Question

How do you use the trapezoidal rule to approximate integral of `e^-3x^2 dx` between [0,1]?

Answer 1

Divide the domain `[0,1]` into smaller trapezoids; calculate the area of each trapezoid; and sum to find an approximation.
Approximate area: `0.5e^(-3) ~~ 0.24894`

Explanation:

Since `e^(-3)` is a constant and
since `int c*f(x) dx = c int f(x) dx` for any constant `c`

I will find an approximate value for `int_0^1 x^2 dx` using trapezoids
and then multiply the result by `e^(-3)`

For demonstration purposes I will divide the domain into 4 trapezoidal areas, each with a width of `0.25`

Trapezoid 1
from `0` to `0.25`
has an area `=(0.25^2-0^2)/2*0.25 = 0.03125`
`color(white)("XXX")`(i.e. average height times width)

Similarly
Trapezoid 2 (from `0.25` to `0.5`)
`color(white)("XXX")`has an area of `0.09375`
Trapezoid 3 (from `0.5` to `0.75`)
`color(white)("XXX")`has an area of `0.15625`
Trapezoid 4 (from `0.75` to `1.0`)
`color(white)("XXX")`has an area of `0.21875`

Total area of all 4 trapezoids
`color(white)("XXX")= 0.03125+0.09375+0.15625+0.21875`
`color(white)("XXX")=0.5`

So `int_0^1 x^2 dx ~~ 0.5`

and since `int_0^1e^(-3)x^2 dx = e^(-3)int_0^1x^2dx`

`int_0^1 e^(-3)x^2 dx ~~ 0.5e^(-3)`

If you wish to carry this further, note that `e^(-3) ~~0.049787` (according to my calculator)