When the trapezium rule is used to estimate the integral

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

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

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

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

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

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

The trapezium rule says this:

Curve with lines drawn down from it to the x axis at regular intervals, a width of h apart. The first is at x = x_0.

The area under the curve \(f(x)\) between the points \(x_0\) and \(x_n\), with this interval divided into \(n\) smaller intervals of width \(h = \dfrac{(x_n - x_0)}{n}\) is given by

\[\displaystyle\int_{x_0}^{x_n} \! f(x) \, \mathrm{d}x \approx \dfrac{1}{2} h \left[ (y_0 + y_n) + 2(y_1 + y_2 + \dotsb + y_{n-1} ) \right],\]

where \(y_i = f(x_i)\).

So dividing the interval \(0 \le x \le 1\) into \(N\) subintervals for our problem gives \(h=\dfrac{1}{N}\), and the trapezium rule gives

\[\begin{align*} \displaystyle\int_0^1 \! 2^x \, \mathrm{d}x \hspace{2mm} & \approx \dfrac{1}{2} \dfrac{1}{N} \left[ (2^0 + 2^1) + 2(2^{\frac{1}{N}}+2^{\frac{2}{N}}+ ... + 2^{\frac{N-1}{N}}) \right] \\ \end{align*}\]

We have what looks like a series inside the brackets. Do we know any ways of simplifying series?

\[2^{\frac{1}{N}}+2^{\frac{2}{N}}+ ... + 2^{\frac{N-1}{N}}\]

is a geometric series with a first term of \(2^{\frac{1}{N}}\), a common ratio of \(2^{\frac{1}{N}}\), and \(N-1\) terms so we can write it as,

\[\dfrac{2^{\frac{1}{N}}(1 - 2^{\frac{N-1}{N}})}{1-2^{\frac{1}{N}}}\]

using the formula for the sum of a geometric progression.

The whole expression can now be written as

\[\dfrac{1}{2N} \left[3 + 2 \left( \dfrac{2^\frac{1}{N}(1-2^{\frac{N-1}{N}})}{1-2^{\frac{1}{N}}}\right )\right].\]

There are many ways to rearrange this expression, but three answers have a \(1\) in them, so we can use that as a starting point.

\[\begin{align*} \dfrac{1}{2N} \left[3 + 2 \left( \dfrac{2^\frac{1}{N}(1-2^{\frac{N-1}{N}})}{1-2^{\frac{1}{N}}}\right )\right] &= \dfrac{1}{2N} \left[1 + 2\left(\dfrac{1-2^{\frac{1}{N}}}{1-2^{\frac{1}{N}}}\right) + 2 \left( \dfrac{2^\frac{1}{N}(1-2^{\frac{N-1}{N}})}{1-2^{\frac{1}{N}}}\right )\right]\\ &= \dfrac{1}{2N} \left[1 + \dfrac{2(1-2^{\frac{1}{N}}) + 2(2^{\frac{1}{N}}-2)}{1-2^{\frac{N-1}{N}}}\right]\\ &= \dfrac{1}{2N} \left(1+\dfrac{2}{2^{\frac{1}{N}}-1} \right) \end{align*}\]

So the answer is (b).