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Step-by-Step Solution
Step 1: Interpret the Integral and the Floor Function
We are given:
$$\int_{0}^{a} e^{x - [x]} \, dx = 10e - 9,$$
where $[x]$ denotes the greatest integer less than or equal to $x$ (floor function). Let $n = [a]$. Then we can write:
$$
a = n + \{a\},
$$
where $\{a\}$ is the fractional part of $a$ and $n$ is a non-negative integer (because $a > 0$).
Step 2: Break the Integral into Integer and Fractional Intervals
Consider the function $e^{\,x - [x]}$ on integer intervals. Specifically, between any two integers $k$ and $k+1$, the expression $x - [x]$ is simply $x - k$, which ranges from $0$ to $1$ as $x$ goes from $k$ to $k+1$. Hence:
1. From $x = 0$ to $x = n$, there are $n$ full intervals of length 1 each (assuming $n$ is a whole number).
2. From $x = n$ to $x = a$, we have part of an interval from $n$ up to $n + \{a\} = a$.
Step 3: Integrate Over Each Part
Integral from 0 to n:
For each unit interval $[k, k+1)$, $x - [x] = x - k$ goes from $0$ to $1.$ Therefore,
$$
\int_{k}^{k+1} e^{x - k} \, dx
= \int_{0}^{1} e^{t} \, dt
= \left[e^t\right]_{0}^{1} = e - 1.
$$
There are $n$ such intervals from $0$ to $n$, hence the total contribution is
$$
n (e - 1).
$$
Integral from n to a:
Here $x$ ranges from $n$ to $a = n + \{a\}$, so $x - [x] = x - n$ runs from $0$ to $\{a\}$. Thus,
$$
\int_{n}^{a} e^{x - n} \, dx
= \int_{0}^{\{a\}} e^{t} \, dt
= \left[e^t\right]_{0}^{\{a\}}
= e^{\{a\}} - 1.
$$
Step 4: Apply the Given Condition
Summing these parts, the total integral is
$$
n(e - 1) + \bigl(e^{\{a\}} - 1\bigr).
$$
According to the problem statement, this sum equals $10e - 9$:
$$
n(e - 1) + \bigl(e^{\{a\}} - 1\bigr) = 10e - 9.
$$
Step 5: Solve for n and {a}
We look for integer $n$ and fractional part $\{a\}$ satisfying the equation. Suppose $n = 10$; then:
$$
10(e - 1) + \bigl(e^{\{a\}} - 1\bigr) = 10e - 10 + e^{\{a\}} - 1.
$$
We want this to be $10e - 9,$ so set
$$
10e - 10 + e^{\{a\}} - 1 = 10e - 9.
$$
This simplifies to
$$
e^{\{a\}} - 1 = 1 \quad \Rightarrow \quad e^{\{a\}} = 2 \quad \Rightarrow \quad \{a\} = \ln(2).
$$
Thus $n = 10$ and $\{a\} = \ln(2).$
Step 6: Conclude the Value of a
Since $a = n + \{a\},$ we get
$$
a = 10 + \ln(2).
$$
Hence, the correct choice is
$$
10 + \ln(2).
$$
