# Using Simpson's Rule to approximate $\int_0^3{\sqrt{9-x^2}dx}$

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?

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compare it with ${\pi 3^2 \over 4}$ ... I think you need more smaller step sizes – Santosh Linkha Jul 30 '12 at 18:53
Yes, that looks correct (I did not check your arithmetic). Note though, to nitpick, you should write e.g., $4f(3/2)\approx 10.3923$. – David Mitra Jul 30 '12 at 18:54
experiment, what do you mean exactly? – user69 Jul 30 '12 at 18:56
The answers (in 32 bit arithmetic, with 5 digits) are $n=6$: $\approx 6.9978$, $n=60$: $\approx 7.0664$, $n=600$: $\approx 7.0685$, $n=6000$: $\approx 7.0686$, and same for $n=60000$. Your computations above are off a little, I believe. – copper.hat Jul 30 '12 at 19:14
I'm bored and feeling generous: You should have $3.0000, 11.8322, 5.6569, 10.3923, 4.4721, 6.6332, 0.0000$. – copper.hat Jul 30 '12 at 19:28

Since $h=\dfrac {b-a}n$ and $a=0, b=3$:
$$\int_a^b{f(x)dx}\approx S_n = \frac{1}{3}h[f(a)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+...+4f(x_{n-1})+f(b)]\\=\frac 16 ( 3+ 11.8321+ 5.6568+10.3923+4.4721+6.6332+0)\\=6.9977$$