Question

Write the Riemann sum to find the area under the graph of the function f(x) = x^2 from x = 1 to x = 5:

(I'm going to post what I wrote and I wanted to see if it was correct)

Answer 1

\[\lim_{x \rightarrow \infty} \sum_{i \rightarrow 1}^{n} (\frac{ i }{ n })^{2}(\frac{ 4 }{ n })\]

(now let me just explain why I think it's this (though please tell me ahead of time if it's wrong))

Answer 2

Fun problem. Are you planning to use "left end points" or "right end points?"

Answer 3

it's close, but not quite there

Answer 4

I wrote my answer based on the idea of using left end points, but I'm not completely sure that's what they wanted. Also I don't have the whole riemann sum and limits thing down.

Answer 5

endpoints make no difference
what does make a difference is that you are starting at \(1\)

Answer 6

also, slight notation note, but see attached

Answer 7

anyways, if you were starting at x = 0, then you'd have the correct answer

Answer 8

If you're using a finite number of x values, then the use of left- or right-end-points does make a difference. However, if you intend to find the exact value of the area defined by the curve, then the choice makes no difference.

Answer 9

Okay. Sorry if I'm being a bit slow.

Answer 10

\[\lim_{x \rightarrow \infty} \sum_{i \rightarrow 1}^{n} (\frac{ i }{ n })^{2}(\frac{ 4 }{ n })\]

Answer 11

should be \[\lim_{n \rightarrow \infty} \sum_{i \rightarrow 1}^{n} (\frac{ i }{ n })^{2}(\frac{ 4 }{ n })\]

Answer 12

I'm trying to understand what I'm doing

Answer 13

...where I've changed the 'x' to an 'n.'

Answer 14

Oh yeah, that was just an accident

Answer 15

Could you identify what's not yet clear for you?
I'm sure it was just an accident.

Answer 16

Okay, first- I know I'm supposed to use the limit, one because they don't ask for an approximation and two because this is what the lesson is about. I know the (4/n) part is correct because the width get's infinity smaller/ approaching zero, and that would give me the exact area. The middle part that's neither that nor the limit I sort of took from my notes but I'm not entirely sure how that functions.

Answer 17

Suggestion: If you haven't already, set up a formula for the x values. It will look like\[x _{i}=a+i \frac{ b-a }{ n }\]

Answer 18

Left Endpoint Riemman Sum

\[\Large A = \lim_{n \to \infty}\left[\sum_{i=0}^{n-1}f(x_i)*\Delta x\right]\]


Right Endpoint Riemman Sum

\[\Large A = \lim_{n \to \infty}\left[\sum_{i=1}^{n}f(x_i)*\Delta x\right]\]

If you don't include the limit as n goes to infinity, the two Riemman sums will be different. If you do let n go to infinity, then the two areas are equal.


A: area under the curve from x = a to x = b
i: index value
n: number of rectangles
\(\Large x_i = a + \Delta x*n\)
\(\Large \Delta x = \frac{b-a}{n}\)

Answer 19

Here your a=1 and your b=5.

Answer 20

oops typo, I meant to say

\(\Large x_i = a + \Delta x*i\)

Answer 21

Oh okay, do you guys mind if I give it another go
thanks a ton btw

Answer 22

that really helps

Answer 23

Of course. Thanks for your persistence and your willingness to try things on your own.
Jim, many thx for your contributions here.

Answer 24

np, you'll also use these summation identities

https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/summationdirectory/Summation.html

mainly the first three in this case

Answer 25

\[\lim_{n \rightarrow \infty} \sum_{i=1}^{n} (1 + \frac{ 4i }{ n })^{2}(\frac{ 4 }{ n })\]

Answer 26

Nice job of expressing x_sub_1!

Answer 27

(thanks so much, the majority of the stuff you guys gave me is going in my notes)

Answer 28

Truly happy for you.

Answer 29

Work on your parentheses, though.

( \ d f r a c { i } { n } ) ^ { 2 } produces \((\dfrac{i}{n})^{2}\)

\ l e f t ( \ d f r a c { i } { n } \ r i g h t ) ^ { 2 } produces \(\left(\dfrac{i}{n}\right)^{2}\)

Just add the words "left" and "right".

Answer 30

Okay thanks. Do you mind giving mathmale or jim_thompson a medal because I can only give one and honestly they both helped me @tkhunny

Answer 31

I'll give the other one a medal too

Answer 32

No worries.