like the title says: is that possible? In my case, I have a plot of an integral that is evaluated numerically. Maybe using Interpolate with the points used by mathematica to draw the graph or something?
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$\begingroup$ Do you only have the plot? not the code making it? $\endgroup$– Ruud3.1415Commented Jun 26, 2018 at 14:40
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$\begingroup$ These may be helpfule: stackoverflow.com/q/5364088/4712538, mathematica.stackexchange.com/q/125222/9490, $\endgroup$– Jason B.Commented Jun 26, 2018 at 14:49
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1 Answer
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Starting with the plots
plts = Plot[{Sin[x], Sin[2 x], Sin[3 x]}, {x, 0, 2 Pi},
PlotLegends -> "Expressions"]
Extracting the sets of points from the plot
pts = Cases[plts, Line[pts_] :> pts, Infinity];
Interpolating each set of points
funcs = Interpolation /@ pts;
Plotting the interpolations for comparison
Plot[Evaluate[#[x] & /@ funcs], {x, 0, 2 Pi},
PlotLegends -> Automatic]
EDIT: Addressing the additional questions in the comments.
To evaluate one of the three functions at x == 1 use Part ( [[...]] ). For example, to evaluate the second interpolated function
funcs[[2]][1]
(* 0.909298 *)
To integrate the product of each function with Sin[x] over the interval {0, 1}
NIntegrate[#[x]*Sin[x], {x, 0, 1}] & /@ funcs
(* {0.272675, 0.397216, 0.321925} *)
Or for a single function
NIntegrate[funcs[[2]][x]*Sin[x], {x, 0, 1}]
(* 0.397216 *)
answered Jun 26, 2018 at 15:00
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$\begingroup$ Thanks! One question: how could I integrate the interpolated function? Say I wanted to integrate func Sin[x] between x=0 and x=1, would the following code be correct? (func is like your funcs) NIntegrate[Sin[x] Evaluate[#[x] & /@ func], {x,0,1}] $\endgroup$– FisherCommented Jun 26, 2018 at 20:07