Question

How do you use the Trapezoidal Rule and the Simpson's Rule when n=4 when approximating the integral `(5t + 6) dt` from [3,6]?

Answer 1

`int_3^6(5t+6)dt~~85.5`

Explanation:

`int_a^bf(t)dt~~(Deltax)/2(f(t_0)+2f(t_1)+2f(t_2)+...+2f(t_(n-1))+f(t_n))`
Where `Deltat=(b-a)/2`
We have `a=3,b=6,n=4`
Therefore, `Deltat=(6-3)/4=3/4`
Next, we divide [3,6] into four equal intervals with the length `3/4`
This will give us the following endpoints:
`3,15/4,9/2,21/4,6`

Now we simply evaluate the function at these endpoints;
`f(t_0)=f(a)=f(3)=21`
`2f(t_1)=2f(15/4)=99/2=49.5`
`2f(t_2)=2f(9/2)=57`
`2f(t_3)=2f(21/4)=129/2=64.5`
`f(t_4)=f(b)=f(6)=36`

Now, we just sum up the values and multiply by `(Deltat)/2=3/8`
`=3/8(21+49.5+57+64.5+36)=85.5`

Hope this helps :)