Question

How do you find the area using the trapezoidal approximation method, given `f(x)=x^2 -1`, on the interval [2,4] with n=8?

Answer 1

Trapezium rule gives:

`int_2^4 \ x^2-1 \ dx ~~ 16.69` (2dp)

Explanation:

The values of `f(x)=x^2-1` are tabulated as follows (using Excel) working to 6dp

Using the trapezoidal rule:
`int_a^bydx ~~ h/2{(y_0+y_n)+2(y_1+y_2+...+y_(n-1))}`

We have:

`int_2^4 \ x^2-1 \ dx`
`" " ~~ 0.25/2 { (3 + 15) + 2(4.0625 + 5.25 + 6.5625 +`
`" " 8 + 9.5625 + 11.25 + 13.0625) }`
`" " = 0.125 { 18 + 2(57.75) }`
`" " = 0.125 { 18 + 115.5 }`
`" " = 0.125 { 133.5 }`
`" " = 16.6875`

Let's compare this to the exact value:

`int_2^4 \ x^2-1 \ dx = [x^3/3-x]_2^4`
`" " = (64/3-4)-(8/3-2)`
`" " = 50/3`
`" " = 16.666666 ...`