When the trapezium rule is used to estimate the integral

\[\displaystyle\int_0^1 \! 2^x \, \mathrm{d}x\]

by dividing the interval \(0 \le x \le 1\) into \(N\) subintervals the answer achieved is

\(\dfrac{1}{2N} \left(1+\dfrac{1}{2^{1/N}+1} \right)\),

\(\dfrac{1}{2N} \left(1+\dfrac{2}{2^{1/N}-1} \right)\),

\(\dfrac{1}{N} \left(1-\dfrac{1}{2^{1/N}-1} \right)\),

\(\dfrac{1}{2N} \left(\dfrac{5}{2^{1/N}+1} -1\right)\).