Question

How do you find the area using the trapezoidal approximation method, given `sqrtx`, on the interval [1,4] with n=3?

Answer 1

`int_1^4 sqrt(x) \ dx = 4.646` (4dp)

(For comparison the exact value is `4.6666dot6`)

Explanation:

The values of `f(x)=sqrt(x)` are tabulated as follows (using Excel) working to 6dp.

The Trapezium Rule:

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

uses a series of two consecutive ordinates and a best fit straight line to form trapeziums to approximate the area under a curve, It will have 100% accuracy if `y=f(x)` is a straight line and typically provides a good estimate provided `n` is chosen appropriately.

So,

`int_1^4 sqrt(x) \ dx = 1/2 { (1 + 2) + 2(1.414213 + 1.73205) }`
`" "= 0.5 { + 3 + 2(3.146264) }`
`" "= 0.5 { + 3 + 6.292528 }`
`" "= 0.5 { + 9.292528 }`
`" "= 4.646264`

Exact Value:

`int_1^4 sqrt(x) \ dx = [2/3x^(3/2)]_1^4`
`" "=2/3{4^(3/2)-1^(3/2)}`
`" "= 2/3(8-1)`
`" "= 14/3`
`" "= 4.6666dot6`