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Mathematics 27 Online
OpenStudy (anonymous):

Calculus question?

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

a)What is the average value of \[f(x)=\sqrt{1-x^2}\] over the interval \[0\le x \le 1\]? b) How can you tell whether this average value is more or less than 0.5 without doing any calculations?

OpenStudy (anonymous):

(a)The average value of f(x) is the integral over the interval, so the area under sqrt(1-x^2) from 0 to 1 divided by 1. (b) Since this graph (from 0 to 1) is basically the quarter of a circle of radius one, we just use (1/4)pi(1^2) for the average, which is greater than 0.5

OpenStudy (anonymous):

I got 0.785 for part (a), not sure if it is correct..

OpenStudy (anonymous):

It is.

OpenStudy (anonymous):

Great. Appreciate it!

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