# The lower and upper sums for the Darboux integral of $x^2$

I can't quite get the upper and lower sums of $f(x)=x^2$ from $[a,b]$ (where we are defining the function) .

Here I'm talking about Darboux sums of the form $U(f, P_n) = \displaystyle\sum_{i=1}^n M_i(t_i-t_{i-1})$ and $L(f,P_n)=\displaystyle\sum_{i=1}^n m_i(t_i-t_{i-1})$ where $m_i=inf\{f(x): t_{i-1} \le x \le t_i\}$ and $M_i=sup\{f(x): t_{i-1} \le x \le t_i\}$. (Also, we assume a partition $P_n=\{t_{\mathbb{0}},t_1,t_2,...t_n\}$)

I was thinking that $t_i= \frac{i(b-a)}{n}$, but if we do this in the end the sum is multiplied by $\frac{(b-a)^3}{n^3}$ which is a bit different from the expected $\frac{b^3-a^3}{n^3}$ that easy way to calculate the integral yields. (where the $\frac {1}{n^3}$ cancels out with the limit of the sum that ends up as $\displaystyle\sum_{i=1}^n=i^2$).

So if anyone could shine some light on how to work this out (pick a better $t_i$ or if I'm doing the sum incorrectly) it would be very much appreciated thanks.

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To make things easier, consider $[0,a]$. The function $x^2$ is increasing over $[0,a]$ and it is continuous there. Thus, if we partition $[0,a]$ with $$P=\{t_0,\dots,t_n\}$$ we will have $$M_i=t_i^2$$

and $$m_i=t_{i-1}^2$$

since a continuous functions attains it maximum and minimum in a closed interval. Thus

$$L(f,P)=\sum_{i=1}^n t_{i-1}^2(t_i-t_{i-1})$$

$$U(f,P)=\sum_{i=1}^n t_{i}^2(t_i-t_{i-1})$$

As you say, consdier the partition where $$t_i=\frac{ia}{n}$$

Then \eqalign{ & L(f,P) = \sum\limits_{i = 1}^n {{{\left( {i - 1} \right)}^2}\frac{{{a^2}}}{{{n^2}}}} \frac{a}{n} = \frac{{{a^3}}}{{{n^3}}}\sum\limits_{i = 1}^n {{{\left( {i - 1} \right)}^2}} \cr & U(f,P) = \sum\limits_{i = 1}^n {{i^2}\frac{{{a^2}}}{{{n^2}}}} \frac{a}{n} = \frac{{{a^3}}}{{{n^3}}}\sum\limits_{i = 1}^n {{i^2}} \cr}

Since $$\sum\limits_{i = 1}^n {{i^2}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$$ we get \eqalign{ & L(f,P) = \frac{{{a^3}}}{{{n^3}}}\sum\limits_{i = 0}^{n - 1} {{i^2}} = \frac{{{a^3}}}{3}\frac{{n\left( {n - 1} \right)\left( {2n - 1} \right)}}{{2{n^3}}} \cr & U(f,P) = \frac{{{a^3}}}{{{n^3}}}\sum\limits_{i = 1}^n {{i^2}} = \frac{{{a^3}}}{3}\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{2{n^3}}} \cr}

from where \eqalign{ & L(f,P) \to \frac{{{a^3}}}{3} \cr & U(f,P) \to \frac{{{a^3}}}{3} \cr}

Then $$\int\limits_a^b {{x^2}dx} = \int\limits_0^b {{x^2}dx} - \int\limits_0^a {{x^2}dx} = \frac{{{b^3}}}{3} - \frac{{{a^3}}}{3}$$

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Oh, this is a very clever way to make it work, thank you, I had not thought of this. – Casquibaldo Nov 18 '12 at 19:50

Your $f(x) = x^2$ and the partition $P_n$ is $$P_n = \left \{t_k = a + \dfrac{k(b-a)}n : k=0,1,2,\ldots,n \right \}$$ Then we have that $$L(f,P_n) = \sum_{k=0}^{n-1} m_k (t_{k+1} - t_k) = \sum_{k=0}^{n-1} t_k^2 \dfrac{(b-a)}{n} = \dfrac{(b-a)}{n} \sum_{k=0}^{n-1} \left(a + \dfrac{k(b-a)}n \right)^2$$ similarly, $$U(f,P_n) = \sum_{k=0}^{n-1} M_k (t_{k+1} - t_k) = \sum_{k=0}^{n-1} t_{k+1}^2 \dfrac{(b-a)}{n} = \dfrac{(b-a)}{n} \sum_{k=1}^{n} \left(a + \dfrac{k(b-a)}n \right)^2$$ Now $$\sum_{k=0}^{n-1} \left(a + \dfrac{k(b-a)}n \right)^2 = n a^2 + \dfrac{2a(b-a)}{n} \dfrac{n(n-1)}{2} + \left(\dfrac{(b-a)}{n} \right)^2 \dfrac{n(n-1)(2n-1)}{6}$$ Similarly, $$\sum_{k=1}^{n} \left(a + \dfrac{k(b-a)}n \right)^2 = n a^2 + \dfrac{2a(b-a)}{n} \dfrac{n(n+1)}{2} + \left(\dfrac{(b-a)}{n} \right)^2 \dfrac{n(n+1)(2n+1)}{6}$$ Now you should be able to finish it off. The point you missed is that $$t_k = a + \dfrac{k(b-a)}n$$ and is not $$t_k = \dfrac{k(b-a)}n$$

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