find the average height of y=(2/3)*(sqrt (9 - x^2) ) over the interval [-3,3] i used the average height formula but i get the answer as 0 while the correct answer is supposed to be Pi/2 can someone help me out?
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$\displaystyle \frac{1}{3-(-3)} \int_{-3}^3 \frac{2}{3} \sqrt{9 - x^2} \, dx $ $\displaystyle \frac{1}{9} \int_{-3}^3 \sqrt{9 - x^2} \, dx$ $\displaystyle \frac{1}{9} \cdot \frac{\pi \cdot 3^2}{2}$
how did you integrate to get to the last step? i think thats where im erring
you don't need calculus to integrate, only geometry ... the graph of $\displaystyle y = \sqrt{9 - x^2}$ from -3 to 3 is a semicircle of radius 3.
oh i see but is it possible to use calculus and find the area because thats what the question wants i think
possible, but a bit more involved. use the trig substitution $\displaystyle x = 3\sin{\theta}$
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