How to calculate the lower and upper riemann sum

[FallArk]

I ran into some issues when trying to calculate the lower Riemann sum of $$f\left(x\right)={x}^{3}$$, $$x\in[0,1]$$
I am asked to use the standard partition $${P}_{n}$$ of $$[0,1]$$ with n equal subintervals and evaluate $$L(f,{P}_{n})$$ and $$U(f,{P}_{n})$$
What I did:
$$L(f,{P}_{n}) = \sum_{i=1}^{n}{m}_{i}{\delta}_{{x}_{i}} = \sum_{i=1}^{n}{(\frac{i-1}{n})}^{3}\cdot\frac{1}{n} = \frac{1}{{n}^{4}} \sum_{i=1}^{n} {(i-1)}^{3} = \frac{1}{{n}^{4}} \sum_{i=1}^{n}({i}^{3}-3{i}^{2}+3i-1) = \frac{1}{{n}^{4}}[\sum_{i=1}^{n}{i}^{3} - 3\sum_{i=1}^{n}{i}^{2} + 3\sum_{i=1}^{n}i - \sum_{i=1}^{n}i]$$
Then I know that the sum of $${n}^{3}$$ is $$\frac{{n}^{2}{(n+1)}^{2}}{4}$$, sum of $${n}^{2}$$ is $$\frac{n(n+1)(2n+1)}{6}$$, sums of $$n$$ and $$1$$ are just $$\frac{n(n+1)}{2}$$ and $$n$$
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 $$f\left(x\right)={x}^{3}$$, $$x\in[0,1]$$
I am asked to use the standard partition $${P}_{n}$$ of $$[0,1]$$ with n equal subintervals and evaluate $$L(f,{P}_{n})$$ and $$U(f,{P}_{n})$$
What I did:
$$L(f,{P}_{n}) = \sum_{i=1}^{n}{m}_{i}{\delta}_{{x}_{i}} = \sum_{i=1}^{n}{(\frac{i-1}{n})}^{3}\cdot\frac{1}{n} = \frac{1}{{n}^{4}} \sum_{i=1}^{n} {(i-1)}^{3} = \frac{1}{{n}^{4}} \sum_{i=1}^{n}({i}^{3}-3{i}^{2}+3i-1) = \frac{1}{{n}^{4}}[\sum_{i=1}^{n}{i}^{3} - 3\sum_{i=1}^{n}{i}^{2} + 3\sum_{i=1}^{n}i - \sum_{i=1}^{n}i]$$
Then I know that the sum of $${n}^{3}$$ is $$\frac{{n}^{2}{(n+1)}^{2}}{4}$$, sum of $${n}^{2}$$ is $$\frac{n(n+1)(2n+1)}{6}$$, sums of $$n$$ and $$1$$ are just $$\frac{n(n+1)}{2}$$ and $$n$$
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 $$U(f,{P}_{n})$$ is almost the same since $${M}_{i}$$ is $$\frac{i}{n}$$! which further proves that this function is integrable