If $f''(x)$: conti- on $[a,b]$, then there exist $c$ ($c\in[a,b]$) s.t $\int_{a}^{b}f(x)dx=\frac{b-a}{2}(f(b)+f(a))-\frac{(b-a)^3}{12}f''(c)$

Question

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Q: Proof that

If $f''(x)$: conti- on $[a,b]$, then there exist $c$ ($c\in[a,b]$) s.t $$\int_{a}^{b}f(x)dx=\frac{b-a}{2}(f(b)+f(a))-\frac{(b-a)^3}{12}f''(c)$$ It is MVT?

Answer 1

Put $$m:={a+b\over2},\quad h:={b-a\over2}\ ,$$ and do two partial integrations with clever choice of the integration constants: $$\int_a^b f(x)\ dx=(x-m)f(x)\biggr|_a^b -\int_a^b(x-m) f'(x)\ dx\ .$$ Here the fist part evaluates to $h\bigl(f(a)+f(b)\bigr)$, and for the second part we obtain $$\left.\left({h^2-(x-m)^2\over 2}f'(x)\right)\right|_a^b-{1\over2}\int_a^b \bigl(h^2-(x-m)^2\bigr) f''(x)\ dx\ .$$ This time the first part vanishes. For the second part note that $h^2-(x-m)^2\geq0$ for $a\leq x\leq b$. Therefore by the weighted form of the mean value theorem there is a $c\in[a,b]$ with $${1\over2}\int_a^b \bigl(h^2-(x-m)^2\bigr) f''(x)\ dx={1\over2} f''(c)\int_a^b \bigl(h^2-(x-m)^2\bigr)\ dx\ .$$ Since the last integral evaluates to ${4\over3}h^3$ the claim follows.