DerivationIntegrationintermediate

Derivation: simple midpoint

Replacing the function by a central height to obtain the centred rectangle rule and its error.

Centred rectangle

abmf(m)(b-a) f(m)function fmidpoint
The base is b−ab-a and the height is taken at m=(a+b)/2m=(a+b)/2.
Expand diagram

The base is b−ab-a and the height is taken at m=(a+b)/2m=(a+b)/2.

Rule and error

  1. Define the centre of the interval:

    m=a+b2m=\frac{a+b}{2}

  2. Approximate the function by the constant f(m)f(m):

    ∫abf(x) dx≈∫abf(m) dx\int_a^b f(x)\,dx\approx\int_a^b f(m)\,dx

  3. Since f(m)f(m) is constant with respect to xx, the integral is base times height:

    M1=(b−a)f(m)=(b−a)f ⁣(a+b2)M_1=(b-a)f(m)=(b-a)f\!\left(\frac{a+b}{2}\right)

  4. The linear Taylor term around mm contributes no net area by symmetry. The first remaining term depends on f′′f'':

    ∫abf(x) dx−M1=(b−a)324f′′(ξ)\int_a^b f(x)\,dx-M_1=\frac{(b-a)^3}{24}f''(\xi)