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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.

CoffeeToTheorems
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4 Answers4

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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.

Jack D'Aurizio
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Why don't you just divide $[0,1]$ uniformly and write down the associated Darboux sums?

Boggie Georgiev
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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 $$ L=\frac1N\sum_{k=0}^{N-1}\Bigl(\frac{k}{N}\Bigr)^2,\quad U=\frac1N\sum_{k=1}^{N}\Bigl(\frac{k}{N}\Bigr)^2. $$

Julián Aguirre
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  • That makes so much sense, thank you. How do I prove these are equal? It's not obvious to me by looking at it that they're equal. – CoffeeToTheorems Dec 15 '15 at 11:43
  • Take $1/N^2$ out of the sum and you are left with $\sum_{k=0}^{N-1}k^2$ and $\sum_{k=1}^{N}k^2$. You know the formula for these sums. – Julián Aguirre Dec 15 '15 at 11:46
  • Yes but then that results in ${1 \over 6}(N(N + 1)(2N + 1) = {1 \over 6}(N(N - 1)(2N - 1))$ which simplifies to $2{N^3} + 3{N^2} + N = 2{N^3} - 3{N^2} + N$ which in turn simplifies to $ - {N^3} = {N^3}$ which isn't true for any number other than 0... They aren't equal. – CoffeeToTheorems Dec 15 '15 at 12:21
  • $L$ and $N$ are not equal: $L<\int_0^1x^2,dx<U$. But their limit as $N\to\infty$ is the same and equal to the integral. – Julián Aguirre Dec 15 '15 at 12:23
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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.

Hirshy
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