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?