Hi folks,

This one looks easy, but I'm not getting there.

The integral $ \int_{0}^{1} e^x dx$ is approximated using the trapezium rule, dividing the range into intervals of length h = 1/n.

Show that the result is $\dfrac{(e - 1)(p + 1)}{2n (p - 1)}$ where $p = e^h$ ..................................(1)

so the ordinates are:

\( \begin{array}{llllllll}

x & 0 & 1/n & 2/n & 3/n & ... & n-1/n & 1 \\

& & & & & & & \\

f(x) & 1 & e^\frac{1}{n} & e^\frac{2}{n} & e^\frac{3}{n} & ... & e^\frac{n-1}{n} & e

\end{array} \)

for a range between 0 and 1 a single interval (slice) is 1/n = h

the trapezium rule is

$A = h/2 \; (y_{0} + y_{n} + 2 (y_{1} + y_{2} + y_{3} \; ... \; y_{n-1} ) )$

so

$A = 1/2n \; (1 + e + 2(e^\frac{1}{n} + e^\frac{2}{n} + e^\frac{3}{n} + \; ... \; + e^\frac{n-1}{n}) )$

there are n - 1 terms in the series (after the 2).

Since $a \frac{m}{n} = \sqrt[n] {a^m}$ where a, m and n are constants.

$A = 1/2n \; (1 + e + 2e^\frac{1}{n} (e + e^2 + e^3 + ... + e^{n-1} ) )$ .....................................(2)

so, we have a geometric progression with n - 1 terms. The sum of this series is given by $S_{n} = b \frac{r^n - 1}{r - 1}$ for $r \ge 1$. Where b is the first term and r the common ratio.

In our case b = e and r = e.

$S_{n} - S_{n - 1} = b. r^{n - 1} = e^n$

so

$S_{n-1} = \dfrac{e(e^n - 1)}{e - 1} - e^n = \dfrac {e^n - e}{e - 1} $

put this back into (2)

$A = 1/2n \; (1 + e + 2 e^\frac{1}{n} \frac {e^n - e}{e - 1} )$ and we have $p = e^h = e^\frac{1}{n} $

Now, I was hoping some straightforward algebra would get me to (1) but I cannot get this into the form required.

Can anyone help?