# Area under hyperbola

#### regdude

Hi!
Today I took a test and saw an interesting thing while I was cheating It said that natural logarithm of x is the area under hyperbola 1/x starting from 1 to x.
Is there an easy way to do this for other hyperbolas like 1/x^2, 2/x etc. ?
Or is there a better way for doing this then using natural logarithms?

#### ebaines

Indeed the natural logarithm of A is defined as the area under the curve f(x) = 1/x from x=1 to x equals the value A:

$$\displaystyle ln(A) = \int_1 ^A \frac {dx} x$$

Note that this is defined only for A>0.

To find the area under the curve 1/x^2 you simply integrate:

$$\displaystyle \int \frac 1 {x^2} dx = -\frac 1 x$$

So the area under the curve 1/x^2 from x = 1 to x = A is:

$$\displaystyle 1 - \frac 1 A$$

For the area under 2/x, it's simply two times the natural logarithm.

You posted this question in a pre-calculus forum, so if this look like hieroglyphics to you, don't worry - you'll get to in school soon enough!

#### regdude

Thanks!
It's ok, I understand the basics of integrals and antiderivatives.
The 1/x^2 bit confused me, but then I remembered that derivative of x^n = nx^(n-1).
The sad thing is that I will never study this at school, because of the education programme, in the University we all need to already know this.
This is a lot better way then finding in physics class the area under the curve of isothermic process graph then calculating small parts and summing them.