Question

Using n=4 trapezoids, how do you approximate the value of `int sqrt(x+1) dx` from [1,3]?

Answer 1

To approximate `int_1^3sqrt(x+1) dx` by a series of 4 trapezoids
we need to evaluate `sqrt(x+1)` at 5 points `x={1.0, 1.5, 2.0, 2.5, 3.0}`
(using a constant `Delta x` of `0.5`)

This gives us 4 trapezoidal areas
`A_(T1) = ((sqrt(2)+sqrt(2.5))/2)xx0.5`
`A_(T2) = ((sqrt(2.5)+sqrt(3.0))/2)xx0.5`
`A_(T3) = ((sqrt(3.0)+sqrt(3.5))/2)xx0.5`
`A_(T4) = ((sqrt(3.5)+sqrt(4.0))/2)xx0.5`

Adding together the 4 trapezoidal areas (with the aid of a calculator) gives an approximation of the integral `=3.445563`