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anonymous

  • one year ago

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

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  1. jim_thompson5910
    • one year ago
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    Average value (A) of a function f(x) over the interval from x = a to x = b \[\Large A = \frac{1}{b-a} \int_{a}^{b} f(x) dx\]

  2. anonymous
    • one year ago
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    pi/2?

  3. jim_thompson5910
    • one year ago
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    In your case, you have a = 0 b = pi f(x) = sin(x) \[\Large A = \frac{1}{b-a} \int_{a}^{b} f(x) dx\] \[\Large A = \frac{1}{\pi-0} \int_{0}^{\pi} \sin(x) dx\] \[\Large A = \frac{1}{\pi} \int_{0}^{\pi} \sin(x) dx\] \[\Large A = ???\]

  4. jim_thompson5910
    • one year ago
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    what's the integral of sin(x) ?

  5. anonymous
    • one year ago
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    -cos

  6. jim_thompson5910
    • one year ago
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    yes -cos(x) + C we can drop the +C because we have a definite integral

  7. jim_thompson5910
    • one year ago
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    evaluate -cos(x) when x = pi evaluate -cos(x) when x = 0 subtract the results to get ???

  8. anonymous
    • one year ago
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    2

  9. anonymous
    • one year ago
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    I got pi over 2 but someone told me just 2

  10. anonymous
    • one year ago
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    or sorry 2 over pi

  11. jim_thompson5910
    • one year ago
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    yeah the integral portion should evaluate to 2

  12. jim_thompson5910
    • one year ago
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    which is why A = 2/pi

  13. anonymous
    • one year ago
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    okay thank you very much for clearing that up

  14. jim_thompson5910
    • one year ago
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    np

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