Find the area enclosed between the curves: `y = log_e (x + e) , x = log_e (1/y)` & the x-axis.

Explanation

Given curves are y = `log_(e) ( x + e) " "… (1)`
and ` x = log_(e) ((1)/(y))` or x = `-log_(e)y`
or `y = e^(-x)` or `y = ((1)/(e))^(x) " " … (2)`
The two curves cut at x = 0
`therefore` Graphs of curves (1) and (2) are as shown in the figure .
Required area = shaded area
`= overset(1)underset(0)(int) (x_(1) - x_(2))`dy
`= underset(0)overset(1)(int) [-log_(e) y- (e^(y) - e) ] dy`
`= [-(ylog y ) - e^(y) + ey]_(0)^(1)`
`[y-ylog y - e^(y) + ye]_(0)^(1)`
`= (1 - 0 - e + e) - (0 - 0 - 1 + 0)`
` = 1 + 1 `
= 2 sq. units