The graph of the function f(x)=[tex]\sqrt{a^2-x^2}[/tex] (-a<= x <=a) is a semicircle, centre the origin and bounding diameter the portion of the x-axis between the points A(-a,0) and B(a,0). Show that the mean value with respect to x of the ordinates of the semicircle is [tex]\pi[/tex]a/4. ( This question can be done without using calculus.)(adsbygoogle = window.adsbygoogle || []).push({});

Does ordinates mean the x-coordinates only of the points?

My attempt, mean value =[tex]\frac{1}{b-a}\int_a^{b}f(x)dx = \frac{1}{2a}\int_{-a}^{a}(\sqrt{a^2-x^2})dx = \frac{1}{4a}\int_{-a}^{a}\ln({a^2-x^2})dx[/tex].

I'm interested to know how it can be done without using calculus? Thanks for the help.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Mean value ordinate

**Physics Forums | Science Articles, Homework Help, Discussion**