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# past paper question core 3 watch

1. on jan 2012,q5, part (iii) , it says to use simpson's rule to work the question out. i stated what paper it is and which question it is since i couldn't include the table or the graph here.

Anyways, when i tried to do this, i got:

2/3(1+26) + 4(14+23)+ 2(8+25+19) , which is wrong according to the mark scheme? Can someone elaborate why please?
2. Which exam board, which course specification and which paper?
3. (Original post by old_engineer)
Which exam board, which course specification and which paper?
Ocr core 3 the old spec
4. OK, I've had a quick look. In the most basic form of Simpson's rule (three values, two strips), the coefficients are (1, 4, 1). For seven values and six strips you are effectively applying this three times, to (x0, x1, x2), (x2, x3, x4) and (x4, x5, x6). Note the overlap for x2 and x4. As a result, the pattern followed by the coefficients is (1,4,2,4,2,4,1), not (1,2,4,2,4,2,1).
5. (Original post by elvss567)
on jan 2012,q5, part (iii) , it says to use simpson's rule to work the question out. i stated what paper it is and which question it is since i couldn't include the table or the graph here.

Anyways, when i tried to do this, i got:

2/3(1+26) + 4(14+23)+ 2(8+25+19) , which is wrong according to the mark scheme? Can someone elaborate why please?
6. (Original post by old_engineer)
OK, I've had a quick look. In the most basic form of Simpson's rule (three values, two strips), the coefficients are (1, 4, 1). For seven values and six strips you are effectively applying this three times, to (x0, x1, x2), (x2, x3, x4) and (x4, x5, x6). Note the overlap for x2 and x4. As a result, the pattern followed by the coefficients is (1,4,2,4,2,4,1), not (1,2,4,2,4,2,1).
hi, i'm not sure what you mean by coefficients during simpson's rule?
7. (Original post by Notnek)
will do!

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