Can we use the trapezium rule to estimate this integral?

Can we use the trapezium rule to estimate this integral?

Question

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

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

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

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

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

Suggestion

The notation $\displaystyle\int_0^1 \! 2^x \, \mathrm{d}x$ means the area between the curve $y = 2^x$ and the $x$-axis, from $x=0$ to $x = 1$.

Using the trapezium rule will give us a long sum that will have to be rewritten as a fraction. Can you think of another situation where we know how to rewrite a finite sum of numbers in fractional form?

Solution

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

---

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

---

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

---

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

The trapezium rule says this:

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

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

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).