Let $P_n := [0= x_0, x_1, ... , x_{n-1}, x_n = 5]$, $x_k= \frac{5k}{n}$. Compute L(f,$P_n$).
For reference, this is from Abbot's Understanding Analysis, 7.2.2, we have a different function.
I know how to find the $x_k - x_{k-1}$ from the book, I am just a little stuck on calculating $m_k$, which is the infinum of the function for $x \in [x_{n-1}, x_n]$. I know it is the infinum of the function within each smaller interval in the partition, which leads be to believe that for $x \in [x_{n-1}, x_n]$, the infinum will be $x_{n-1}^2$. Would this imply I would use the same formula for $x_k$ to find the function's infinum?
$f $ is increasing at each $[x_k,x_{k+1}] $ thus
$$m_k=\inf_{[x_k,x_{k+1}]}f=f (x_k)=x_k^2$$
thus
$$L (f,P_n)=\frac {5}{n}\sum_{k=0}^{n-1}x_k^2$$
$$=\frac {5}{n}\sum_{k=0}^{n-1}(\frac {5k}{n})^2$$
$$=\frac {125}{n^3}(1+4+9+... (n-1)^2) $$
$$=\frac {125}{n^2}\frac {(n-1)(2n-1)}{6} $$
The exact value is $$\int_0^5 x^2dx=\frac {125}{3}$$ When $n\to+\infty $, we get the exact value.