# Let $f(x)\in C^2[0,2]$, prove that there exists a $\xi\in(0,2)$ such that $\int_0^2f(x)dx=2f(1)+\frac{f''(\xi)}{3}$.

Let $$f(x)$$ be a twice continuously differentiable function over $$[0,2]$$, i.e. $$f(x)\in C^2[0,2]$$, prove that there exists a $$\xi\in(0,2)$$ such that $$\int_0^2f(x)dx=2f(1)+\frac{f''(\xi)}{3}.$$

First, my naive idea is to use Taylor expansion with Lagrange remainder $$f(x)=f(1)+f'(1)(x-1)+\frac{f''(\xi)}{2}(x-1)^2.$$ Then integrate both sides over $$[0,2]$$ we have that $$\int_0^2f(x)dx=\int_0^2f(1)dx+\int_0^2f'(1)(x-1)dx+\int_0^2\frac{f''(\xi)}{2}(x-1)^2dx=2f(1)+\frac{f''(\xi)}{3}.$$ When I finished writing this, I realized that I was wrong, because in Lagrange remainder, $$\xi$$ is depend on $$x$$ and it is not a constant, for different $$x$$ we get different $$\xi$$, so we can not deduce that $$\int_0^2\frac{f''(\xi)}{2}(x-1)^2dx=\frac{f''(\xi)}{3}.$$ Then I tried many different ideas and failed until I came up with taylor's theorem with Cauchy remainder $$f(x)=f(1)+f'(1)(x-1)+\frac{1}{2!}\int_1^x(x-t)^2f''(t)dt.$$ Integrate both sides over $$[0,2]$$ we have that $$\int_0^2f(x)dx=2f(1)+\int_0^2\left(\int_1^x(x-t)^2f''(t)dt\right)dx.$$ And $$\int_0^2\left(\int_1^x(x-t)^2f''(t)dt\right)dx =\int_0^1\left(\int_1^x(x-t)^2f''(t)dt\right)dx+\int_1^2\left(\int_1^x(x-t)^2f''(t)dt\right)dx.$$ Since $$\int_0^1\left(\int_1^x(x-t)^2f''(t)dt\right)dx= -\int_0^1\left(\int_x^1(x-t)^2f''(t)dt\right)dx=-\int_0^1\left(\int_0^t(x-t)f''(t)dx\right)dt=\int_0^1f''(t)\frac{t^2}{2}dt=\int_0^1f''(x)\frac{x^2}{2}dx.$$ Similarly, we have that $$\int_1^2\left(\int_1^x(x-t)^2f''(t)dt\right)dx=\int_1^2f''(x)\frac{(2-x)^2}{2}dx.$$ Thus we have the following formula $$\int_0^2f(x)dx=2f(1)+\int_0^1f''(x)\frac{x^2}{2}dx+\int_1^2f''(x)\frac{(2-x)^2}{2}dx.$$ Use this and the first mean value theorem of definite integral, there exists $$\xi_1\in(0,1)$$ and $$\xi_2\in(1,2)$$ such that $$\int_0^1f''(x)\frac{x^2}{2}dx=f''(\xi_1)\int_0^1\frac{x^2}{2}dx=\frac{f''(\xi_1)}{6}.$$ $$\int_1^2f''(x)\frac{(2-x)^2}{2}dx=f''(\xi_2)\int_1^2\frac{(2-x)^2}{2}dx=\frac{f''(\xi_2)}{6}.$$ Since $$f''(x)$$ is continuous, by the intermediate value theorem there exists $$\xi\in(0,2)$$ such that $$\frac{f''(\xi_1)}{6}+\frac{f''(\xi_2)}{6}=\frac{f''(\xi)}{3}.$$ Thus we finish our proof.

My Question： (1) Is there a simple way to prove this exercise?

(2) Whether the formula $$\int_0^2f(x)dx=2f(1)+\int_0^1f''(x)\frac{x^2}{2}dx+\int_1^2f''(x)\frac{(2-x)^2}{2}dx$$ is a special case of a general integral formula, like the Euler-Maclaurin formula or Trapezoidal rule?

(answer for the first part) What about if you argue like this: take $$R=\int_0^2\frac{f''(\xi(x))}{2}(x-1)^2dx, \xi(x) \in [0,2]$$

Let $$m=\min_{x \in [0,2]}f''(x), M=\max_{x \in [0,2]}f''(x)$$, so $$m \le f''(\xi(x)) \le M$$

You can multiply with $$(x-1)^2/2 \ge 0$$ and integrate so you get $$m/3 \le R \le M/3$$ so by the intermediate value theorem for $$f''$$, there is $$\xi \in [0,2]$$ st $$R=\frac{f''(\xi)}{3}$$ and you are done!

• Oh yeah, why didn't I think of that
– HGF
Commented Dec 14, 2023 at 2:15

The result holds even if $$f$$ is only assumed to be twice differentiable, the continuity of $$f''$$ is not needed.

The function $$F(x) = \int_{1-x}^{1+x} f(t) \, dt - 2 x f(1)$$ is thrice differentiable with $$F(0) = F'(0) =F''(0) = 0$$ and $$F'''(x) = f''(1+x) + f''(1-x) \, .$$ Taylor's theorem (with the Lagrange remainder) applied to $$F$$ gives $$\int_0^2 f(x) \, dx - 2 f(1) = F(1) = \frac{F'''(\eta)}{3!} = \frac 13 \left( \frac{f''(1+\eta)+f''(1-\eta)}{2}\right)$$ for some $$\eta \in (0, 1)$$.

(Remark: This is similar to what you got as $$\frac{f''(\xi_1)}{6}+\frac{f''(\xi_2)}{6}$$, but without assuming that $$f''$$ is continuous or even integrable.)

Since $$f''$$ has the intermediate value property (Darboux's theorem) is $$\frac{f''(1+\eta)+f''(1-\eta)}{2} = f''(\xi)$$ for some $$\xi \in [1-\eta, 1+\eta] \subset (0, 2)$$, and that completes the proof.

• Very beautiful construction proof.
– HGF
Commented Dec 14, 2023 at 12:31
• $f''$ has Darboux (Intermediate value theorem) property even if discontinuous so the proof I gave also applies though one has to work with infimum and supremum here, and I agree that the result holds then too; cool proof here though Commented Dec 14, 2023 at 14:01