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# TRAPEZIUM question! watch

1. 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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3. 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.
4. (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?!
5. (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?!
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?
6. Wow I'm gonna have a fun maths lesson today. We're doing the trapezium rule I think.
7. It's not difficult once you get used to it.
8. (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
9. (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*
10. 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.
11. (Original post by habosh)
God bless you hun,Now I remember ,stupid me,|I forgot it's 4 intervals,*LOL*
No worries, anytime.
12. (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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