Find the average value of f(x) = sin(x) over the interval [0, π].

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Find the average value of f(x) = sin(x) over the interval [0, π].

Mathematics
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Average value (A) of a function f(x) over the interval from x = a to x = b \[\Large A = \frac{1}{b-a} \int_{a}^{b} f(x) dx\]
pi/2?
In your case, you have a = 0 b = pi f(x) = sin(x) \[\Large A = \frac{1}{b-a} \int_{a}^{b} f(x) dx\] \[\Large A = \frac{1}{\pi-0} \int_{0}^{\pi} \sin(x) dx\] \[\Large A = \frac{1}{\pi} \int_{0}^{\pi} \sin(x) dx\] \[\Large A = ???\]

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Other answers:

what's the integral of sin(x) ?
-cos
yes -cos(x) + C we can drop the +C because we have a definite integral
evaluate -cos(x) when x = pi evaluate -cos(x) when x = 0 subtract the results to get ???
2
I got pi over 2 but someone told me just 2
or sorry 2 over pi
yeah the integral portion should evaluate to 2
which is why A = 2/pi
okay thank you very much for clearing that up
np

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