Answer: (a) \(\displaystyle \int_0^1 (2x-x^2)\,dx\lt U_n\), where

\(\displaystyle U_n=\left(1+\frac{1}{n}\right)\left(\frac{2}{3}-\frac{1}{6n}\right)=\frac{2}{3}+\frac{1}{2n}-\frac{1}{6n^2}.\)

(b) A lower bound is

\(\displaystyle L_n=\left(1-\frac{1}{n}\right)\left(\frac{2}{3}+\frac{1}{6n}\right)=\frac{2}{3}-\frac{1}{2n}-\frac{1}{6n^2}.\)

(c) \(\displaystyle U_n-L_n=\frac{1}{n}\), so

\(\displaystyle \lim_{n\to\infty}(U_n-L_n)=0.\)

Let \(f(x)=2x-x^2\) for \(0\le x\le 1\).

Since \(f'(x)=2-2x\ge 0\) on \([0,1]\), the function is increasing on this interval. Hence right-endpoint rectangles give an upper bound and left-endpoint rectangles give a lower bound.

(a) Divide \([0,1]\) into \(n\) intervals of width \(\frac{1}{n}\). Using right endpoints \(x=\frac{r}{n}\),

\(\displaystyle U_n=\sum_{r=1}^n \frac{1}{n}f\!\left(\frac{r}{n}\right)=\sum_{r=1}^n \frac{1}{n}\left(2\frac{r}{n}-\left(\frac{r}{n}\right)^2\right).\)

Therefore

\(\displaystyle \int_0^1(2x-x^2)\,dx\lt U_n.\)

Now

\(\displaystyle U_n=\frac{2}{n^2}\sum_{r=1}^n r-\frac{1}{n^3}\sum_{r=1}^n r^2.\)

Using \(\displaystyle \sum_{r=1}^n r=\frac{n(n+1)}{2}\) and \(\displaystyle \sum_{r=1}^n r^2=\frac{n(n+1)(2n+1)}{6}\),

\(\displaystyle U_n=\frac{2}{n^2}\cdot\frac{n(n+1)}{2}-\frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}.\)

So

\(\displaystyle U_n=1+\frac{1}{n}-\frac{1}{6}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)=\left(1+\frac{1}{n}\right)\left(\frac{2}{3}-\frac{1}{6n}\right).\)

Hence

\(\displaystyle \int_0^1(2x-x^2)\,dx\lt \left(1+\frac{1}{n}\right)\left(\frac{2}{3}-\frac{1}{6n}\right).\)

(b) For the lower bound, use left endpoints. The first rectangle has height \(f(0)=0\), so

\(\displaystyle L_n=\sum_{r=1}^{n-1} \frac{1}{n}f\!\left(\frac{r}{n}\right)=\sum_{r=1}^{n-1} \frac{1}{n}\left(2\frac{r}{n}-\left(\frac{r}{n}\right)^2\right).\)

Since the function is increasing,

\(\displaystyle L_n\lt \int_0^1(2x-x^2)\,dx.\)

Now

\(\displaystyle L_n=\frac{2}{n^2}\sum_{r=1}^{n-1} r-\frac{1}{n^3}\sum_{r=1}^{n-1} r^2.\)

Using \(\displaystyle \sum_{r=1}^{n-1} r=\frac{(n-1)n}{2}\) and \(\displaystyle \sum_{r=1}^{n-1} r^2=\frac{(n-1)n(2n-1)}{6}\),

\(\displaystyle L_n=\frac{2}{n^2}\cdot\frac{(n-1)n}{2}-\frac{1}{n^3}\cdot\frac{(n-1)n(2n-1)}{6}.\)

So

\(\displaystyle L_n=\left(1-\frac{1}{n}\right)-\frac{1}{6}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)=\left(1-\frac{1}{n}\right)\left(\frac{2}{3}+\frac{1}{6n}\right).\)

Thus a lower bound is

\(\displaystyle L_n=\left(1-\frac{1}{n}\right)\left(\frac{2}{3}+\frac{1}{6n}\right).\)

(c) Subtracting,

\(\displaystyle U_n-L_n=\left(\frac{2}{3}+\frac{1}{2n}-\frac{1}{6n^2}\right)-\left(\frac{2}{3}-\frac{1}{2n}-\frac{1}{6n^2}\right)=\frac{1}{n}.\)

Since \(\displaystyle \frac{1}{n}\to 0\) as \(n\to\infty\),

\(\displaystyle \lim_{n\to\infty}(U_n-L_n)=0.\)