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Let $\,f$ be three times differentiable on the interval $[a, b]$. Show that there is a $c$ such that: $$f(b) - f(a) = \frac{b-a}{2}(f'(a)+f'(b)) - \frac{(b-a)^3}{12}f'''(c).$$ Answered in the comments.

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Henrik
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    This looks similar to a Taylor's Theorem problem, but you need to be clever, since the second-derivative term is missing from above. But there should be a way around that. – ShallowBlue Nov 05 '14 at 22:06
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    I am surprised to see the simple trapezoidal rule for integration coming up in so many forms. Put $g(x)=f'(x)$ and then you have $$\int_{a}^{b}g(x),dx=\frac{b-a}{2}\cdot{g(a)+g(b)} -\frac{(b-a)^{3}}{12}g''(c)$$ and a proof is available at http://math.stackexchange.com/a/535304/72031 – Paramanand Singh Nov 06 '14 at 12:20
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    In reality it is $f'''(s_1)$ and $f'''(s_2)$, and not two times $f'''(s)$ with the same $s$. – Christian Blatter Nov 06 '14 at 12:25
  • @ParamanandSingh Thanks! – Henrik Nov 07 '14 at 10:03

1 Answers1

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Here is a method that uses Rolle's theorem. Let

\begin{equation} g(x) = f(a) + \frac{x - a}{2}(f'(a) + f'(x)) - R(x - a)^3, \quad x\in [a,b]. \end{equation}

Here, $R$ is the constant that makes $f(b) = g(b)$. Note that $g(a) = f(a)$, $g'(a) = f'(a)$ and $g''(a) = f''(a)$. Set $h(x) = f(x) - g(x)$, for all $x\in [a,b]$. Then $h(a) = h(b) = 0$, hence $h'(c_1) = 0$ for some $c_1\in (a,b)$. Since $h'(a) = 0$ and $h'(c_1) = 0$, $h''(c) = 0$ for some $c\in (a,c_1)$. Thus $f''(c) = g''(c)$, or

\begin{equation} f''(c) = f''(c) + \frac{c - a}{2}f'''(c) - 6R(c - a). \end{equation}

Simplifying this equation results in

\begin{equation} R = \frac{f'''(c)}{12}. \end{equation}

Thus

\begin{equation} f(b) - f(a) = g(b) - f(a) = \frac{b - a}{2}(f'(a) + f'(b)) - \frac{(b - a)^3}{12}f'''(c). \end{equation}

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