Let a be a positive real number such that ${\int }_0^a{e}^{x-\left[x\right]}dx=10e-9$ where $\left[x\right]$ is the greatest integer less than or equal to x then a is equal to :

${\int }_0^{\mathrm{a}} {\mathrm{e}}^{\mathrm{x}-\left\{\mathrm{x}\right\}} \mathrm{dx}=10\mathrm{e}-9$

Since  $\mathrm{a}\in \mathrm{R},$

Let  $\mathrm{a}=\mathrm{n}+\mathrm{\lambda }$

            where  $0\le \mathrm{\lambda }<1$

$\begin{array}{rcl}{\int }_0^{\mathrm{a}} {\mathrm{e}}^{\left\{\mathrm{x}\right\}} \mathrm{dx}& =& {\int }_0^{\mathrm{n}} {\mathrm{e}}^{\left\{\mathrm{x}\right\}} \mathrm{dx}+{\int }_{\mathrm{n}}^{\mathrm{n}+\mathrm{\lambda } }{\mathrm{e}}^{\left\{\mathrm{x}\right\}} \mathrm{dx}\\ & =& \mathrm{n} {\int }_0^1 {\mathrm{e}}^{\left\{\mathrm{x}\right\}} \mathrm{dx}+{\int }_0^{\mathrm{\lambda }} {\mathrm{e}}^{\left\{\mathrm{x}\right\}} \mathrm{dx}\end{array}$

$$\begin{gathered} =\mathrm{n} {\int }_0^1 {\mathrm{e}}^{\mathrm{x}} \mathrm{dx}+{\int }_0^{\mathrm{\lambda }} {\mathrm{e}}^{\mathrm{x}} \mathrm{dx} \\ =\mathrm{n}\left({\mathrm{e}}^1-1\right)+\left({\mathrm{e}}^{\mathrm{\lambda }}-1\right) \\ ={\mathrm{e}}^{\mathrm{\lambda }}+\mathrm{ne}-(\mathrm{n}+1) \\ =10\mathrm{e}-9 \\ \therefore \mathrm{n}=10, {\mathrm{e}}^{\mathrm{\lambda }}-(\mathrm{n}+1)=-9 \\ {\mathrm{e}}^{\mathrm{\lambda }}-11=-9 \\ {\mathrm{e}}^{\mathrm{\lambda }}=2 \\ \mathrm{\lambda }=\log 2 \\ \mathrm{a}=10+\log 2 \end{gathered}$$