Koshila D.

asked • 07/27/15

reasons for different answers when finding area using Simpsons rule and numerical integration

I have a function √(x2(x+40)) to be integrated from 0 upto -4. Using Simpson's will give me 19.02 but using normal numerical methods giving me -19.5 !! What's the reason behind this difference in signs.........??
Also the 

Koshila D.

I have used (Δx/3)((y1+y9)+4(y2+y4+y6+y8)+2(y1+y3+y5+y7+y9))
Area is divided into 8 parts
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07/27/15

Jon P.

tutor
What do you have Δx equal to?
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07/27/15

Koshila D.

(0-(-4))/8 
So that gives me a positive answer overall. Since we have to find an area, I hope that the answer getting from Simpson's is correct. But the numerical integration from 0 to (-4) gives the negative.....I'm so confused in this answer.
 
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07/28/15

Jon P.

tutor
That's the problem. The limits of integration are from 0 to -4, so Δx is -4 - 0 = -4.  Not +4.  
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07/28/15

Koshila D.

Oh........all right I see...so that's the problem. I have been given only the diagram of the function. And the function stretches from zero towards negative direction. 
So when I need to find the area using normal integration method, do I have to integrate from 0 to (-4) OR from (-4) to 0.........???
 
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07/29/15

Jon P.

tutor
Well, it all depends on how the actual problem is stated.  You originally said "I have a function √(x2(x+40)) to be integrated from 0 up to -4".  I took that to mean literally that you knew that the limits of the integration were 0 to -4.  It is certainly possible to do an integration in the negative direction, and the results are just the negative of the same integration with the limits switched.
 
However...  If you what you were actually asked to do was to find the area under the curve, then you would definitely want to do the integration in the positive direction, from -4 to 0.  In that case, your original integration was NOT correct, but your use of the Simpson's rule was OK.
 
So to answer your specific question, when you need to find the area under a curve, you integrate in the positive direction, in this case from -4 to 0.
 
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07/29/15

Koshila D.

So that's the problem behind this change of signs. 
Thank you very much sir.I really do appreciate your move to help me out of this problem.
 
And one more question sir...!!1
What's the reason for slightest change in the two answers...?? Simpson's giving me 19.02 and integration giving 19.5
 
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07/29/15

Jon P.

tutor
Simpson's rule is by its nature an approximation.  The correct answer is given by the integral. 
 
Simpson's rule essentially breaks the area under the curve into a finite number of rectangles and uses them to estimate the area.  Integration starts by doing the same thing, but the difference is that integration takes this to the limit where the rectangles are infinitesimally narrow, whereas Simpson's rule uses a finite number of actual rectangles.  If you extended Simpson's rule to break the width of the interval into more and more slices, you would see that in the end it approaches the same result as the integral itself.  But as long as you use a finite number of slices, Simpson's rule will always be off somewhat from the actual area.
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07/30/15

1 Expert Answer

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Jon P. answered • 07/27/15

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I've seen different formulation of Simpson's rule, so I can't be sure exactly which one you are using.
 
But... I suspect that the reason is that you are being confused by the fact that the integral interval is from 0 to -4, that is, in the negative direction.  Since the function is always positive throughout the interval, the integral would normally be positive, but the fact that the range goes in the negative direction means that the result comes out negative.
 
In Simpson's rule you have to multiply a weighted average of function values at several points times a factor related to the size of the interval.  You think of that factor as being related to the distance between the end points of the interval, and distance is generally thought of as a positive number.  But it's not actually the distance between the end points -- it has to have a sign that is consistent with the negative direction of the interval, and in this case the direction is negative.
 
Again, if I saw your specific calculations I could be more specific in describing this, but take a look at that multiplier and see if you can tell what I mean.
 
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