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sauravshakya

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

  • one year ago
  • one year ago

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

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

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

    • one year ago
  4. sauravshakya
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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|

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

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

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

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

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

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

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

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

    • one year ago
  13. experimentX
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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.

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

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

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

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

    • one year ago
  18. experimentX
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    yep!!

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

    • one year ago
  20. sauravshakya
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    thanx again

    • one year ago
  21. experimentX
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    yw

    • one year ago
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