Obtention and estimate error of trapezoid rule: Why was necessary to mention that $f''(\xi)$ is constant?

Question

I'm considering the following integral

$$I=\int_a^b[f(a)+\Delta f(a)\alpha+\frac {f''(\xi)}{2}\alpha(\alpha-1)h^2]dx$$

where $a\le x\le b, h=b-a, \Delta f(a)=f(b)-f(a), \xi$ is a number between $a$ and $b$. If $\alpha=\frac{x-a}{h}$ then $dx=hd\alpha$. Thus

$$I=h\int_0^1[f(a)+\Delta f(a)\alpha+\frac {f''(\xi)}{2}\alpha(\alpha-1)h^2]d\alpha$$

If one assumes that for $h$ small, $f''(\xi)$ is approximately constant, then

$$I=h\left[\alpha f(a)+\frac{\alpha^2}{2}\Delta f(a)+\left(\frac{\alpha^3}{6}-\frac{\alpha^2}{4}\right)+f''(\xi)h^2\right]_0^1d\alpha$$

My question is, why is the author making this assumption of "If one assumes that for $h$ small, $f''(\xi)$ is approximately constant"? Wasn't already a constant?

Thank you in advance

Answer 1

The $\xi$ in the remainder term depends on the value of $\alpha$ (and $x$), so in principle it actually should be part of the integration, and is not a constant factor that can be pulled out of the integral. If however for all $\alpha$ (or $x$) in the relevant range, the quantity $f''(\xi)$ is bounded, then you can bound how large the integral of the last term might be.

Answer 2

In the remainder of a linear interpolation at $a$ and $b$ you get $$ f(x)=p_1(x)+(x-a)(x-b)r(x). $$ On the one hand, one can show that for twice differentiable $f$ the function $r$ is continuous in $x$. On the other hand, as you used it, it can be shown that there is a $\xi_x$ in the hull of the points $a,b,x$ so that $r(x)=\frac{f''(\xi_x)}2$.

For the integration you can now take the first version to apply the mean value theorem, $$ \int_a^b(x-a)(x-b)r(x)dx=r(\eta)\int_a^b(x-a)(x-b)dx. $$ After that use the connection to the second derivative with $\xi=\xi_\eta$. So there is no need to consider $h$ small enough or $\xi$ nearly constant or such.