I have to use Simpson's rule:
$$\int_a^b f(x) \, dx \approx S_n = \frac{1}{3}h[f(a)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+\cdots+4f(x_{n-1})+f(b)]$$
when $n=6$ to approximate the integral:
$$\int_0^3{\sqrt{9-x^2}dx}$$
to four decimal places.
I've gotten $$f(0) = 3$$ $$ 4f\left(\frac{1}{2}\right)=11.831$$ $$2f(1)=5.6569$$ $$4f\left(\frac{3}{2}\right)=10.3923$$ $$2f(2)=4.4721$$ $$4f\left(\frac{5}{2}\right)=6.6332$$ $$f(3)=0$$ and then.. $$\int_0^3{\sqrt{9-x^2} \, dx}\approx S_6 = \left (\frac{1}{3}\right )\left(\frac{1}{2}\right )\left (41.9857\right ) = 6.9976 $$
Did I do this right and is this the correct answer?