Review question

# Can we use the trapezium rule to estimate this integral? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6592

## Suggestion

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

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 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?