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