Calculus doubt in a research paper I am trying to read the research paper here and have a small doubt in it. On page 177 the following integration is carried out:
$$\int_{t_n-\tau}^{t_{n+1}-\tau}z^{\alpha-2}dz=h(z^*_1(\tau))^{\alpha-2}$$
Here $h>0$, $t_i=ih,\tau\in(0,t_n)$ and $z^*_1(\tau)\in (0,t_2)$.
It is not clear to me why this equality is true. My idea is that it has something to with the mean value theorem of integral calculus but I am not sure as to how it applies. Can someone help please?
Thanks
 A: On page 177, the paper deals with integrals of the form
$$
\int_0^{t_1}\left(\int_{t_n-\tau}^{t_{n+1}-\tau}z^{\alpha-2}\,\mathrm dz\right)f(\tau,x(\tau))\,\mathrm d\tau,\quad\ldots\quad\int_{t_{n-1}}^{t_n}\left(\int_{t_n-\tau}^{t_{n+1}-\tau}z^{\alpha-2}\,\mathrm dz\right)f(\tau,x(\tau))\,\mathrm d\tau.
$$
So let's pick a $k\in \{1,\ldots,n\}$ and consider the integral
$$
\int_{t_{k-1}}^{t_k}\left(\int_{t_n-\tau}^{t_{n+1}-\tau}z^{\alpha-2}\,\mathrm dz\right)f(\tau,x(\tau))\,\mathrm d\tau
$$
with $t_0=0$. 
Let $\tau\in (t_{k-1},t_k)$ be fixed. By the mean value theorem you mention, we have the existence of a $z_k^*(\tau)\in (t_n-\tau,t_{n+1}-\tau)$ such that
$$
\int_{t_n-\tau}^{t_{n+1}-\tau}z^{\alpha-2}\,\mathrm dz=h\cdot z_k^*(\tau)^{\alpha-2},
$$
with $h=t_{n+1}-\tau-(t_n-\tau)=t_{n+1}-t_n$. Let's take a closer look at where $z_k^*(\tau)$ can vary for a fixed $\tau$. Since 
$$
t_n-\tau< z_k^*(\tau)< t_{n+1}-\tau\quad \text{and}\quad t_{k-1}<\tau<t_k
$$
we have that 
$$
t_n-t_k<z_k^*(\tau)<t_{n+1}-t_{k-1}.
$$
Now, on page 176 it says that $t_n=nh$ and so $t_{n+1}-t_{k-1}=t_{n-k+2}$ and $t_n-t_k=t_{n-k}$. Thus
$$
z_k^*(\tau)\in (t_{n-k},t_{n-k+2}).
$$
Note that the $z_k^*$'s in the paper are reversed compared to mine, i.e. $z_k^*$ here is $z_{n-k}^*$ in the paper.
