Simpson's Rule Question! Very hard!

May 2010
8
0
Hi

Simpson's Rule i dont understand anything. its really hard.

Please help me with this question:

1. A vehicle starts from rest and its velocity is measured every 8 seconds, with values as follows:

Time t (s) =Velocity v (ms-1)
0 = 0
1.0 =0.4
2.0 =1.0
3.0 =1.7
4.0 =2.9
5.0 =4.1
6.0 =6.2
7.0 =8.0
8.0 =9.4

The distance traveled in 8.0 seconds is given by


[FONT=&quot]
8.0 ON THE TOP AND 0 ON THE BOTTOM SORRY ABOUT THE CONFUSION
[/FONT]

vdt. Estimate this distance using simpson's rule, giving your answer to 3 significant figures.


please help me answer this really hard question.

thank you very much
 

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MHF Hall of Honor
Mar 2010
2,340
821
Chicago
Hi

Simpson's Rule i dont understand anything. its really hard.

Please help me with this question:

1. A vehicle starts from rest and its velocity is measured every 8 seconds, with values as follows:

Time t (s) =Velocity v (ms-1)
0 = 0
1.0 =0.4
2.0 =1.0
3.0 =1.7
4.0 =2.9
5.0 =4.1
6.0 =6.2
7.0 =8.0
8.0 =9.4

The distance traveled in 8.0 seconds is given by


[FONT=&quot]
8.0 ON THE TOP AND 0 ON THE BOTTOM SORRY ABOUT THE CONFUSION
[/FONT]

vdt. Estimate this distance using simpson's rule, giving your answer to 3 significant figures.


please help me answer this really hard question.

thank you very much
Simpson's rule is an approximation, and basically there is a formula and all you have to do is plug in numbers.

For this problem I would think the question designer might ask for the composite Simpson's rule because of the number of data points, but since this is not explicitly asked for, I would just go with ordinary Simpson's rule.

You do have some reference material to give you the formula, right? It's on Wikipedia too.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
No, problems involving Simpson's rule are NOT hard. It is simply a matter of memorizing Simpson's rule (or knowing where to look it up- I'll bet it's in your textbook) and then plugging in numbers.

Simpson's rule says that
\(\displaystyle \int_a^b f(x) dx= \frac{b-a}{3}(f(x_0)+ 4f(x_1)+ 2f(x_2)+ 4f(x_3)+ \cdot\cdot\cdot\)\(\displaystyle + 2f(x_{n-2})+ 4f(x_{n-1})+ f(x_n))\) where \(\displaystyle x_0= a\), \(\displaystyle x_n= b\) and the \(\displaystyle x_i\) are equally spaced between a and b (that is, \(\displaystyle x_i- x_{i-1}= \frac{b-a}{n}\)). In order to make the "4", "2" alternation work out, there must be an odd number of points (n even since we start counting at 0).

In this particular problem, you have 9 points with n= 8. a= 0, b= 8.0 so b- a= 8.0.
\(\displaystyle \frac{8}{3}(0+ 4(.4)+ 2(1.0)+ 4(1.7)+ 2(2.9)+ 4(4.1)+ 2(6.2)+ 4(8.0)+ 9.4)\)
 
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