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Estimating area under a curve in log scale 
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#1
Jun1110, 08:07 AM

#2
Jun1110, 10:42 AM

Mentor
P: 21,297

Do you mean the area under the curves as shown in the graph? You need a proper description of the region whose area you want. For example, it might be "For each function, find the area of the region bounded by the graph of the function and the xaxis, between x = 10^{3} and x = 10^{2}.
I would estimate each area using vertical rectangular strips. You could use trapezoids, but it would be a bit more work. For each rectangular strip you need the rectangle's height and width. You can read the height from the function's graph, and the width of each strip varies as x increases from left to right. For example, between x = 10^{3} and x = 10^{2}, each tickmark is 1/1000 or .001. Between x = 10^{2} and x = 10^{1}, each tickmark is 1/100 or .01. 


#3
Jun1110, 11:08 AM

P: 771

Print the image on two pieces of paper, cut out the paper under each curve down to the Xaxis and weigh each piece.
Cut a third piece of paper along the Xaxis, along the right and left edges of the graphed area, and along the 10^0 horizontal line to give you a reference weight. You'll have to consider how to account for the log scales. If you don't have a sufficiently sensitive scale, paste the images on cardboard before cutting. 


#4
Jun1110, 11:42 AM

P: 3,016

Estimating area under a curve in log scale
A really quick estimate would be given if divide each area into simple shapesy sum up the individual areas of them. I did a quick outline (really quick) of the solid curve. The red lines are the estimates. The more time you put into it the better the estimate.



#5
Jun1110, 01:06 PM

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P: 41,096

The xaxis is arbitrary on a log scale. The area is infinite. It's a trick question



#6
Jun1310, 07:02 PM

Sci Advisor
P: 4,032

You can also ignore anything below 10^2 because this is less than 1 % of the peak value.
Given the other errors, 1% is an unlikely precision. Especially concentrate on that peak at the top. Try to get that right because the values in there are much higher than the rest of the graph and so the errors would be greatest there also. Redraw the graph converting both scales to linear values, otherwise you can't take rectangles and calculate areas. 


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