If lnx is the exponent we raise e to--- how is it the area under the curve of 1/x of the formular of lnx = 1/x?

The area under the curve 1/x, is the definite integral of
f(x) - another curve or line and between the limits x = a and x =
b.

Since there are not specified the limits x = a and x =
b, also it is not specified the other curve or line, we'll calculate the indefinite
integral of 1/x and not the area under the curve.

The
indefinite integral of f(x) = 1/x is:

Int f(x) = Int
dx/x

Int dx/x = ln x + C

C -
family of constants.

To understand the family of constants
C, we'll consider the result of the indefinite integral as the function
f(x).

f(x) = ln x + C

We'll
differentiate f(x):

f'(x) = (ln x +
C)'

f'(x) = 1/x + 0

Since C
is a constant, the derivative of a constant is
cancelling.

So, C could be any constant, for
differentiating f(x), the constant will be zero.

Now, we'll
calculate the area located between the curve 1/x, x axis, x=a and
x=b:

Integral [f(x) - ox]dx, x = a to x =
b

Int f(x) dx = F(b) - F(a), using Leibniz-Newton
formula

Int dx/x = ln b - ln
a

Since the logarithms have matching bases, we'll transform
the difference into a product:

Int dx/x = ln
|b/a|

The area located between the curve 1/x,
x axis, x=a and x=b
is:

Int dx/x = ln
|b/a|