Question

How do you find the area using the trapezoidal approximation method, given `(x²-6x+9) dx`, on the interval [0,3] with n=3?

Answer 1

`int_0^3 \ x^2-6x+9 \ dx ~~ 9.5`

Explanation:

We have:

`y = x^2-6x+9`

We want to estimate `int \ y \ dx` over the interval `[0,3]` with `3` strips; thus:

`Deltax = (3-0)/3 = 1`

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_0^3 \ x^2-6x+9 \ dx`
`\ \ \ ~~ 1/2 * { 9 - 0 + 2*(4 + 1) }`
`\ \ \ = 0.5 * { 9 + 2*(5) }`
`\ \ \ = 0.5 * { 9 + 10 }`
`\ \ \ = 0.5 * 19`
`\ \ \ = 9.5`

Actual Value

For comparison of accuracy:
`A = int_0^3 \ x^2-6x+9 \ dx`

`\ \ \ = [x^3/3-3x^2+9x]_0^3`
`\ \ \ = 9-27+27`
`\ \ \ = 9`