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anonymous

  • one year ago

Average value of a function on an interval. Will post equstion and work in a comment

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  1. anonymous
    • one year ago
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    \[\frac{ 1 }{ \pi - 0 } \int\limits\limits_{0}^{\pi} \sin (x) dx\] \[\int\limits_{0}^{\pi} \sin(x) dx = [-\cos] _{0}^{\pi}\] \[[-\cos \pi] - [-\cos 0] => [-(-1)] - [-1] = 1 + 1 = 2\] Average value = \[\frac{ 1 }{ \pi } (2) = \frac{ 2 }{ \pi }\] I'm asked to find the x value that corresponds to 2/pi sin(x) = 2/pi using the arcsin on the calculator I get ~ 0.690 But there is a second x value that corresponds to 2/pi, which is 2.451, how do I solve for the 2nd x value?

  2. anonymous
    • one year ago
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    Can I find the 2nd x value algebraically, using calculus, or using a graphing calculator?

  3. javalos01
    • one year ago
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    @rushwer

  4. javalos01
    • one year ago
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    @Rushwr

  5. IrishBoy123
    • one year ago
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    it's going to be symmetrical about \(\pi/2\)|dw:1443181191167:dw|

  6. anonymous
    • one year ago
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    @IrishBoy123 Thank you. How do I calculate it?

  7. IrishBoy123
    • one year ago
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    |dw:1443192745454:dw|

  8. IrishBoy123
    • one year ago
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    |dw:1443192978309:dw|

  9. anonymous
    • one year ago
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    Ok. This is what I did. Please let me know if I over complicated it, or if there's a simpler way. Since the interval is [0, pi] and the first co-ordinate is 0 + arcsin(2/pi) the second co-ordinate is pi - arcsin(2/pi) = 2.451

  10. IrishBoy123
    • one year ago
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    seems fair enough

  11. anonymous
    • one year ago
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    @IrishBoy123 How did you calculate it?

  12. IrishBoy123
    • one year ago
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    the calculator gives you the first answer, but you know from the curve that there are two. the symmetry of the sine curve gives you the second, which was what i was getting at with this: |dw:1443210395870:dw| but from knowing \(\sin(\pi - x) = \sin \pi \cos x - \sin x \cos \pi = \sin x\) you migt hav egot a result from another direction but i saw the symmetry of the curve....if that answers your question

  13. anonymous
    • one year ago
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    I get the symmetry of the sine curve. I don't understand sin(π−x)=sinπcosx−sinxcosπ=sinx

  14. IrishBoy123
    • one year ago
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    second part was just saying/showing that \(sin (\pi - x) = \sin x\)

  15. anonymous
    • one year ago
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    ok thanks for your help

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