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Romberg Integration is used to approximate $\int_2^3 \! f(x) \, \mathrm{d} x$ If $f(2)=.51342, f(3)=.36788, R_{3,1}=.43687 , R_{3,3}=.43662$ Find $f(2.5)$

I am quite confused how to proceed since I don't even know what the function is. The only peice of information that may be helpful is $R_{1,1}= { h \over 2}(f(2)+f(3))={1 \over 2}(.51342+.36788)=.44065$
Maybe putting this into a piece-wise function may be helpful

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    @joriki I have a few questions that have not been answered. When I have the time I will answer them so that others can have the answer2012-10-14

1 Answers 1

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$R_{1,1}$

$R_{2,1},R_{2,2}$

$R_{3,1}, R_{3,2}, R_{3,3}$

We need to use the recursive formulas. The first formula is simple and easy: $R_{1,1}={h \over 2}( f(a)+f(b))={1 \over 2}(.51342+.36788)=.44065$ $R_{3,3}={16R_{3,2}-R_{2,2} \over 15}=.43662$ $R_{3,2}={4R_{3,1}-R_{2,1} \over 3}={4(.43687)-R_{2,1} \over 3}=.58249-{ 1 \over 3}R_{2,1}$ $R_{2,2}={4R_{2,1}-R_{1,1} \over 3}={4r_{2,1}-.44065\over 3}={4 \over 3}R_{2,1}-.14688$

Now we have to subsitute $R_{3,2}$ and $R_{2,2}$ into $R_{3,3}$ $R_{3,3}={1 \over 15}(16(.58249-{1 \over 3}R_{2,1})-({4 \over 3}R_{2,1}-.14688)=.43662$ Now that we only have one variable its simple to solve and $R_{2,1}=.43761$ Now we can solve $R_{2,1}$ $R_{2,1}={ 1\over 2}R_{1,1}+h_1f(a+h_2)={ 1\over 2}R_{1,1}+h_1f(2.5)$ $f(2.5)={2R_{2,1}-R_{1,1} \over 2h}={2(.43761)-(.44065) \over 2(.5)}=.43457$ So we can conclude that $f(2.5)=.43457$