How do you find the average value of #sinx# as x varies between #[0,pi]#?

Question

Christopher Allers · Accepted Answer

#2/pi#

The average value of a function #f# on the interval #[a,b]# is found through the integral expression

#1/(b-a)int_1^bf(x)dx#

Here, this gives us an average value of

#1/(pi-0)int_0^pisin(x)dx#

The antiderivative of #sin(x)# is #-cos(x)#:

#=1/pi[-cos(x)]_0^pi#

#=1/pi(-cos(pi)-(-cos(0)))#

#=1/pi(-(-1)-(-1))#

#=2/pi#

Axel Alvarez · Answer

undefined To find the average value of $ \sin(x) $ as $ x $ varies between $ 0 $ and $ \pi $, you would first integrate $ \sin(x) $ over the given interval, which is $ [0, \pi] $, and then divide the result by the length of the interval. The integral of $ \sin(x) $ with respect to $ x $ is $ -\cos(x) $. Evaluating this integral from $ 0 $ to $ \pi $ gives $ -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 - (-1) = 2 $. Since the length of the interval $ [0, \pi] $ is $ \pi - 0 = \pi $, the average value of $ \sin(x) $ over this interval is $ \frac{2}{\pi} $.