Integral Problem

[limber dawn]

Is g(x) given? @blissful hull

[blissful hull]

no for some reason

[limber dawn]

okay wait you dont need g(x)

So it’s asking for a right and left riemann sum correct

[blissful hull]

i believe so yes

[limber dawn]

for the left riemann sum (left endpoint) its going to be:

g(-2) + g(-1) + g(0)…

all the way to g(3)

[blissful hull]

but what about g(4)

can I show you what work I already did? it was an exam question he is letting me revise

[limber dawn]

yep

[blissful hull]

he gave me that comment but I am not sure if the rest of my answer is correct

[limber dawn]

its a left riemann sum, so -2,-1,0,1,2,3 = 6 sub intervals

hmm how do i explain this

okay, for this interval its -2 to 4 correct

[blissful hull]

yep

[limber dawn]

so theres 7 numbers i believe

[blissful hull]

yes

[limber dawn]

if you do a left riemann sum, you choose the left most number of the area under the curve you’re looking for:

ex: from -2 to -1, you would do g(-2)

correct so far?

[blissful hull]

ok I agree yea

[limber dawn]

so now since its 6 subintervals you do that 6 times, -1 to 0, 0 to 1…

[blissful hull]

so from g(-2) I move over 6 times regardless of what's after it

[limber dawn]

correct, i believe the equation would end up being

delta x • (g(-2)+g(-1)+g(0)+g(1)+g(2)+g(3))

see how theres 6 subintervals there without the 4

[blissful hull]

yeaaa

so with the right, we start at the very right one which is 4 and move over to the left 6 times right?

[limber dawn]

for the right riemann sum, it’d be the right most number

no you’d start at -1

i think visualizing it with a table would be easier tbh

[blissful hull]

for the right is it -1 because it's to the right of -2?

[blissful hull]

true

[limber dawn]

yes and so forth, so -1 0 1 2 3 4

you always start at the lowest number

[blissful hull]

Well I thought since it was asking for the right side, I start with the 4 at the end and move 6 to the left

[limber dawn]

you would still end with -1 i think yes?

its 6 including g(4)

if you’d like to do it that way it works out too

personally i’ve only seen riemann sums going from lowest x value to highest x value

[blissful hull]

wait I think theres a misunderstanding. The boxes are moved 2 times because every 1 jump is 2 boxes

[limber dawn]

regardless of endpoints

[blissful hull]

from x=0 to x=1, there is two jumps

[limber dawn]

but from 1 to 2 theres 1

lemme sketch a quick table for you

[blissful hull]

it's just the way he made the graph, every 1 we move over, it's two jumps if we look at the boxes

[limber dawn]

for riemann sums you usually don’t wanna look at the jumps

just the value at the x value

[blissful hull]

oh yea i forgot about that

[limber dawn]

this is an accurate table of the graph yes?

[blissful hull]

yep looks better actually

so my left endpoints would start at -2 and end at 3

and my right endpoints would start at 4 and end at -1

[limber dawn]

for the left riemann sum focus it like this:

right riemann sum like this

[limber dawn]

correct yes

[blissful hull]

so for left I would do 1 (0 + 1.5 + 0 - 1.5 + 0.5 - 1)

[limber dawn]

i agree on that yes

[blissful hull]

and for right I would do 1 ( 0.5 - 1 + 0.5 - 1.5 + 0 + 1.5 )

[limber dawn]

uhhh

yes thats correct

preferably you would go the other way, 1.5 to 0.5

because an integral goes from a to b not b to a

[blissful hull]

ohh I understand now

I appreciate your help, once I revise this with the correct answer I master this objective.

[limber dawn]

No problem!