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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.)

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.

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.

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