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.