Question

Estimate the value of the integral from negative 1 to 3 of x squared, dx by using the Trapezoidal Rule with n = 4?

Answer 1

The integral

`I= int_(-1)^3 x^2dx`

With `n = 4` will have the intervals

`[-1, 0]`
`[0, 1]`
`[1, 2]`
`[2, 3]`

Imagine drawing trapezoids on the graph. Integration is finding the area, so we have to add the areas of each of these trapezoids up.

Trapezoid 1

This will have one base of length `1^2` and the other length `0^2`. It will have a height of `0.5`. Thus the area will be

`((1 + 0)1)/2 = 0.50`

Repeat this for each to get

`A_2 = 0.50`
`A_3 = 2.50`
`A_4= 6.50`

It follows that

`A_"total" = 0.50 + 0.50 + 2.50 + 6.50`

`A_"total" = 10`

Therefore, by the trapezoidal rule, with `n = 4`, `int_(-1)^3 x^2dx =10`.

The actual value is `[1/3x^3]_(-1)^3 = 1/3(3)^3 - 1/3(-1)^3 = 9 + 1/3 = 9.33`

So our approximation isn't too bad.

Hopefully this helps!