Answer: (a) \(\displaystyle \int_0^1 x^3\,dx \lt U_n\), where \(\displaystyle U_n=\left(\frac{n+1}{2n}\right)^2\).

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

(c) \(\displaystyle U_n-L_n=\frac1n\), so \(\frac1n\lt 10^{-3}\Rightarrow n\gt 1000\).

Hence the least integer value is \(\boxed{1001}\).

The curve \(y=x^3\) is increasing on \([0,1]\), so rectangles using right-hand endpoints give an upper bound and rectangles using left-hand endpoints give a lower bound.

(a) Each rectangle has width \(\frac1n\). Using right-hand endpoints, the heights are

\(\left(\frac1n\right)^3,\left(\frac2n\right)^3,\dots,\left(\frac{n}{n}\right)^3.\)

Therefore

\(\displaystyle \int_0^1 x^3\,dx\lt \frac1n\left[\left(\frac1n\right)^3+\left(\frac2n\right)^3+\cdots+\left(\frac{n}{n}\right)^3\right]=\frac1{n^4}\sum_{r=1}^n r^3.\)

Using \(\displaystyle \sum_{r=1}^n r^3=\left(\frac{n(n+1)}2\right)^2\),

\(\displaystyle \frac1{n^4}\sum_{r=1}^n r^3=\frac{n^2(n+1)^2}{4n^4}=\left(\frac{n+1}{2n}\right)^2.\)

So \(\displaystyle \int_0^1 x^3\,dx\lt U_n\), where \(\displaystyle U_n=\left(\frac{n+1}{2n}\right)^2\).

(b) Using left-hand endpoints, the heights are

\(0^3,\left(\frac1n\right)^3,\left(\frac2n\right)^3,\dots,\left(\frac{n-1}{n}\right)^3.\)

Hence

\(\displaystyle \int_0^1 x^3\,dx\gt \frac1n\left[\left(\frac1n\right)^3+\left(\frac2n\right)^3+\cdots+\left(\frac{n-1}{n}\right)^3\right]=\frac1{n^4}\sum_{r=1}^{n-1} r^3.\)

Now

\(\displaystyle \sum_{r=1}^{n-1} r^3=\left(\frac{(n-1)n}{2}\right)^2,\)

so

\(\displaystyle L_n=\frac1{n^4}\sum_{r=1}^{n-1} r^3=\frac{(n-1)^2n^2}{4n^4}=\left(\frac{n-1}{2n}\right)^2.\)

(c) We need

\(\displaystyle U_n-L_n\lt 10^{-3}.\)

Now

\(\displaystyle U_n-L_n=\left(\frac{n+1}{2n}\right)^2-\left(\frac{n-1}{2n}\right)^2=\frac{(n+1)^2-(n-1)^2}{4n^2}=\frac{4n}{4n^2}=\frac1n.\)

So

\(\displaystyle \frac1n\lt 10^{-3}\Rightarrow n\gt 1000.\)

Therefore the least value of \(n\) is \(\boxed{1001}\).