A-level Mathematics/OCR/FP2/Complex Integration

Midpoint Rule

The Midpoint Rule is more accurate than the Trapezium Rule. It works by finding the mid-points of rectangles drawn to the curve. The Midpoint Rule is:

$\underset{a}{\overset{b}{\int }}f\left(x\right)dx\approx =h[f\left(x_1\right)+f\left(x_2\right)+\dots +f\left(x_n\right)]$

Where: $h=\frac{b-a}{n}$ n is the number of strips.

and $x_i=\frac{1}{2}[(a+\{i-1\}h)+(a+ih)]$

Example

Use the Midpoint Rule to evaluate $\underset{1}{\overset{5}{\int }}{x}^2+2xdx$ using 4 strips.

Firstly, we work out h.

$h=\frac{5-1}{4}=\frac{1}{1}=1$

Now we begin to set up the Midpoint Rule.

$\underset{1}{\overset{5}{\int }}{x}^2+2xdx\approx 1[f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+f\left(x_4\right)]$

$x_1=\frac{1}{2}[(1+\{1-1\}1)+(1+1\times 1)]=\frac{1}{2}[\left(1\right)+\left(2\right)]=1.5$

$x_2=\frac{1}{2}[(1+\{2-1\}1)+(1+2\times 1)]=\frac{1}{2}[\left(2\right)+\left(3\right)]=2.5$

$x_3=\frac{1}{2}[(1+\{3-1\}1)+(1+3\times 1)]=\frac{1}{2}[\left(3\right)+\left(4\right)]=3.5$

$x_4=\frac{1}{2}[(1+\{4-1\}1)+(1+4\times 1)]=\frac{1}{2}[\left(4\right)+\left(5\right)]=4.5$

$\underset{1}{\overset{5}{\int }}{x}^2+2xdx\approx 1[f(1.5)+f(2.5)+f(3.5)+f(4.5)]$

Now we need to solve f(n)

$\underset{1}{\overset{5}{\int }}{x}^2+2xdx\approx 1[5.25+11.3+19.3+29.3]$

$\underset{1}{\overset{5}{\int }}{x}^2+2xdx\approx 65$

As you can see the resultant from the midpoint rule is closer to the true value $65\frac{1}{3}$ than the trapezium rule, but worse than Simpson's Rule.