$f(x) = x^2$. For each positive integer $n$, let $P_n$ be the partition $P_n = \{0, \frac{1}{n}, \frac{2}{n},..., \frac{n-1}{n},1 \}$ of $[0,1]$.

Question

Let $f(x) = x^2$. For each positive integer $n$, let $P_n$ be the partition $P_n = \{0, \frac{1}{n}, \frac{2}{n},..., \frac{n-1}{n},1 \}$ of $[0,1]$.

Show that $S(P)=\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^2}$ .Find a similar expression for $s(P)$.

I know that $S(P)=\sum_{i=1}^{n}M_j(x_j-x_{j-1})$ where $M_j=\sup_{[x_{j-1},x_j]}f(x)$. I'm not sure how to go from there.

Answer 1

For $j=1,2...n\;$, we have

$$x_j=j\frac{1}{n},$$

$$x_{j}-x_{j-1}=\frac{1}{n},$$

$$\displaystyle{M_j=sup_{x\in[x_{j-1},x_j]}f(x)=f(x_j)=x_j^2=\frac{j^2}{n^2}},$$

since $f$ is increasing.

and

$$S(P)=\sum_{j=1}^n(x_j-x_{j-1})M_j=\frac{1}{n^3}\sum_{j=1}^n j^2$$

$$=\frac{1}{n^3}\frac{n(n+1)(2n+1)}{6}$$

$$=\frac{1}{6n^2}(2n^2+3n+1)$$

$$=\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^2}.$$

if $n\to+\infty$, $\; S(P)$ goes to

$$\int_0^1 x^2dx=\frac{1}{3}.$$

Answer 2

Hint

$$S(P)=\dfrac{1}{n}\sum_{k=1}^n\frac{k^2}{n^2}$$ and $$\sum_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}{6}.$$

Answer 3

To add $s(P)$ is the same except $M_j$ is instead

$$m_j = \inf_{[x_{j-1}, x_j]} f(x) = f(x_{j-1}) = \left(\frac{j-1}{n}\right)^2$$

Now $$\sum_{j=1}^n \left(\frac{j-1}{n}\right)^2 = \sum_{k=0}^{n-1} \left(\frac{k}{n}\right)^2$$