Riemann Lower Sum by definition

Question

I have asked a similar question before, and didn't get the answer I was looking for, so i'll try to be more clear here. If necessary here is the link: Riemann sums upper and lower sums question

Okay I am trying to find the lower Riemann sum.

The question is $f(x)=(x-2)^{2} +1, [a,b]=1,3,$ find the lower sum with $n=3.$

An answer given by someone: ''' f′(x)=$2(x−2)$

For $n=3$, the partition of $[1,3]$ is

the 3 sub intervals $[1,5/3]$ and $[5/3,7/3]$ and $[7/3,3]$

f is decreasing at $[1,2]$ and increasing at $[2,3]$.

in $[5/3,7/3]$ the minimum of f(x) is f($2$)= $1$

the lower sum is $2/3$*(($(5/3 −2)^2 +1)$+$1$+($(7/3 −2)^2) +1)$

'''

I have two major questions:

1. I see the derivative was taken, was this to see where f was increasing and decreasing? If so how do I do this, just plug in the partitioned values in the derivative?

2. By definition of a lower sum our delta x was 2/3, which implies our x_k=a+k*(delta x) = 1+(2/3)k. We did not use this x_k anywhere, why is it necessary to calculate this(my text does it), I know in a right sum we would plug this x_k anywhere their is an x, but we didn't do that here.

Answer 1

In its current state this question is almost unreadable, but I assume the question asks to find the lower sum of $f(x)=(x-2)^2+1$ on the interval $[1,3]$ when it is partitioned into $3$ parts.

First step would be to partition $[1,3]$ into $3$ parts of equal length: $\frac{3-1}{3}=\frac{2}{3}$, so we can use the intervals: $[1,1+2/3],[1+2/3,1+4/3],[1+4/3,1+6/3]$, or equivalently, $[1,\frac53],[\frac53,\frac73],[\frac73,3]$. Call this partition $P$.

Apply the definition of the lower sum: $L(f,P)=\sum_{k=1}^3m_k(x_k-x_{k-1})$, where $m_k$ is the infimum of the function $f$ over the interval $[x_{k-1},x_k]$. Here the $x_i$ represent the bounds of the intervals of the partition. Since we created partitions of equal length, we have that $(x_k-x_{k-1})=\frac{2}3$ for all partitions, so we reduce the problem to finding the value of $\frac23\sum_{k=1}^3m_k$.

To find the $m_k$, you would need to compute the infimums of the function in each interval. Luckily, the function is continuous and differentiable on a compact set, so by some theorems in analysis, the minimum is the infimum and it must be on the boundary of the set, or on a stationary point.

I will compute the first $m_k$ for reference, the rest you can do. So compute the minimum of $f$ on $[1,5/3]$ by finding the values at the boundaries: $f(1)=2, f(5/3)=10/9$. Finally look for stationary points in the interval: $f'(x)=2(x-2)=0\implies x=2$. $x=2$ is not in the interval $[1,5/3]$ so $m_1=10/9$.