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Numerical integration and the Plot of a function give different results?

Luther Nayhm
Posted 10 years ago

The attached worksheet describes the issue.

The numerical integrated value does not match the values in the plotted graphic over the same range.

I am likely making a novice mistake, but any help would be appreciated.

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POSTED BY: Luther Nayhm
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Luther Nayhm
Luther Nayhm
Posted 10 years ago

You are absolutely correct! See, I told you I probably did something stupid.

Thanks for your help. I get in a hurry and skip steps I know not to skip.
POSTED BY: Luther Nayhm

You may have forgotten to subtract the value of f1 at b=0:

Clear[f1];
f1[b_] = Integrate[(b Sqrt[1 - b^2])/(1 - b)^2, b];
ListPlot[Table[f1[b], {b, 0, .13, .01}]]
ListPlot[Table[f1[b] - f1[0], {b, 0, .13, .01}]]
POSTED BY: Gianluca Gorni
Luther Nayhm
Luther Nayhm
Posted 10 years ago

Not sure what you are repling to.

If I integrate my expression, I get an analytical function that has a value of 3 when the variable goes to zero, but the ListPlotTable[NIntegrate]] for that same integrated function shows the correct value....0....when the variable goes to zero. The analytical function is biased upward by 3 units...and the curve that is generated is the same as the ListPlot only biased upward by 3 units.

My conclusion is that Mathematica has incorrectly performed the integration on the function and has given me an incorrect analytical expression. I can live with the numerical integration but the disconnect with the analytical expression was unexpected.
POSTED BY: Luther Nayhm
xavi pirlo
xavi pirlo
Posted 10 years ago

y1 = 0; y2 = 1; n = 50; y = Range[y1,y2, (y2-y1)/(n-1.)]; f[x, y] := y Sin[x] If your function is Listable you can do this

points = {y, NIntegrate[f[x, y], {x, 0, Pi}]} // Transpose; ListLinePlot[points] If your function is not Listable:

points = Map[{#, NIntegrate[f[x, #], {x, 0, Pi}]} &, y] ListLinePlot[points] enter image description here
POSTED BY: xavi pirlo
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