$. So, $x - [x] = (n+u) - (n+0) = u$. $\int_n^a {{e^{x - [x]}}} dx = \int_0^\alpha {{e^u}} du = [e^u]_0^\alpha = e^\alpha - e^0 = e^\alpha - 1$

Step 5: Combine the parts and set up the equation Now, we combine the results from Step 3 and Step 4: $\int_0^a {{e^{x - [x]}}} dx = n(e - 1) + (e^\alpha - 1)$ We are given that this integral is equal to $10e - 9$. $n(e - 1) + e^\alpha - 1 = 10e - 9$ $ne - n + e^\alpha - 1 = 10e - 9$ $ne + e^\alpha = 10e - 8 + n$

Step 6: Solve for n and $\alpha$ by comparing coefficients We have $a = n + \alpha$, where $n$ is a non-negative integer and $0 \le \alpha < 1$. The equation is $ne + e^\alpha = 10e - 8 + n$. Since $n$ is an integer and $0 \le \alpha < 1$, we can equate the coefficients of $e$ and the constant terms. Comparing the coefficients of $e$: $n = 10$

Now substitute $n=10$ into the equation: $10e + e^\alpha = 10e - 8 + 10$ $10e + e^\alpha = 10e + 2$ $e^\alpha = 2$

Step 7: Find $\alpha$ and then 'a' From $e^\alpha = 2$, we take the natural logarithm of both sides: $\ln(e^\alpha) = \ln(2)$ $\alpha = \ln(2)$

Since $0 \le \ln(2) < 1$ (because $1 \le 2 < e$), our value of $\alpha$ is valid. Now we find $a$ using $a = n + \alpha$: $a = 10 + \ln(2)$

Step 8: Express the answer in the required format The value of $a$ is $10 + \ln(2)$. We need to check which option matches this. The options are given in terms of $\log_e$. So, $\ln(2)$ is the same as $\log_e(2)$. Thus, $a = 10 + \log_e(2)$.

Common Mistakes & Tips

• Incorrectly handling the fractional part: Ensure that $0 \le \{x\} < 1$ is always maintained. When splitting the integral, the upper limit of the second part must be strictly less than 1 after the substitution.
• Algebraic errors when comparing coefficients: Be careful when equating the terms. Since $n$ is an integer and $\alpha$ is a fractional part, the structure of the equation $ne + e^\alpha = 10e - 8 + n$ allows for direct comparison.
• Mistaking the base of the logarithm: The problem uses $e$, so natural logarithm ($\ln$ or $\log_e$) is implied.

Summary

The problem involves integrating a periodic function $e^{\{x\}}$. We decomposed the integral into parts corresponding to full integer intervals and the remaining fractional part of the upper limit $a$. By equating the integral to the given value $10e - 9$ and comparing coefficients after expressing $a$ as $n+\alpha$, we found $n=10$ and $\alpha = \ln(2)$. This led to $a = 10 + \ln(2)$.

The final answer is \boxed{10 + {\log _e}2}.