Answer: (a) \(\displaystyle \int_0^1 (1-x)^2\,dx \lt U_n\), where \(\displaystyle U_n=\frac{2n^2+3n+1}{6n^2}\).
(b) A lower bound is \(\displaystyle L_n=\frac{2n^2-3n+1}{6n^2}\), so \(\displaystyle L_n \lt \int_0^1 (1-x)^2\,dx\).
(c) \(\displaystyle U_n-L_n=\frac{1}{n}\), hence \(\displaystyle \lim_{n\to\infty}(U_n-L_n)=0\).
The curve is \(y=(1-x)^2\), which is decreasing on \([0,1]\). Therefore rectangles using left endpoints give an upper bound, and rectangles using right endpoints give a lower bound.
(a) Using left-end ordinates \(x=0,\frac1n,\frac2n,\dots,\frac{n-1}{n}\),
\(\displaystyle \int_0^1 (1-x)^2\,dx \lt \frac1n\left(1-\frac0n\right)^2+\frac1n\left(1-\frac1n\right)^2+\cdots+\frac1n\left(1-\frac{n-1}{n}\right)^2.\)
So
\(\displaystyle \int_0^1 (1-x)^2\,dx \lt \frac1n+\frac1{n^3}\sum_{r=1}^{n-1}(n-r)^2.\)
Now
\(\displaystyle \frac1n+\frac1{n^3}\sum_{r=1}^{n-1}(n-r)^2=\frac1n+\frac1{n^3}\sum_{r=1}^{n-1}(n^2-2nr+r^2).\)
Using \(\sum_{r=1}^{n-1} r=\frac{n(n-1)}{2}\) and \(\sum_{r=1}^{n-1} r^2=\frac{(n-1)n(2n-1)}{6}\),
\(\displaystyle \frac1n+\frac{n^2(n-1)}{n^3}-\frac{2n}{n^3}\cdot \frac{n(n-1)}{2}+\frac1{n^3}\cdot \frac{(n-1)n(2n-1)}{6}.\)
This simplifies to
\(\displaystyle \frac1n+\frac{(n-1)(2n-1)}{6n^2}=\frac{2n^2+3n+1}{6n^2}.\)
Hence
\(\displaystyle \int_0^1 (1-x)^2\,dx \lt U_n, \qquad U_n=\frac{2n^2+3n+1}{6n^2}.\)
(b) Using right-end ordinates \(x=\frac1n,\frac2n,\dots,\frac{n}{n}\),
\(\displaystyle \int_0^1 (1-x)^2\,dx \gt \frac1n\left(1-\frac1n\right)^2+\frac1n\left(1-\frac2n\right)^2+\cdots+\frac1n\left(1-\frac{n-1}{n}\right)^2.\)
Thus
\(\displaystyle \int_0^1 (1-x)^2\,dx \gt \frac1{n^3}\sum_{r=1}^{n-1}(n-r)^2.\)
Again,
\(\displaystyle \frac1{n^3}\sum_{r=1}^{n-1}(n-r)^2=\frac1{n^3}\sum_{r=1}^{n-1}(n^2-2nr+r^2),\)
so
\(\displaystyle \frac{n^2(n-1)}{n^3}-\frac{2n}{n^3}\cdot \frac{n(n-1)}{2}+\frac1{n^3}\cdot \frac{(n-1)n(2n-1)}{6}=\frac{2n^2-3n+1}{6n^2}.\)
Therefore a lower bound is
\(\displaystyle L_n=\frac{2n^2-3n+1}{6n^2}.\)
(c) Now
\(\displaystyle U_n-L_n=\frac{2n^2+3n+1}{6n^2}-\frac{2n^2-3n+1}{6n^2}=\frac{6n}{6n^2}=\frac1n.\)
Since \(\displaystyle \frac1n\to 0\) as \(n\to\infty\),
\(\displaystyle \lim_{n\to\infty}(U_n-L_n)=0.\)
