$f$ is second order diffierentiable over $[a,b]$, $f(a)=f(b)=0$, $|f''(x)|\leqslant M$, Show that $\left|\displaystyle\int_a^b f(x) dx\right|\leqslant \dfrac{M}{12}(b-a)^3$?

To make out the coefficient $\dfrac{1}{12}$, I tried this way:

Write that $c=\dfrac{a+b}{2}$, according to the Taylor's formula, $$f(x)=f(c)+f'(c)(x-c)+\dfrac{f''(\eta)}{2}(x-c)^2,$$ Putting $x=a, x=b$, notice that $f(a)=f(b)=0$, then $$ \begin{split} 0&=f(c)+f'(c)(a-c)+\dfrac{f''(\eta_1)}{2}(a-c)^2,\\ 0&=f(c)+f'(c)(b-c)+\dfrac{f''(\eta_2)}{2}(b-c)^2,\\ \end{split} $$ Then according to Darboux theorem, there exists $\zeta$ such that $$f(c)=-\dfrac{(b-a)^2}{16}(f''(\eta_1)+f''(\eta_2))=-\dfrac{(b-a)^2}{8}f''(\zeta).$$ Then $$f(x)=\color{red}{{-\dfrac{(b-a)^2}{8}f''(\zeta)}}+\color{blue}{f'(c)(x-c)}+\color{green}{\dfrac{f''(\eta)}{2}(x-c)^2}.$$

Integrate both sides, the red part gives $-\dfrac{1}{8}$, the blue part gives zero, and the green part gives $\dfrac{1}{24}$, then the coeffcient $\dfrac{1}{12}$ shows up.

However, I cannot get a proper inequality from this.

  • $\begingroup$ For $M>0,\,a<b$, the choice $f(x)=f_0(x):=\frac{M}{2}(x-a)(b-x)$ obtains $\int_a^b f_0(x)dx=\frac{M}{12}(b-a)^3$. Can you see how this bounds the general case? $\endgroup$
    – J.G.
    Sep 11, 2018 at 16:11
  • 1
    $\begingroup$ Probably this can help you with the inequality: Integral form of the remainder from Taylor's Remainder Theorem $$R_{k}(x)=\int _{a}^{x}{\frac {f^{(k+1)}(t)}{k!}}(x-t)^{k}\,dt$$ If the upper bound is $M$: $$|f^{(k+1)}(x)|\leq M$$ Then, we have: $$|R_{k}(x)|\leq M{\frac {|x-a|^{k+1}}{(k+1)!}}$$ $\endgroup$
    – paulplusx
    Sep 11, 2018 at 16:34
  • 1
    $\begingroup$ @paulplusx This is not useful here. For one thing it doesn't take into account the information that $f(a) = f(b) = 0$ so it does not give the best prefactor $\frac{1}{12}$. $\endgroup$
    – Winther
    Sep 11, 2018 at 16:54
  • $\begingroup$ @Winther Understood. Thanks for pointing it out. $\endgroup$
    – paulplusx
    Sep 11, 2018 at 17:00

1 Answer 1


Let $c:=\dfrac{a+b}{2}$. Integration by parts (using $f(a)=0$ and $f(b)=0$) gives $$\int_a^b\,f(x)\,\text{d}x=-\int_a^b\,(x-c)\,f'(x)\,\text{d}x\,.$$ By the Mean Value Theorem, $$f'(x)=f'(c)+(x-c)\,f''\big(\xi(x)\big)\text{ for all }x\in[a,b]\,,$$ where $\xi(x)$ is a number (inclusively) between $x$ and $c$. That is, we obtain $$\left|\int_a^b\,f(x)\,\text{d}x\right|=\left|\int_a^b\,(x-c)\,f'(c)\,\text{d}x+\int_a^b\,(x-c)^2\,f''\big(\xi(x)\big)\,\text{d}x\right|\,.$$ Since $\displaystyle \int_a^b\,(x-c)\,f'(c)\,\text{d}x=0$ and $\big|f''(t)\big|\leq M$ for all $t\in[a,b]$, we conclude that $$\left|\int_a^b\,f(x)\,\text{d}x\right|\leq M\,\left|\int_a^b\,(x-c)^2\,\text{d}x\right|=M\,\left(\frac{(b-a)^3}{12}\right)\,.$$ The inequality becomes an equality if and only if

  • $f(x)=+M(x-a)(b-x)$ for all $x\in[a,b]$, or
  • $f(x)=-M(x-a)(b-x)$ for all $x\in[a,b]$.

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