Question

How do you use the trapezoidal rule with n=4 to approximate the area between the curve `y=1/(x-1)^2` from 2 to 3?

Answer 1

`int_2^3 \ 1/(x-1)^2 \ dx ~~ 0.509`

Explanation:

We have:

`y = 1/(x-1)^2`

We want to estimate `int \ y \ dx` over the interval `[2,3]` with `4` strips; thus:

`Deltax = (3-2)/3 = 1/4`

Note that we have a fixed interval (strictly speaking a Riemann sum can have a varying sized partition width). The values of the function are tabulated as follows;

Trapezium Rule

`A = int_2^3 \ 1/(x-1)^2 \ dx`
`\ \ \ ~~ 0.25/2 * { 1 + 0.25 + 2*(0.64 + 0.4444 + 0.3265) }`
`\ \ \ = 0.125 * { 1.25 + 2*(1.411) }`
`\ \ \ = 0.125 * { 1.25 + 2.822 }`
`\ \ \ = 0.125 * 4.072`
`\ \ \ = 0.509`

Actual Value

For comparison of accuracy:

`A= int_2^3 \ 1/(x-1)^2 \ dx`
`" \ \ \ = [-1/(x-1)]_2^3`
`" \ \ \ = [1/(x-1)]_3^2`
`" \ \ \ = 1/1-1/2`
`" \ \ \ = 0.5`