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

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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|**

The area bounded the curve f(x) and xaxis , and x = a and b is given by:

Area = Integral f(x)dx from x = a to x = b.

If Integral f(x) dx = F(x), then {Integral f(x) dx from x= a to b} = F(b)-F(a).

Therefore in the present case the between f(x) = 1/x and x axis is

Area = Integral f(x) dx = Integral (1/x) dx from from x = a to b .

But Integral (1/x)dx = lnx.

Therefore F(x) = lnx.

Therefore the area under the curve 1/x between from x=a to x= b = F(b)- F(a) = lnb - lna.

Also ln(x) is not defined for a = 0 or a < 0. So a is always a positive quantity.