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sauravshakya

  • one year ago

y=sinx find the average value of y in the interval 0 to 90

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  1. sauravshakya
    • one year ago
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    I mean find lim x-->0 {sin0 +sindx +sin2dx +sin3dx +...+sin(ndx)/(n+1)} where ndx=90

  2. sirm3d
    • one year ago
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    mean value theorem for integrals?

  3. sauravshakya
    • one year ago
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    No not mean value theorem... Here I just want to find the avereage value

  4. sauravshakya
    • one year ago
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    ndx=90 n=90/dx Now, lim x-->0 {sin0 +sindx +sin2dx +sin3dx +...+sin(ndx)/(n+1)} lim x-->0 {sin0 +sindx +sin2dx +sin3dx +...+sin(ndx)/(90/dx +1)} lim x-->0 {(sin0 +sindx +sin2dx +sin3dx +...+sin(ndx))dx/(90+dx)} |dw:1359042392858:dw|

  5. sauravshakya
    • one year ago
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    Now, from there I will get 1/90 Which surely not CORRECT

  6. sauravshakya
    • one year ago
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    Now, If I do ndx =pi/2 then I will get 2/pi

  7. sauravshakya
    • one year ago
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    @experimentX can u PLZ help

  8. experimentX
    • one year ago
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    this should be close to zero.|dw:1359042684509:dw|

  9. shubhamsrg
    • one year ago
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    Well 90 is degree, you surely can't use that in these calculations ? pi/2 is appropiate.

  10. experimentX
    • one year ago
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    hmm ... is 90 really degree?

  11. sauravshakya
    • one year ago
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    And yes that was what I was going to ask... Why does it matter

  12. experimentX
    • one year ago
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    degree is just a notation made for the sake of ease ... radian is the actual value.

  13. experimentX
    • one year ago
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    you might have noted lim x->0 sin(x)/x = 1 or ... as approx .. when 'x' is very small, sin(x) ~ x ... this never works for degree ... it sounds funny.

  14. sauravshakya
    • one year ago
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    So actually sin(pi/2 radian) =1

  15. sauravshakya
    • one year ago
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    And we supposed pi/2 radian =90 degree

  16. experimentX
    • one year ago
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    yes!! radian is more natural measure of angles.

  17. sauravshakya
    • one year ago
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    So, is the correct answer 2/pi

  18. experimentX
    • one year ago
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    yep!!

  19. experimentX
    • one year ago
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    compare these with the area bound by the curves with x axis an 0 and pi/2

  20. sauravshakya
    • one year ago
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    thanx again

  21. experimentX
    • one year ago
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    yw

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