Question

How do you use the trapezoidal rule with n=4 to approximate the area between the curve `lnx` from 1 to 3?

Answer 1

`int_(1)^(3) \ ln x \ dx ~~ 1.28 " " (2dp)`

Explanation:

The values of `y=lnx` are tabulated as follows (using Excel) working to 5dp

Using the trapezoidal rule:

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

We have:

`int_(1)^(3) \ ln x \ dx ~~ 0.5/2 { (0 + 1.098612) + 2(0.405465 + 0.693147 + 0.91629) }`
`0.25 { + 1.098612 + 2(2.014903) }`
`0.25 { + 1.098612 + 4.029806 }`
`0.25 { + 5.128418 }`
`1.282104`

It is sometimes interesting to compare against the actual exact answer. Using `int \ lnx dx = xlnx - x` we get

`int_(1)^(3) \ ln x \ dx = [xlnx - x]_(1)^(3)`
`" " = (3ln3-3) - (1ln1-1)`
`" " = (3ln3-3) - (-1)`
`" " = 3ln3-2`
`" " = 1.295836...`