I want to evaluate the Riemann integral $\int_0^1 {{x^2}dx} $ I want to find upper and lower estimates of the form: $$U \ge {1 \over {6{N^3}}}(N(N + 1)(2N + 1))$$ $$L \ge {1 \over {6{N^3}}}(N(N - 1)(2N - 1))$$
Then show they're equal and then evaluate the mentioned Riemann integral. I think the fact that $\sum\limits_{x = 1}^n {{x^2}} = {1 \over 6}(n(n + 1)(2n + 1))$ must be involved.
On
Well, $x^2$ is a convex function on $[0,1]$, hence by the Cauchy-Hadamard inequality we have:
$$ \frac{1}{4n^2}\left(2k+1\right)^2\leq n\int_{k/n}^{(k+1)/n}x^2\,dx \leq \frac{1}{2n^2}\left(k^2+(k+1)^2\right)\tag{1}$$ so, by exploiting the identity $\sum_{k=1}^{n}k^2 = \frac{n(n+1)(2n+1)}{6}$ and summing $(1)$ over $k=0,1,2,\ldots,(n-1)$ we get:
$$ \frac{1}{3}-\frac{1}{12 n^2}\leq \int_{0}^{1}x^2\,dx \leq \frac{1}{3}+\frac{1}{6n^2}\tag{2} $$ for any $n\geq 1$. By letting $n\to +\infty$ we have $\int_{0}^{1}x^2\,dx =\frac{1}{3}$ without resorting to the fundamental theorem of Calculus. We may deal with the lower and upper Riemann sums in a similar fashion.
On
Hint: we divide the intervall $[0,1]$ into $n$ rectangles with length $\Delta x=\frac{1}{n}$. Using this partition you can get upper and lower estimates for the integral via $$U_n=\Delta x\cdot \left(f(0)+f(\Delta x)+f(2\cdot \Delta x)+\ldots +f((n-1)\cdot \Delta x)\right) \\ O_n= \Delta x\cdot \left(f(\Delta x)+f(2\cdot\Delta x)+f(3\cdot\Delta x)\ldots +f(n\cdot \Delta x)\right).$$
Using the identity $\sum\limits_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$ will be useful in the process.
Choose a partition of $[0,1]$ into $N$ intervals of length $1/N$. Since $x^2$ is increasing on $[0,1]$, for the lower sum you can take the value of $x^2$ at the left end, and for the upper integral, the value at the right end. This gives $$ L=\frac1N\sum_{k=0}^{N-1}\Bigl(\frac{k}{N}\Bigr)^2,\quad U=\frac1N\sum_{k=1}^{N}\Bigl(\frac{k}{N}\Bigr)^2. $$