# Filling between boundaries

I would like to visualize what it graphically means to integrate between two boundary values. Therefor I'd like to make a Filling between these two values. Is there a way to get this done?

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Something like this (source code is free there) ?demonstrations.wolfram.com/IntegrationIsASum –  Vitaliy Kaurov Aug 21 '12 at 8:03

An alternative is to use Piecewise as follows

Plot[{Sin[x], Piecewise[{{Sin[x], -Pi <= x <= Pi}}, _]}, {x, -2 Pi, 2 Pi},
Filling -> {2 -> {Axis, Yellow}}, PlotStyle -> {Green, Directive[Red, Thick]}]


which gives

Or

Use Show to superimpose two variants (the second one with your choice of the variable bounds -- -Pi and 2Pi in the example below) of the plot:

Show[Plot[Sin[x], {x, -3 Pi, 3 Pi}],
Plot[Sin[x], {x, - Pi, 2 Pi},
Filling -> Axis, FillingStyle -> Yellow]]


Update: Yet another method using ColorFunction with ColorFunctionScaling->False, Mesh and MeshShading,

Plot[Sin[x], {x, -2 Pi, 2 \[Pi]},
Mesh -> {{0}},
MeshShading -> {Directive@{Thick, Blue}}, Filling -> Axis,
ColorFunction -> (If[-Pi <= #1 <= Pi/2, If[#2 > 0, Red, Yellow], White] &),
ColorFunctionScaling -> False]


Update 2: All inside Manipulate:

First, a cool combo control from somewhere in the docs:

 popupField[Dynamic[var_], list_List] :=
Opener[False, Appearance -> Medium]],
InputField[Dynamic[var], Appearance -> "Frameless"]}},
Frame -> All, FrameStyle -> Orange,
Background -> {{Orange, Orange}}]


and, then,

Manipulate[Column[{ Dynamic@Show[ Plot[func[x], {x, -2 Pi, 2 \[Pi]},
Ticks -> {Range[-2 Pi, 2 Pi, Pi/2], Automatic},
Mesh -> {{0}}, MeshShading -> {Directive@{Thick, color0}},
Filling -> Axis,
ColorFunction -> (If[lb <= #1 <= ub, If[#2 > 0, color1, color2], White] &),
ColorFunctionScaling -> False, ImageSize -> {600, 300}],
Graphics[{Gray, Line[{{-2 Pi, 0}, {2 Pi, 0}}],
Orange, PointSize[.02], Dynamic[(Point[{lb = Min[First[pt1], First[pt2]], 0}])],
Brown, PointSize[.02],  Dynamic[(Point[{ub = Max[First[pt1], First[pt2]], 0}])]},
PlotRange -> 1.], PlotLabel -> Style[ "\nArea = " <>
ToString[Quiet@NIntegrate[func[t], {t, lb, ub}]] <> "\n",
"Subsection", GrayLevel[.3]]]},  Center],
Row[{Spacer[30], Rotate[Style["functions", GrayLevel[.3], 12], 90 Degree],
Spacer[5],Control@{{func, Sin, ""},  popupField[#, {Sin, Cos, Sec, Cosh, ArcSinh}] &}
Spacer[15], Rotate[Style["colors", GrayLevel[.3], 12], 90 Degree],
Spacer[5],  Rotate[Style["line", GrayLevel[.3], 10], 90 Degree],
Control@{{color0, Blue, ""}, ColorSlider[#, AppearanceElements -> "Spectrum",
ImageSize -> {40, 40}, AutoAction -> True] &},
Spacer[5], Rotate[Style["above", GrayLevel[.3], 10], 90 Degree],
Control@{{color1, Green, ""}, ColorSlider[#, AppearanceElements -> "Spectrum",
ImageSize -> {40, 40}, AutoAction -> True] &},
Spacer[5], Rotate[Style["below", GrayLevel[.3], 10], 90 Degree],
Control@{{color2, Green, ""}, ColorSlider[#, AppearanceElements -> "Spectrum",
ImageSize -> {40, 40}, AutoAction -> True] &}},Spacer[0]],
{{lb, -Pi}, ControlType -> None},
{{ub, 3 Pi/2}, ControlType -> None},
{{pt1, {-Pi, 0}}, Locator, Appearance -> None},
{{pt2, {3 Pi/2, 0}}, Locator, Appearance -> None},
Alignment -> Center, ControlPlacement -> Top, AppearanceElements -> Automatic]


-

Always search the doc when you don't know what to do. In most cases you'll not only find it possible, but find a thorough documentation along with examples. In this case you have this link.

Here is an example:

Show[
Plot[{Cos[x], Sin[x]}, {x, 0, 2 \[Pi]}],
Plot[{Cos[x], Sin[x]}, {x, \[Pi]/4, 5/4 \[Pi]},
PlotRange -> {{0, 2 \[Pi]}, All} , Filling -> 1 -> {2}]
]


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Thx for the answer, but what I would like to do is to shade just one of these function between two boundary values. –  RMMA Aug 21 '12 at 8:12
@rainer I simply supplied an example. If you want to fill it in alternatively look through the documentation for Filling. –  jVincent Aug 21 '12 at 8:40

Area will be red or blue depending whether b > a or not.

Manipulate[
Plot[{f[x], UnitStep[Sign[b - a] (x - a)] UnitStep[Sign[b - a] (b - x)] f[x]},
{x, -7, 7}, PlotStyle -> {Thick, Thickness[0]}, Filling -> {2 -> 0},
FillingStyle -> Directive[Opacity[.5], If[b - a > 0, Red, Blue]],
PlotLabel -> "AREA = " <> ToString[NIntegrate[f[x], {x, a, b}]]], {{b, 4}, -7, 7},
{{a, -1}, -7, 7}, {f, {Sin, Cos, Tanh, Sech}}]


Also take a look at source code at the Wolfram Demonstration Project.

-

With axis-constrained locators:

DynamicModule[{pts = {{0, 0}, {Pi, 0}}},
LocatorPane[Dynamic[pts, (pts[[1]] = {#[[1, 1]], 0}; pts[[2]] = {#[[2, 1]], 0}) &],
Dynamic[
Framed@Show@
{Plot[Sin@x, {x, 0, 2 Pi}],
Plot[Sin@x, {x, pts[[1, 1]], pts[[2, 1]]}, Filling -> Axis]
}
]]]


Edit

This is the full code, with the label:

DynamicModule[{pts = {{0, 0}, {Pi, 0}}},
LocatorPane[Dynamic[pts, (pts[[1]] = {#[[1, 1]], 0}; pts[[2]] = {#[[2, 1]], 0}) &],
Dynamic[
Framed@Show@
{Plot[Sin@x, {x, 0, 2 Pi},
PlotLabel -> ToString@StandardForm[Integrate[sin[x],
{x, pts[[1, 1]], pts[[2, 1]]}]] <> " = " <>
ToString[Integrate[Sin@x, {x, pts[[1, 1]], pts[[2, 1]]}]]],

Plot[Sin@x, {x, pts[[1, 1]], pts[[2, 1]]}, Filling -> Axis]
}
]]]

-

IntegralPlot[f_, {x_, L_, U_}, {l_, u_}, opts : OptionsPattern[]] :=
Module[{col = ColorData[1, 1]},
Plot[{ConditionalExpression[f, x > l && x < u], f},
{x, L, U},
Prolog -> {{col, Line[{{l, 0}, {l, f /. {x -> l}}}]}, {col,
Line[{{u, 0}, {u, f /. {x -> u}}}]}},
Filling -> {1 -> Axis},
PlotStyle -> col,
opts]]

IntegralPlot[PDF[NormalDistribution[0, 1]][x], {x, -4, 4}, {1, 2}]


IntegralPlot[x^2, {x, 0, 10}, {4, 6}, PlotLabel -> "x^2"]


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Nice idea to capture the functionality within a function. –  Dragan Mrakovic Nov 23 '14 at 15:00