What is the area between x axis and the curve x^-1?

The first step is to re-write the given curve y = x^-1,
using the property of negative power.

x^-1 =
1/x

The area under the curve 1/x, is the definite integral
of y minus integral of 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, we'll calculate the indefinite integral of 1/x and not the area under the
curve.

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

Int f(x) dx = 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 and we'll consider the limit
lines x=a and x=b:

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

We'll apply Leibniz-Newton
formula:

Int f(x) dx = F(b) -
F(a)

 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|