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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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    I'm bored and feeling generous: You should have $3.0000, 11.8322, 5.6569, 10.3923, 4.4721, 6.6332, 0.0000$.2012-07-30

1 Answers 1

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Since $h=\dfrac {b-a}n$ and $a=0, b=3$:

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