Understand the integrand

We need to evaluate $\int_0^a e^{x-[x]}\,dx,$ where $[x]$ is the greatest integer function.

On each interval $[n,n+1)$, we have $[x]=n$, so $x-[x]=x-n.$ Hence, $e^{x-[x]}=e^{x-n}.$

Integral over one unit interval

For any integer $n$, $\int_n^{n+1} e^{x-n}\,dx = \int_0^1 e^t\,dt = e-1.$

So each complete interval of length $1$ contributes exactly $e-1$.

Write $a$ in the form $a=n+r$

Let $a=n+r,$ where $n=[a]$ is an integer and $0\le r<1$.

Then $\int_0^a e^{x-[x]}dx = \sum_{k=0}^{n-1}\int_k^{k+1} e^{x-k}dx + \int_n^{n+r} e^{x-n}dx.$

Therefore, $\int_0^a e^{x-[x]}dx = n(e-1) + \int_0^r e^t\,dt.$

So, $\int_0^a e^{x-[x]}dx = n(e-1) + (e^r-1).$

Use the given value

Given, $n(e-1) + (e^r-1) = 10e-9.$

Rearranging, $n(e-1) + e^r = 10e-8.$

Find the integer part $n$

Since $0\le r<1$, we have $1\le e^r<e.$

Thus, $n(e-1)+1 \le 10e-9 < n(e-1)+(e-1?)$ but it is easier to test nearby integers.

Try $n=9$: $9(e-1) + (e^r-1) = 9e-9 + e^r -1 = 9e-10 + e^r.$ Setting equal to $10e-9$ gives $9e-10+e^r = 10e-9,$ $e^r = e+1,$ which is impossible because $e^r<e$.

Try $n=10$: $10(e-1) + (e^r-1) = 10e-10 + e^r -1 = 10e-11+e^r.$ Setting equal to $10e-9$ gives $10e-11+e^r = 10e-9,$ $e^r = 2.$ This is possible, since $1<2<e$.

Hence, $r=\ln 2.$

Compute $a$

Therefore, $a=n+r=10+\ln 2.$

Match with options

$a=10+\log_e 2,$ which is Option B.