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I asked a similar question yesterday, but I didn't really get the info I wanted so maybe if I post a question and get an answer I will understand this concept better. Just some background info the topic is using Riemann sums approximation to find upper/lower sums.

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

Here is my attempt: $\sum_{1}^{3}$, by Riemann's definition: $\Delta x= (b-a)/n,$ so our $\Delta\,x$ is $2/3.$ $x*k= a+\Delta\,x*k= 2/3k+1$ for $k\in \{0,1,2,3\}$ according to my book, why does it include 0 is my question?

So I have my $\Delta x$ and my $x∗k.$ Now I have to check where the function given is decreasing and increasing on the intervals given, correct?

So my 3 sub intervals are $[1,5/3]$ and $[5/3,7/3]$ and $[7/3,3]$

If I plug $1$ in $(x-2)^{2} +1$ I get $2,$ plugging in $5/3$ I get $10/9.$ so It appears from $[1,5/3]$ we are decreasing. So for this first interval my lower sum is 5/3.

Next interval is from $[5/3,7/3]$. Plugging in $5/3$ I get $10/9.$ Plugging in $7/3$ I get $10/9.$ In this interval I have no lower sum, so I add the two and divide by 2, giving me a lower sum of 2.

Final interval is from $[7/3,3]$.Plugging in $7/3$ I get $10/9.$ Plugging in $3$ I get $2.$ In this interval the lower sum is 7/3.

Now here is where I get lost. The answer according to me text is 2/3*(((5/3 -2)^2 +1) +((2-2)^2 +1) + ((7/3 -2)^2 +1)). So what I understand what they did is they expanded the summation 3 times, I get that.

They plugged x as the lower for each expansion, ie. 5/3 for the first expansion, 2 for the secound expansion, and 7/3 for the last expansion. I get this.

Then they multiplied the whole thing by $\Delta x$

My main concern is where did we use 2/3K +1. What was the point of even figuring this out, wouldn't we have been fine with just $\Delta x$. I thought we would plug in 2/3K +1 where x is in each expansion like right/left sums.

EDIT: Can someone answer the questions in my post please, I get how to do most of it


marked as duplicate by Namaste calculus Jul 29 '17 at 22:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.



$f'(x)=2 (x-2) $

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

$\sigma=(x_0=1 , x_1=\frac {5}{3} , x_2=\frac {7}{3} , x_3=3)$

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

in $[\frac {5}{3},\frac {7}{3}] $, the minimum of $f (x) $ is $f (2)=0^2+1=\color {red}{1}$.

the lower sum is $$L(f,\sigma)=\frac {2}{3}\Bigl ((\frac {5}{3}-2)^2+1+\color {red}{1}+(\frac {7}{3}-2)^2+1\Bigr) $$ You can finish.


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