trying to find a way to calculate the upper sum $U(f,P)$ of $f(x)=x^{2}$ for $x \in [0,1]$ without the well known rules of integrals, just by partitioning the interval $[0,1]$ and performing a Riemann Sum. In the problem it tells me that I should use the formula for the sum of the squares, i.e.
$$\sum_{i=1}^{n}i^{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$$
The following shows how far I've gone,
$$U(f,P)=\sum_{i=1}^{n}M_{i}(t_{i}-t_{i-1})$$ where $M_{i}=\sup\{f(x_{i}): x_{i} \in [t_{i-1},t_{i}]\}$, and $(t_{i}-t_{i-1})$ is the interval being evaluated.
By a result shown in class, I know that. If $f$ is continuous then it is integrable, so $f(x)=x^{2}$ is indeed integrable, so we have no problem in finding a partition which fulfills $U(f,P)-L(f,P)<\epsilon \hspace{2mm}\forall\epsilon>0$.
When I start to partition the interval $[0,1]$ in $n$ equally spaced subintervals I get the following partition: $P=\{0,\frac{1}{n},\frac{2}{n},\frac{3}{n},...,\frac{n-1}{n},1\}$ and the values of $M_{i}$ for each of this $n$ intervals in relation with $i$ are $M_{i}=n-(n+i)= i$ is it?
So, if my reasoning is correct, then we have that:
$$U(f,P)=\sum_{i=1}^{n}i^{2}\cdot\frac{1}{n}=\frac{1}{n}\sum_{i=1}^{n}i^{2}=\frac{1}{n}\cdot\frac{n\left(n+1\right)\left(2n+1\right)}{6}=\frac{(n+1)(2n+1)}{6}$$
But, if we take the limit of $U(f,P)$ when $n$ goes to infinity, we get:
$$\lim_{n \to \infty} U(f,P)=\lim_{n \to \infty} \frac{(n+1)(2n+1)}{6}=\infty$$
Which is of course false because using integration rules one can see that $\int_{0}^{1}x^{2}dx=\frac{1}{3}$.
What did I do wrong in my reasoning?
- I have some doubts about my partition $\frac{1}{n}$, because this partition should depend on $i$ ($(t_{i}-t_{i-1})$) and it doesn't.
Anything will be much appreciated.
Thank you very much.
Where you wrote $i^2$ in the expression for $U(f,P)$, it should be $\left(\dfrac i n\right)^2$.