 # The formula for local midpoint rule in the interval [ &#x2212;<!-- − --> Elle Weber 2022-04-07 Answered
The formula for local midpoint rule in the interval
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Step 1
Set
$g\left(h\right)={\int }_{-h/2}^{h/2}f\left(x\right)dx-hf\left(0\right)$
and compute
$g\left(0\right)=0\phantom{\rule{0ex}{0ex}}{g}^{\prime }\left(h\right)=\frac{1}{2}\left(f\left(h/2\right)+f\left(-h/2\right)\right)-f\left(0\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{g}^{\prime }\left(0\right)=0\phantom{\rule{0ex}{0ex}}$
etc. to find the Taylor expansion of g.

We have step-by-step solutions for your answer! Noelle Wright
Step 1
Put $f\left(x\right)=f\left(0\right)+x{f}^{\prime }\left(0\right)+{x}^{2}{f}^{″}\left(0\right)/2+...$ so ${\int }_{-h/2}^{h/2}f\left(x\right)dx=hf\left(0\right)+{h}^{3}{f}^{″}\left(0\right)/\left(2\ast 3\ast 8\right)+...$ .

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