Answer:
Step-by-step explanation:
Well, since it was not given the interval let's use the interval [0,5] with n=10
So now, for the Trapezoidal Rule to approximate the area enclosed by the Integral of: [tex]f(x)=7\pi \sin(x)[/tex]
[tex]T_{10}=\frac{b-a}{2n}[f(a)+2f(x_1)+ ....2f(x_{n-1})+f(b)][/tex] Plugging in:
[tex]T_{10}=\frac{5-0}{2*10}[f(0)+2f(\frac{1}{2})+2f(1)+2f(\frac{3}{2})+2f(2)+2f(5/2)+2f(3)+2f(7/2)+2f(4)+2f(9/2) +f(5)][/tex]
[tex]T_{10}=\frac{1}{4}[0+21.086+37+43.87+39.99+26.322+6.20-15.43-33.285-42.99-21.087][/tex]
[tex]T_{10}\approx 15.419[/tex]
Now the same area according to Simpson rule:
[tex]S_{10}=\frac{b-a}{3n}[f(a)+4f(x_{1})+2f(x_{2})+4f(x_{3} )+2f(x_{4})+4f(x_{5})+2f(x_{6})+4f(x_{7})+2f(x_{8})+4f(x_{9})+f(b)]\\S_{10}=\frac{5}{3*10}[0+74.01+43.87+79.98+26.322+12.413-15.43-66.571-42.99-21.08]\approx 15.085[/tex]
[tex]S_{10}\approx 15.0585[/tex]
