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:

$\int _{a}^{b}f\left(x\right)\,dx\approx =h\left[f\left(x_{1}\right)+f\left(x_{2}\right)+\ldots +f\left(x_{n}\right)\right]$

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

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

Example

Use the Midpoint Rule to evaluate $\int _{1}^{5}x^{2}+2x\ dx$ 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.

$\int _{1}^{5}x^{2}+2x\ dx\approx 1\left[f\left(x_{1}\right)+f\left(x_{2}\right)+f\left(x_{3}\right)+f\left(x_{4}\right)\right]$

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

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

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

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

$\int _{1}^{5}x^{2}+2x\ dx\approx 1\left[f\left(1.5\right)+f\left(2.5\right)+f\left(3.5\right)+f\left(4.5\right)\right]$

Now we need to solve f(n)

$\int _{1}^{5}x^{2}+2x\ dx\approx 1\left[5.25+11.3+19.3+29.3\right]$

$\int _{1}^{5}x^{2}+2x\ dx\approx 65$

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.