$R_{1,1}$
$R_{2,1},R_{2,2}$
$R_{3,1}, R_{3,2}, R_{3,3}$
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 $f(2.5)=.43457$