Question

b) Determine the upper sums and lower sums of \( \Large f(x) = x^{3}\) over \(\Large I = [0,1]\) for equidistant partitions.

Answer 1

you mean compute the reimann sum?

Answer 2

i am not sure but i have a) part of question and there was a sum .. induction to prove etc..

Answer 3

should i post part a) too here ?

Answer 4

divide the interval in to \(n\) equal parts of length \(\frac{1}{n}\) then put
\(x_0=0\)
\(x_1=\frac{1}{n}\),
\(x_2=\frac{2}{n}\) and in general \(x_k=\frac{k}{n}\)

Answer 5

you want
\[\sum_{k=0}^n(\frac{k}{n})^3\times \frac{1}{n}\] or
\[\frac{1}{n^4}\sum_{k=0}^nk^3\]

Answer 6

ok..

Answer 7

look up the formula for the sum of n cubes, which is the square of the formula for the sum of \(n\)
namely
\[\left(\frac{n(n+1)}{2}\right)^2\]

Answer 8

ok..

Answer 9

so you get
\[\frac{1}{n^4}\times \left(\frac{n(n+1)}{2}\right)^2\]

Answer 10

take the limit as \(n\to \infty\) and you will get \(\frac{1}{4}\) because both numerator and denominator are polynomials of degree 4 and the ratio of the leading coefficients is \(\frac{1}{4}\)

Answer 11

ancient arab mathematicians new this formula, i think around 1100

Answer 12

*knew

Answer 13

wow you know even all math where the root is coming from and which time :)

Answer 14

@satellite73 thank you very much ..

Answer 15

yw
you might want to put in more details, but i think the idea is all there
good luck

Answer 16

hmm i hope i get points with this solution :) it should be enough..

Answer 17

@satellite73
kindly help me please

this question has got me stuck for more than a month

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