Answer: (a) \(\displaystyle \int_0^1 2^x\,dx\lt U_N\), where \(\displaystyle U_N=\frac{2^{1/N}}{N\left(2^{1/N}-1\right)}\).

(b) A lower bound is \(\displaystyle L_N=\frac{1}{N\left(2^{1/N}-1\right)}\).

(c) \(\displaystyle U_N-L_N=\frac{1}{N}\), so we need \(\frac{1}{N}\lt 10^{-4}\). Hence \(N\gt 10^4\), and the least integer value is \(\boxed{10001}\).

The function \(y=2^x\) is increasing on \([0,1]\). With \(N\) rectangles of width \(\frac1N\):

• right-endpoint rectangles give an upper bound,
• left-endpoint rectangles give a lower bound.

(a)

Using right-hand endpoints, the rectangle heights are

\(2^{1/N},\;2^{2/N},\;\dots,\;2^{N/N}.\)

So the total rectangle area is

\(\displaystyle \frac1N2^{1/N}+\frac1N2^{2/N}+\cdots+\frac1N2^{N/N}.\)

Since these rectangles lie above the curve,

\(\displaystyle \int_0^1 2^x\,dx\lt \frac1N\sum_{n=1}^N\left(2^{1/N}\right)^n.\)

This is a geometric series with ratio \(r=2^{1/N}\), so

\(\displaystyle \sum_{n=1}^N r^n=\frac{r(r^N-1)}{r-1}.\)

Hence

\(\displaystyle \int_0^1 2^x\,dx\lt \frac1N\cdot\frac{2^{1/N}\left((2^{1/N})^N-1\right)}{2^{1/N}-1}.\)

Since \((2^{1/N})^N=2\), this becomes

\(\displaystyle \int_0^1 2^x\,dx\lt \frac{2^{1/N}}{N\left(2^{1/N}-1\right)}.\)

So \(\displaystyle U_N=\frac{2^{1/N}}{N\left(2^{1/N}-1\right)}\).

(b)

Using left-hand endpoints, the rectangle heights are

\(2^0,\;2^{1/N},\;2^{2/N},\;\dots,\;2^{(N-1)/N}.\)

So the total rectangle area is

\(\displaystyle \frac1N+\frac1N2^{1/N}+\cdots+\frac1N2^{(N-1)/N}.\)

These rectangles lie below the curve, so

\(\displaystyle \int_0^1 2^x\,dx\gt \frac1N\sum_{n=0}^{N-1}\left(2^{1/N}\right)^n.\)

Using

\(\displaystyle \sum_{n=0}^{N-1} r^n=\frac{r^N-1}{r-1},\)

with \(r=2^{1/N}\),

\(\displaystyle \int_0^1 2^x\,dx\gt \frac1N\cdot\frac{(2^{1/N})^N-1}{2^{1/N}-1}.\)

Since \((2^{1/N})^N=2\),

\(\displaystyle \int_0^1 2^x\,dx\gt \frac{1}{N\left(2^{1/N}-1\right)}.\)

So \(\displaystyle L_N=\frac{1}{N\left(2^{1/N}-1\right)}\).

(c)

\(\displaystyle U_N-L_N=\frac{2^{1/N}}{N(2^{1/N}-1)}-\frac{1}{N(2^{1/N}-1)}=\frac{2^{1/N}-1}{N(2^{1/N}-1)}=\frac1N.\)

We require

\(\displaystyle \frac1N\lt 10^{-4},\)

so

\(\displaystyle N\gt 10^4.\)

Therefore the least integer value is \(\boxed{10001}\).