Given an interval and a basic hyperbola function (like: $y^2 – x^2 = 1$), how would you go about finding the average vertical distance between the two hyperbola branches?
So far, I've been looking at the distance and length formulas for hyperbolas, but I'm stuck on how exactly to relate this to the average value calculations with integrals.
I think you're supposed to use the average value integral formula here, but I can't figure out what to integrate
Best Answer
Due to the symmetry of the two branches with respect to the $x$ axis, just consider the average of the distance between the $x$ axis and the upper branch with equation $y=\sqrt{1+x^2}$ (and finaly multiply by 2) The average you are looking for is therefore obtained by considering the equivalence of two areas :
$$\int_{a}^{b}\sqrt{1+x^2} dx = M \times (b-a)$$
Explanation : On the left hand side is the area under the curve on $[a,b]$ ; on the right hand side, it is the equivalent area of rectangle with sides $[a,b]$ and $[0,M]$.
Therefore, the average length is :
$$M=\dfrac{1}{b-a}\int_{a}^{b}\sqrt{1+x^2} dx$$
(I leave you the final calculation)