Simpson's Rule question (1 Viewer)

youngsky

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Hey guys, I'm having trouble with this question:

"Use Simpson's Rule and 4 sub-intervals to approximate the area bounded by the curve y = e-x2 , the x-axis, and the lines x = -2 and x = 2."

It seems using the formula A = (b-a)/6 {f(a) + f(b) + 4[f(a+b)/2]...} from Cambridge gives me double the actual answer? I got 3.32 :/
 

braintic

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Without seeing your working, it is not possible to see where you went wrong.

But just checking: you need two applications of the rule - for each application you have b-a=2, yes?
 

Trebla

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Hey guys, I'm having trouble with this question:

"Use Simpson's Rule and 4 sub-intervals to approximate the area bounded by the curve y = e-x2 , the x-axis, and the lines x = -2 and x = 2."

It seems using the formula A = (b-a)/6 {f(a) + f(b) + 4[f(a+b)/2]...} from Cambridge gives me double the actual answer? I got 3.32 :/
That formula is for two subintervals only. You need to apply it across -2 < x < 0 then apply it again across 0 < x < 2 before adding the two areas.
 

youngsky

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Without seeing your working, it is not possible to see where you went wrong.

But just checking: you need two applications of the rule - for each application you have b-a=2, yes?
Alright, so I basically did it like this:

h = b-a/n, where h is the width of each subinterval and n is the number of subintervals
= 2-(-2)/6
= 2/3

Drew a table of values with the x coordinates -2, -1, 0, 1, 2 and y values e^-4, e^-1, 1, e^-1, e^-4

Then I used the formula A ~ (b-a)/6 {f(a)+f(b)+4[f(a+b)/2]...} and I ended but getting around 3.3

That formula is for two subintervals only. You need to apply it across -2 < x < 0 then apply it again across 0 < x < 2 before adding the two areas.
Hence the "..." in the formula I quoted. But my question is, why does it work if you apply it across the 2 domains, but not one, single domain? (i.e. -2 to 2)
 

Trebla

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Hence the "..." in the formula I quoted. But my question is, why does it work if you apply it across the 2 domains, but not one, single domain? (i.e. -2 to 2)
You need to understand how the formula is derived. See link below:
http://en.m.wikipedia.org/wiki/Simpson's_rule

Basically you are approximating an area by a parabola which is chosen such that it intersects the curve at three points with one of those points being halfway between the two other points in terms of x-values. This approach consequently leads to two sub-intervals being constructed for the purposes of the approximation and the formula is derived from that.

If you apply that formula just from x = -2 to x = 2 you are effectively only creating two sub-intervals between x = -2 and x = 2 rather than 4 sub-intervals.
 

braintic

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Alright, so I basically did it like this:

h = b-a/n, where h is the width of each subinterval and n is the number of subintervals
= 2-(-2)/6
= 2/3

Drew a table of values with the x coordinates -2, -1, 0, 1, 2 and y values e^-4, e^-1, 1, e^-1, e^-4

Then I used the formula A ~ (b-a)/6 {f(a)+f(b)+4[f(a+b)/2]...} and I ended but getting around 3.3



Hence the "..." in the formula I quoted. But my question is, why does it work if you apply it across the 2 domains, but not one, single domain? (i.e. -2 to 2)
Why are you setting n=6?
h is not equal to 2/3.
 

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