Let $$a$$ be a positive real number such that $$\int_0^a e^{x-[x]} dx = 10e - 9$$ where $$[x]$$ is the greatest integer less than or equal to $$x$$. Then, $$a$$ is equal to:

Since $$x - [x] = \{x\}$$ (the fractional part of $$x$$), we have $$e^{x-[x]} = e^{\{x\}}$$, which is periodic with period 1.

On each interval $$[n, n+1)$$, $$e^{\{x\}} = e^{x-n}$$, so $$\int_n^{n+1} e^{\{x\}} dx = \int_0^1 e^t\,dt = e - 1$$.

Write $$a = 10 + \delta$$ where $$0 < \delta < 1$$. Then: $$\int_0^a e^{\{x\}}\,dx = 10(e-1) + \int_{10}^{10+\delta} e^{x-10}\,dx = 10(e-1) + \left[e^{x-10}\right]_{10}^{10+\delta} = 10(e-1) + (e^\delta - 1).$$

Setting this equal to $$10e - 9$$: $$10e - 10 + e^\delta - 1 = 10e - 9 \implies e^\delta = 2 \implies \delta = \log_e 2.$$

Therefore $$a = 10 + \log_e 2$$.