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How can I calculate the area under a curve after plotting discrete data as per below?

(*Plot of Power against α° with Mathematica v 8.0.0.0*)
powerdata = {{-360, 0}, {-350, 2160.018}, {-340, 3217.711}, {-330, 4238.57}, {-320, 4605.528}, {-310, 5389.598}, {-300, 5921.861}, {-290, 6520.71}, {-280, 6765.366}, {-270, 6402.162}, {-260, 5873.491}, {-250, 5100.044}, {-240, 4205.911}, {-230, 3352.66}, {-220, 2609.472}, {-210, 1926.244}, {-200, 1277.845}, {-190, 656.4205}, {-180, 0}, {-170, 695.2516}, {-160, 1439.652}, {-150, 2240.824}, {-140, 3154.999}, {-130, 4281.635}, {-120, 5719.457}, {-110, 7541.721}, {-100, 9868.855}, {-90, 12838.96}, {-80, 16579.34}, {-70, 21116.01}, {-60, 26655.91}, {-50, 33219.76}, {-40, 40198.46}, {-30, 45864.48}, {-20, 47093.57}, {-10, 41730.7}, {0, 0}, {10, 95148.07}, {20, 148445.8}, {30, 150025.8}, {40, 130629.4}, {50, 107084.8}, {60, 85503.33}, {70, 67427.1}, {80, 52707.47}, {90, 40988.6}, {100, 31697.96}, {110, 24364.54}, {120, 18551.79}, {130, 13618.07}, {140, 9413.644}, {150, 5864.437}, {160, 3208.192}, {170, 1424.567}, {180, 0}, {190, 1479.933}, {200, 2878.621}, {210, 4439.735}, {220, 6607.909}, {230, 8777.864}, {240, 10524.45}, {250, 12753.58}, {260, 15041.08}, {270, 15794.93}, {280, 16654.41}, {290, 17244.18}, {300, 16388.56}, {310, 15911.84}, {320, 13940.65}, {330, 11507.29}, {340, 8191.162}, {350, 4223.378}, {360, 0}};

(*Input data suppressed*)
dataPlot = ListPlot[powerdata, PlotRange -> {{-360, 360}, {0, 155000}}, AxesOrigin -> {-360, 0},
    PlotStyle -> PointSize[.02], PlotMarkers -> Automatic, Joined -> True,
    AxesLabel -> {Alpha, Power}, PlotLabel -> "Power versus Alpha"]

Mathematica graphics

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  • $\begingroup$ 10 Total[powerdata[[All, 2]]] $\endgroup$
    – David G. Stork
    Commented Jan 4, 2015 at 15:12
  • $\begingroup$ properly, you should only take half of the end values for the trapezoidal rule. this works because the end values are zero.. $\endgroup$
    – george2079
    Commented Jan 4, 2015 at 15:53
  • $\begingroup$ See this answer, as well as the deprecated ListIntegrate, which gives similar code for doing what ListIntegrate used to do. $\endgroup$
    – Michael E2
    Commented Jan 4, 2015 at 17:04

1 Answer 1

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Integrate[Interpolation[powerdata][x], {x, -360, 360}]
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  • $\begingroup$ These two answers agree to within 0.1%. $\endgroup$
    – David G. Stork
    Commented Jan 4, 2015 at 15:36
  • $\begingroup$ Immediately after posting this query (Jan 4 at 15:59) I added : "Since posting I did use Interpolation and Integrate and got 1.55954*10^7. Which seems reasonable. I had used -360 to 360. " Being new to this site I mistakenly posted this comment as an answer. Quite rightly, it was deleted at 18:29. Apologies to all. I shall try to be more careful in the future. $\endgroup$
    – Jim McE
    Commented Jan 18, 2015 at 19:25

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