1. ## Area under 1/x

Can anyone see a simple way of explaining why the area of the triangle ( regardless of where the tangent is drawn) is always the same? In other words why does it work? I can justify it algebraically.

2. ## Re: Area under 1/x

suppose the point of tangency is $x_0$

then the tangent line will have slope $-\dfrac {1}{x_0^2}$ and will include the point $\left(x_0, \dfrac{1}{x_0}\right)$

Given this line see if you can figure out where the x and y intercepts occur.

The area of the triangle is then trivial to compute.

Spoiler:
you should find that $A = \dfrac 1 2$

3. ## Re: Area under 1/x

Yes i can see an algebraic way as you describe above. Is there a geometric way of seeing the result.
I thought what if we fitted a rectangle under it? Its area would always be one.. can't see why the triangle will always be fixed ?

4. ## Re: Area under 1/x

Originally Posted by rodders
Yes i can see an algebraic way as you describe above. Is there a geometric way of seeing the result.
I thought what if we fitted a rectangle under it? Its area would always be one.. can't see why the triangle will always be fixed ?
It is the reciprocal function of $y=x$. This gives it tremendous symmetry across that line. What you are seeing with this area is one manifestation of that symmetry. I am not sure what you are looking for as a "geometric way of seeing the result". But, I would start with that line of symmetry.