How to calculate the lower and upper riemann sum

[FallArk]

I ran into some issues when trying to calculate the lower Riemann sum of \(\displaystyle f\left(x\right)={x}^{3}\), \(\displaystyle x\in[0,1]\)
I am asked to use the standard partition \(\displaystyle {P}_{n}\) of \(\displaystyle [0,1]\) with n equal subintervals and evaluate \(\displaystyle L(f,{P}_{n})\) and \(\displaystyle U(f,{P}_{n})\)
What I did:
\(\displaystyle L(f,{P}_{n}) = \sum_{i=1}^{n}{m}_{i}{\delta}_{{x}_{i}} = \sum_{i=1}^{n}{(\frac{i-1}{n})}^{3}\cdot\frac{1}{n} = \frac{1}{{n}^{4}} \sum_{i=1}^{n} {(i-1)}^{3} = \frac{1}{{n}^{4}} \sum_{i=1}^{n}({i}^{3}-3{i}^{2}+3i-1) = \frac{1}{{n}^{4}}[\sum_{i=1}^{n}{i}^{3} - 3\sum_{i=1}^{n}{i}^{2} + 3\sum_{i=1}^{n}i - \sum_{i=1}^{n}i]\)
Then I know that the sum of \(\displaystyle {n}^{3}\) is \(\displaystyle \frac{{n}^{2}{(n+1)}^{2}}{4}\), sum of \(\displaystyle {n}^{2}\) is \(\displaystyle \frac{n(n+1)(2n+1)}{6}\), sums of \(\displaystyle n\) and \(\displaystyle 1\) are just \(\displaystyle \frac{n(n+1)}{2}\) and \(\displaystyle n\)
But I'm not so sure if my anwsers are correct.

 

[I like Serena]

I ran into some issues when trying to calculate the lower Riemann sum of \(\displaystyle f\left(x\right)={x}^{3}\), \(\displaystyle x\in[0,1]\)
I am asked to use the standard partition \(\displaystyle {P}_{n}\) of \(\displaystyle [0,1]\) with n equal subintervals and evaluate \(\displaystyle L(f,{P}_{n})\) and \(\displaystyle U(f,{P}_{n})\)
What I did:
\(\displaystyle L(f,{P}_{n}) = \sum_{i=1}^{n}{m}_{i}{\delta}_{{x}_{i}} = \sum_{i=1}^{n}{(\frac{i-1}{n})}^{3}\cdot\frac{1}{n} = \frac{1}{{n}^{4}} \sum_{i=1}^{n} {(i-1)}^{3} = \frac{1}{{n}^{4}} \sum_{i=1}^{n}({i}^{3}-3{i}^{2}+3i-1) = \frac{1}{{n}^{4}}[\sum_{i=1}^{n}{i}^{3} - 3\sum_{i=1}^{n}{i}^{2} + 3\sum_{i=1}^{n}i - \sum_{i=1}^{n}i]\)
Then I know that the sum of \(\displaystyle {n}^{3}\) is \(\displaystyle \frac{{n}^{2}{(n+1)}^{2}}{4}\), sum of \(\displaystyle {n}^{2}\) is \(\displaystyle \frac{n(n+1)(2n+1)}{6}\), sums of \(\displaystyle n\) and \(\displaystyle 1\) are just \(\displaystyle \frac{n(n+1)}{2}\) and \(\displaystyle n\)
But I'm not so sure if my anwsers are correct.

Hey FallArk,

Note that we can simplify it a bit - and use the formula you provided:
$$L(f,{P}_{n}) = \frac{1}{{n}^{4}} \sum_{i=1}^{n} {(i-1)}^{3}
= \frac{1}{{n}^{4}} \sum_{j=0}^{n-1} {j}^{3}
= \frac{1}{{n}^{4}} \cdot \frac{{(n-1)}^{2}{n}^{2}}{4}
\to\frac 14
$$
Since
$$\int_0^1 x^3\,dx = \frac 14 x^4 \Big|_0^1 = \frac 14$$
I think we are good. (Smile)

 

[FallArk]

Hey FallArk,

Note that we can simplify it a bit - and use the formula you provided:
$$L(f,{P}_{n}) = \frac{1}{{n}^{4}} \sum_{i=1}^{n} {(i-1)}^{3}
= \frac{1}{{n}^{4}} \sum_{j=0}^{n-1} {j}^{3}
= \frac{1}{{n}^{4}} \cdot \frac{{(n-1)}^{2}{n}^{2}}{4}
\to\frac 14
$$
Since
$$\int_0^1 x^3\,dx = \frac 14 x^4 \Big|_0^1 = \frac 14$$
I think we are good. (Smile)

Ooooo, that is clever! and just so happen that \(\displaystyle U(f,{P}_{n})\) is almost the same since \(\displaystyle {M}_{i}\) is \(\displaystyle \frac{i}{n}\)! which further proves that this function is integrable