anonymous
  • anonymous
Find the average value of f(x) = sin(x) over the interval [0, π].
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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jim_thompson5910
  • jim_thompson5910
Average value (A) of a function f(x) over the interval from x = a to x = b \[\Large A = \frac{1}{b-a} \int_{a}^{b} f(x) dx\]
anonymous
  • anonymous
pi/2?
jim_thompson5910
  • jim_thompson5910
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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jim_thompson5910
  • jim_thompson5910
what's the integral of sin(x) ?
anonymous
  • anonymous
-cos
jim_thompson5910
  • jim_thompson5910
yes -cos(x) + C we can drop the +C because we have a definite integral
jim_thompson5910
  • jim_thompson5910
evaluate -cos(x) when x = pi evaluate -cos(x) when x = 0 subtract the results to get ???
anonymous
  • anonymous
2
anonymous
  • anonymous
I got pi over 2 but someone told me just 2
anonymous
  • anonymous
or sorry 2 over pi
jim_thompson5910
  • jim_thompson5910
yeah the integral portion should evaluate to 2
jim_thompson5910
  • jim_thompson5910
which is why A = 2/pi
anonymous
  • anonymous
okay thank you very much for clearing that up
jim_thompson5910
  • jim_thompson5910
np

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