# Mean Value Theorem with variable bounds?

In the exercise, applying the Mean Value Theorem to the function $f(x)=x^j$ in the interval $[\frac{k-i}{k},\frac{k-i+1}{k}]$, they obtain:

$\left ( \frac{k-i}{k}\right )^j-\left ( \frac{k-i+1}{k}\right )^j= \left (\frac{k-i}{k}-\frac{k-i+1}{k}\right )j x_i^{j-1}=\frac{j}{k}x_i^{j-1}$.

for some $x_i \in [\frac{k-i}{k},\frac{k-i+1}{k}]$.

My question is how they can make $x_i$ as a sequence like this? Is it simply because the bounds of the interval are depending on $i$? or there exists a version of the MVT such that they get such a sequence when the interval bounds are variable?

Thank you.

• Just for context, are you wanting to prove that $x\to x^j$ is riemann integrable? – Git Gud Jun 13 '13 at 9:47
• I don't think so. I'm just asking how $x_i$ is variable? – Aissa_09 Jun 13 '13 at 9:56
• It's, as you say, simply because the bounds of the interval depend on $i$. For each fixed $i$, they find an $x_i$ corresponding to the interval $[{k-i\over k},{k-i+1\over k}]$. – David Mitra Jun 13 '13 at 10:37

$x_i \in [\frac{k-i}{k},\frac{k-i+1}{k}]$ can actually be $x_i \in (\frac{k-i}{k},\frac{k-i+1}{k})$