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]?

1 Answer
Sep 23, 2015

Answer:

#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 :)