Numerical Methods-Simpson's Rule MEI NM
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Numerical Methods-Simpson's Rule MEI NMHi,
http://www.ocr.org.uk/download/pp_07..._jun_l_gce.pdf
I'm having trouble with question 2 in the paper above. I can calculate the trapezium and midpoint rule estimates with h=0.5 fine, but I don't understand how to calculate the Simpson's rule estimate with h=0.25 without calculating more trapezium and midpoint rule values.
The markscheme (http://www.ocr.org.uk/download/ms_07..._l_gce_jun.pdf) says to use the formula Sn=2Mn+Tn/3, and this works, but I don't see why it does without more values.
Can anyone help? -
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Re: Numerical Methods-Simpson's Rule MEI NMYou don't (really) use a step value (of h) when using Simpson's rule. You use a mid-point value.
Simpson's rule is![\int_a^bf(x) dx=\frac{b-a}{6}\left[f(a)+4f(\frac{a+b}{2})+f(b) \right]](http://www.thestudentroom.co.uk/latexrender/pictures/d7/d73630e6bcd98a8baabdff6ac03a783b.png "\int_a^bf(x) dx=\frac{b-a}{6}\left[f(a)+4f(\frac{a+b}{2})+f(b) \right]")
When the question says, use
, that's because the interval 
is 0.5, giving a mid-point value of 0.25.
When it says use a step value of
that means split the interval up into two intervals, each of length 
and mid-point values of 
Now use Simpson's rule to work out the areas of the two sub-intervals and then add them together. -
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Re: Numerical Methods-Simpson's Rule MEI NMThanks(Original post by steve10)
You don't (really) use a step value (of h) when using Simpson's rule. You use a mid-point value.
Simpson's rule is
When the question says, use, that's because the interval is 0.5, giving a mid-point value of 0.25.
When it says use a step value of that means split the interval up into two intervals, each of length and mid-point values of
Now use Simpson's rule to work out the areas of the two sub-intervals and then add them together.
