Question

How to you approximate the integral of `(t^3 +t) dx` from [0,2] by using the trapezoid rule with n=4?

Answer 1

Dividing the range `[0,2]` into 4 trapezoids of equal width gives each trapezoid a width of `1/2`

The area of each trapezoid is
`"(average height) " xx " (width)"`

For `f(x) = (t^3+t)`
the heights of the trapezoids are
`f(0) = 0`
`f(1/2) = 5/8`
`f(1) = 2`
`f(3/2) = 39/8`
`f(2) = 10`

The sum of the areas of the trapezoids is
`A_t = ((0+5/8)/2*1/2) +((5/8+2)/2*1/2) + ((2+39/8)/2*1/2)+((39/8+10)/2*1/2)`

`=1/4*((0+10)+2*(5/8+2+39/8))`

`=6 1/4`

So `int_0^2 (t^3+t) dt`
is approximately `6 1/4`