MHB Can Simpson's Rule be Applied to $$I_{35}$$?

[karush]

do be done by Simpsons Rule for $n=4, 8, 16, 32 $

$$I_{35}=\int_{0}^{4} \left(3{x}^{5}-8{x}^{3}\right)\,dx$$

before I even start on this and seeing the graph how can SR even be done on this?

 

[MarkFL]

First, let's get the exact answer so we know what to expect...W|A gives:

$$\int_0^4 3x^5-8x^3\,dx=1536$$

Okay, next, we need:

[box=green]

Simpson's Rule​

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$$\int_a^b f(x)\,dx\approx S_n$$

where

$$S_n=\frac{b-a}{3n}\left(f\left(x_0 \right)+4f\left(x_1 \right)+2f\left(x_2 \right)+\cdots+2f\left(x_{n-2} \right)+4f\left(x_{n-1} \right)+f\left(x_n \right) \right)$$

$$x_i=a+i\frac{b-a}{n}$$[/box]

We are given $a=0$ and $b=4$...and so let's begin with the first case $S_4$.

i) $n=4$.

With this value for $n$, we find that $x_i=i$. And so:

$$S_4=\frac{1}{3}\left(f(0)+4f(1)+2f(2)+4f(3)+f(4)\right)=\frac{1}{3}(0+4(-5)+2(32)+4(512)+2560)=\frac{4652}{3}=1550.\overline{6}$$

Can you proceed for the other values of $n$?

 

[karush]

$$I_{35 }= \int_0^4 3x^5-8x^3\,dx=1536=S_n$$
$$S_n=\frac{b-a}{3n}\left(f\left(x_0 \right)+4f\left(x_1 \right)+2f\left(x_2 \right)+\cdots+2f\left(x_{n-2} \right)+4f\left(x_{n-1} \right)+f\left(x_n \right) \right)$$$$x_i=a+i\frac{b-a}{n}$$

$\displaystyle a=0 \ \ b=4$

i) $\displaystyle n=4 \ \ x_i=i$

$\displaystyle
S_4=\frac{1}{3}\left(f(0)+4f(1)+2f(2)+4f(3)+f(4)\right)=\frac{1}{3}(0+4(-5)+2(32)+4(512)+2560)=\frac{4652}{3}=1550.\overline{6}$
---------------------------------
ii) $\displaystyle n=8 \ \ x_i=0+i\frac{4-0}{8} =\frac{i}{2}$

$\displaystyle
S_8=\frac{1}{6}
\left[
f(0)+4f\left(\frac{1}{2}\right)
+2f\left(\frac{2}{2}\right)
+4f\left(\frac{3}{2}\right)
+2f\left(\frac{4}{2}\right)
+4f\left(\frac{5}{2}\right)
+2f\left(\frac{6}{2}\right)
+4f\left(\frac{7}{2}\right)
+f\left(\frac{8}{2}\right)
\right]\approx 1537$

Ill stop here since the $n=16$ and $n=32$ would be done the same way