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    I know how to use trapezium rule,but I keep getting a wrong answer and I still can't figure out what is wrong
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    Are you doing:
    (1.first value) + (2.second value) + (2.third value) ...... + (1.last value)
    = Ans,

    Then multiply the express by h/2, where h = (b-a)/(number of intervals) = The gap between each of the coordinates.

    If that express finds M^2, you'd then need to square root it to find M.
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    (Original post by Gaz031)
    Are you doing:
    (1.first value) + (2.second value) + (2.third value) ...... + (1.last value)
    = Ans,

    Then multiply the express by h/2, where h = (b-a)/(number of intervals) = The gap between each of the coordinates.

    If that express finds M^2, you'd then need to square root it to find M.
    I know that I did that already but I found out that the error is in the hight,which is b-a!!!,we have a as a zero and b 1,so 1-0/5=.2,yet the h should be 1.25,confusing,eh?! :eek: :confused:
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    (Original post by habosh)
    I know that I did that already but I found out that the error is in the hight,which is b-a!!!,we have a as a zero and b 1,so 1-0/5=.2,yet the h should be 1.25,confusing,eh?! :eek: :confused:
    5 Values of V is 5 coordinates. 4 intervals! h = (b-a)/(number of intervals) = 0.25.
    Try redo it with the new value of h, correct?
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    Wow I'm gonna have a fun maths lesson today. We're doing the trapezium rule I think.
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    It's not difficult once you get used to it.
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    (Original post by mik1a)
    Wow I'm gonna have a fun maths lesson today. We're doing the trapezium rule I think.
    you'll be fine, all it comes down to in the end is learning a forumula
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    (Original post by Gaz031)
    5 Values of V is 5 coordinates. 4 intervals! h = (b-a)/(number of intervals) = 0.25.
    Try redo it with the new value of h, correct?
    God bless you hun,Now I remember ,stupid me,|I forgot it's 4 intervals,*LOL*
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    Ok cool, just did it. It's pretty simple really, It's the sort of thing I'd have expected to have seen at GCSE - writing down to 6 decimal places and relying on your calculator.

    area = h/2(y(0) + 2[y(1) + y(2) + ... + y(n-1)] + y(n)]
    where h is the interval length.

    does this tequnique become integration when h tends to zero? because that's what you do really.. divide it into infinitely small slithers.
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    (Original post by habosh)
    God bless you hun,Now I remember ,stupid me,|I forgot it's 4 intervals,*LOL*
    No worries, anytime.
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    (Original post by mik1a)
    Ok cool, just did it. It's pretty simple really, It's the sort of thing I'd have expected to have seen at GCSE - writing down to 6 decimal places and relying on your calculator.

    area = h/2(y(0) + 2[y(1) + y(2) + ... + y(n-1)] + y(n)]
    where h is the interval length.

    does this tequnique become integration when h tends to zero? because that's what you do really.. divide it into infinitely small slithers.
    ur quite right...
 
 
 
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