If the area of the region $${(x,y):-1 \leq x \leq 1,0 \leq y \leq a + e^{|x|}-e^{-x},a > 0}$$ is $$\frac{e^{2}+8e+1}{e}$$, then the value of a is :

We need to find a such that the area of the region $$\{(x,y): -1 \leq x \leq 1, 0 \leq y \leq a + e^{|x|} - e^{-x}\}$$ equals $$\frac{e^2 + 8e + 1}{e}$$.

Since this area can be computed by integrating the upper boundary from $$x=-1$$ to $$x=1$$, we set up

$$\text{Area} = \int_{-1}^{1}(a + e^{|x|} - e^{-x})\,dx$$.

On $$x\in[-1,0]$$ we have $$|x|=-x$$, so the integrand becomes $$a + e^{-x} - e^{-x} = a$$. On $$[0,1]$$, $$|x|=x$$ and the integrand is $$a + e^x - e^{-x}$$.

Therefore,

$$\text{Area} = \int_{-1}^{0}a\,dx + \int_{0}^{1}(a + e^x - e^{-x})\,dx = a + [ax + e^x + e^{-x}]_0^1 = a + (a + e + e^{-1}) - (0 + 1 + 1) = 2a + e + \frac{1}{e} - 2.$$

Setting this equal to the given area gives

$$2a + e + \frac{1}{e} - 2 = \frac{e^2 + 8e + 1}{e},$$

or equivalently

$$2a + \frac{e^2 + 1}{e} - 2 = \frac{e^2 + 8e + 1}{e}.$$

Solving for $$a$$ yields

$$2a = \frac{e^2 + 8e + 1}{e} - \frac{e^2 + 1}{e} + 2 = \frac{8e}{e} + 2 = 8 + 2 = 10,$$

so $$a = 5$$.

The correct answer is Option 3: 5.