The area enclosed between the curves y =log_{e} (x + e), x = log_{e}and the x-axis is

y =log_{e} (x + e) and x = log_{e} look like –

The curves intersect at (0, 1)

(Putting x = 0 in the 2 curves, y = log_{e}(e) = 1 and 0 = log_{e}(1/y),

i.e., y = 1/e^{0} = 1)

So, bounds are x = 1 – e to x = 0 for the first curve and then x = 0 to apparently x = ∞ for the second curve.

Therefore,

(Ans)

1