Lower and Upper Riemann sums

[trail locust]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with: ```diff
   +close

[chrome shell]

Geometric intuition

notice that the bars in photo 1 are larger than the area under the graph. (because f(x) = x^a is monotonically increasing)
notice that the bars in photo 2 are smaller than the area under the graph.
go from here @thorny flume

[dusty gulch]

3a

[thorny flume]

thank you but i already knew this

i just want to know what i am supposed to say to prove it

i can get behind proving that L <= U

[chrome shell]

Point out that the lower bound is equivalent to a left sided Riemann sum with dx = 1.
Left sided Riemann sums are always smaller than the actual integral if f(x) is monotonically increasing, thus this holds as a lower bound.

Apply the same logic to the upper bound

[chrome shell]

Left sided Riemann sums are always smaller than the actual integral if f(x) is monotonically increasing
If you need to prove this (which I don't think you need to do as it's a trivial and was immediately pointed out when I read about Riemann sums), do by induction on the amount of partitions, starting with base case 1 partition.

[thorny flume]

@chrome shell So can I just say for a>0, the function is monotonically increasing. leftsided R sum is always smaller than the actual integral, and the right sided R sum is always greater than the actual integral. Therefore left sided R sum is always <= integral <= right sided R sum. for n subintervals dx = n-0/n = 1. left-sided R sum can be defined as 1^a + ... + (n-1)^a and right sided R sum can be defined as 1^a + .. + n^a. would this work?

[chrome shell]

It would!

[chrome shell]

Remember though, if In your course the fact I quoted here isn’t proven in class, maybe prove it yourself to be safe.