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\[\int\limits_{0}^{1} \frac{ 1 }{ x+1 }\] when n=4 trapezoidal rule/simpson's rule error
oops i forgot a dx
\[E _{trapezoidal} =\frac{ (b-a)^3 }{ 12n^2 }(\max \left| f"(x) \right|) \] how would you find f"(x)? and how did they get 2? f"(x) = -2(x+1)^-3
wait, i see what i did wrong...
write f as\[f(x)=(x+1)^{-1} => f'(x)=-1(x+1)^{-2} => f''(x)=-1*-2(x+1)^{-3}\]So\[f''(x)=2(x+1)^{-3}=\frac{ 2 }{ (x+1)^3 }\]
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yeah i saw what i did wrong. i forgot a negative >:(
Cheer up, you've found out yourself what went wrong!
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