Question

How do you use the trapezoid rule to estimate the area under the graph of `y=x^2+2` for values of `x` from 0 to 3?

Answer 1

Trapezium rule gives:

`int_0^3 \ x^2+2 \ dx ~~ 15.18` (2dp)

Explanation:

The values of `f(x)=x^2+2` are tabulated as follows (using Excel) working to 4dp. The number of ordinates is not specified in the question so I have arbitrarily chosen to use `n=5`

Using the trapezoidal rule:

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

We have:

`int_0^3 \ x^2+2 \ dx`
`" " ~~ 0.6/2 { 2 + 11 + 2(2.36 + 3.44 + 5.24 + 7.76) }`

`" " = 0.3 { 13 + 2(18.8) }`
`" " = 0.3 { 13 + 37.6 }`
`" " = 0.3 * 50.6`
`" " = 15.18`

Let's compare this to the exact value:

`int_0^3 \ x^2+2 \ dx = [1/3x^3+2x]_0^3`
`" " = (27/3+6) - (0)`
`" " = 15`