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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 218223, 11170]*) (*NotebookOutlinePosition[ 263652, 12744]*) (* CellTagsIndexPosition[ 263486, 12735]*) (*WindowFrame->Normal*) Notebook[{ Cell[GraphicsData["Bitmap", "\<\ CF5dJ6E]HGAYHf4PAg9QL6QYHg"], "Graphics", ShowCellBracket->False, CellMargins->{{25, 24}, {5, 7}}, ImageSize->{81, 22}, ImageMargins->{{0, 0}, {0, 0}}, ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Accumulation\n", StyleBox[ "Authors: Bill Davis, Horacio Porta and Jerry Uhl ", "Subtitle"], StyleBox["\[Copyright]1999", "Subtitle", FontSize->12], StyleBox["\nProducer: Bruce Carpenter\n", "Subtitle"], StyleBox["Publisher: ", "Subtitle", FontSize->12], StyleBox[ButtonBox["Math Everywhere, Inc.", ButtonData:>{ URL[ "http://www.matheverywhere.com"], None}, ButtonStyle->"MEIHyperlink"], FontSize->12], StyleBox[" Distributor: ", "Subtitle", FontSize->12], StyleBox[ButtonBox["Wolfram Research, Inc.", ButtonData:>{ URL[ "http://www.wolfram.com"], None}, ButtonStyle->"MEIHyperlink", ButtonNote->"Makers of Mathematica!"], FontSize->12] }], "PrefaceTitle", CellMargins->{{Inherited, Inherited}, {Inherited, 0}}], Cell[TextData[{ "2.04 Transforming Integrals \n", StyleBox["Samples", FontSize->16, FontSlant->"Italic"] }], "Title"], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " Initializations" }], "Special2"], Cell[BoxData[{ \(\(Off[General::spell]; \)\), \(\(Off[General::spell1]; \)\), \(\(Off[Plot::plnr]; \)\), \(\(Off[ParametricPlot3D::ppcom]; \)\), \(\(SetOptions[Limit, Analytic \[Rule] True]; \)\), \(Needs["\"]; \n\n If[MemberQ[{"\", \ "\", \ "\"}, \n\ \ \ \ Context[Gray]], \ Remove["\<*`Gray\>"]]; \n<< "\"; \n Remove["\<*`Gray\>"]; \n<< "\"; \n Graphics`Colors`GosiaGreen = RGBColor[0, \ 0.392187, \ 0]; \n \n$Pre\ = \ Chop; \nCMView\ = \ {2.7, \ 1.6, \ 1.2}; \)}], "Input", InitializationCell->True] }, Closed]], Cell["Basic Problem", "Subsubsection"], Cell[CellGroupData[{ Cell["\<\ B.1) Breaking more of the code of the integral: Transforming integrals\ \>", "Subsection", CellTags->"2.04.B1"], Cell["B.1.a) Combining the fundamental formula and the chain rule", "Subsubsection", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[TextData[{ "How do you know that the integrals\n ", Cell[BoxData[ \(\[Integral]\_a\%b\( f'\)[u[x]] \(u'\)[x] \[DifferentialD]x\)]], "\nand\n ", Cell[BoxData[ \(\[Integral]\_\(u[a]\)\%\(u[b]\)\(f'\)[u] \[DifferentialD]u\)]], "\nare equal?" }], "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell["Answer:", "Special1", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "The best way to see why they are equal is to calculate both of them.\nThe \ second is the easier because the fundamental formula tells you that\n ", Cell[BoxData[ RowBox[{ \(\[Integral]\_\(u[a]\)\%\(u[b]\)\(f'\)[u] \[DifferentialD]u\), "=", RowBox[{ RowBox[{\(f[u]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], \(u[a]\), \(u[b]\)]}], "=", \(f[u[b]] - f[u[a]]\)}]}]]], ".\n \nFor the first integral, the chain rule tells you that the \ derivative of \n ", Cell[BoxData[ \(f[u[x]]\)]], " \nis \n ", Cell[BoxData[ \(\(f'\)[u[x]] \(u'\)[x]\)]], ". \nThe fundamental formula steps in to say \n ", Cell[BoxData[ RowBox[{ \(\[Integral]\_a\%b\( f'\)[u[x]] \(u'\)[x] \[DifferentialD]x\), "=", RowBox[{\(f[u[x]]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "a", "b"]}]}]]], "\n ", Cell[BoxData[ \(\( = f[u[b]] - f[u[a]]\)\)]], ".\nNow do you know why \n ", Cell[BoxData[ \(\[Integral]\_a\%b\( f'\)[u[x]] \(u'\)[x] \[DifferentialD]x = \[Integral]\_\(u[a]\)\%\(u[b]\)\(f'\)[u] \[DifferentialD]u\)]], ".\nReason: They are both equal to\n ", Cell[BoxData[ \(f[u[b]] - f[u[a]]\)]], ".\nA point of anxiety might come up: \nIn the integral\n ", Cell[BoxData[ \(\[Integral]\_\(u[a]\)\%\(u[b]\)\(f'\)[u] \[DifferentialD]u\)]], ",\nthe lone symbol ", Cell[BoxData[ \(u\)]], " is treated as a variable and ", Cell[BoxData[ \(u[a]\)]], " and ", Cell[BoxData[ \(u[b]\)]], " are numbers. In the integral,\n ", Cell[BoxData[ \(\[Integral]\_a\%b\( f'\)[u[x]] \(u'\)[x] \[DifferentialD]x\)]], ",\n", Cell[BoxData[ \(u[x]\)]], " is a function, ", Cell[BoxData[ \(x\)]], " is a variable, and ", Cell[BoxData[ \(a\)]], " and ", Cell[BoxData[ \(b\)]], " are numbers.\nThis should not cause you any trouble." }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}] }, Closed]], Cell["B.1.b)", "Subsubsection", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[TextData[{ "Now you know\n ", Cell[BoxData[ \(\[Integral]\_a\%b\( f'\)[u[x]] \(u'\)[x] \[DifferentialD]x = \[Integral]\_\(u[a]\)\%\(u[b]\)\(f'\)[u] \[DifferentialD]u\)]], ".\nWhat practical use is this?" }], "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell["Answer:", "Special1", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "Notational magic allows you to take the more complicated integral and \ replace it with the less complicated but equal integral.\nPair them up as \ follows:\n ", Cell[BoxData[ \(\(f'\)[u[x]]\)]], " <-------> ", Cell[BoxData[ \(\(f'\)[u]\)]], "\n ", Cell[BoxData[ \(\(u'\)[x] \[DifferentialD]x\)]], " <-------> ", Cell[BoxData[ \(\ \[DifferentialD]u\)]], "\n ", Cell[BoxData[ \(\[Integral]\_a\%b\)]], " <----------------> ", Cell[BoxData[ \(\[Integral]\_\(u[a]\)\%\(u[b]\)\)]], ".\nLots of folks like to call this a \"transformation\" of a hard integral \ into an easy integral." }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}] }, Closed]], Cell["B.1.c)", "Subsubsection", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "Here is ", StyleBox["Mathematica", FontSlant->"Italic"], "'s calculation of ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi] Cos[x\^2] 2 x \[DifferentialD]x\)]], ":" }], "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x];\)\), "\n", \(MathematicaCalculation = \[Integral]\_0\%\[Pi]\( Cos[ x\^2]\ 2\ x\) \[DifferentialD]x\)}], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(Sin[\[Pi]\^2]\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Use a transformation to explain where this result comes from.\ \>", "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell["Answer:", "Special1", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "The integral at the center stage is\n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi] Cos[x\^2] 2 x \[DifferentialD]x\)]], ".\nThe key is that ", Cell[BoxData[ \(2 x\)]], " is the derivative of ", Cell[BoxData[ \(x\^2\)]], ", so go with\n ", Cell[BoxData[ \(u[x] = x\^2\)]], ".\nThis gives the pairings\n ", Cell[BoxData[ \(Cos[x\^2]\)]], " <-------------------> ", Cell[BoxData[ \(Cos[u]\)]], "\n ", Cell[BoxData[ \(2 x \[DifferentialD]x = \(u'\)[x] \[DifferentialD]x\)]], " <-------> ", Cell[BoxData[ \(\[DifferentialD]u\)]], "\n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi]\( = \[Integral]\_a\%b\)\)]], " <--------------------> ", Cell[BoxData[ \(\[Integral]\_\(u[a]\)\%\(u[b]\)\( = \[Integral]\_\(0\^2\)\%\(\[Pi]\^2\)\)\)]], "\nSo\n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi] Cos[x\^2] 2 x \[DifferentialD]x = \[Integral]\_\(0\^2\)\%\(\[Pi]\^2\)Cos[u] \[DifferentialD]u\)]], ".\nIn other words, the substitution ", Cell[BoxData[ \(u[x] = x\^2\)]], " transforms\n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi] Cos[x\^2] 2 x \[DifferentialD]x\)]], " into ", Cell[BoxData[ \(\[Integral]\_\(0\^2\)\%\(\[Pi]\^2\)Cos[u] \[DifferentialD]u\)]], ". \nThe second integral is cake. \nYou look for a function ", Cell[BoxData[ \(f[u]\)]], " with ", Cell[BoxData[ \(\(f'\)[u] = Cos[u]\)]], ".\nHere's one:" }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[BoxData[{ \(\(Clear[f, u];\)\), "\n", \(\(f[u_] = Sin[u];\)\)}], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell["Check:", "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(\(f'\)[u]\)], "Input"], Cell[BoxData[ \(Cos[u]\)], "Output"] }, Open ]], Cell[TextData[{ "And now you can say with confidence and authority that\n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi] Cos[x\^2] 2 x \[DifferentialD]x = \[Integral]\_\(0\^2\)\%\(\[Pi]\^2\)Cos[u] \[DifferentialD]u\)]], "\n is given by:" }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(f[\[Pi]\^2] - f[0\^2]\)], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(Sin[\[Pi]\^2]\)], "Output"] }, Open ]], Cell[TextData[{ "Check with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s calculation." }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(MathematicaCalculation\)], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(Sin[\[Pi]\^2]\)], "Output"] }, Open ]], Cell["Got it.", "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}] }, Closed]], Cell["B.1.d)", "Subsubsection", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "Here is ", StyleBox["Mathematica", FontSlant->"Italic"], "'s calculation of \n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi]\( E\^Sin[a\ x]\) Cos[a\ x] \[DifferentialD]x\)]], ":" }], "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x, a];\)\), "\n", \(MathematicaCalculation = \[Integral]\_0\%\[Pi]\( E\^Sin[a\ x]\ Cos[ a\ x]\) \[DifferentialD]x\)}], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(\(-\(1\/a\)\) + \[ExponentialE]\^Sin[a\ \[Pi]]\/a\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Use a transformation to explain where this result comes from.\ \>", "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell["Answer:", "Special1", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "The spotlight is on\n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi]\( E\^Sin[a\ x]\) Cos[a\ x] \[DifferentialD]x\)]], ". \nThe function\n ", Cell[BoxData[ \(a\ Cos[a\ x]\)]], " \nis the derivative of ", Cell[BoxData[ \(Sin[ax]\)]], "; so go with\n ", Cell[BoxData[ \(u[x] = Sin[a\ x]\)]], ".\nThe \"", Cell[BoxData[ \(a\ Cos[a\ x]\)]], "\" term you need is not immediately available; react by rewriting the \ integral as\n ", Cell[BoxData[ RowBox[{ RowBox[{"(", StyleBox[\(1\/a\), FontSize->16], ")"}], \(\[Integral]\_0\%\[Pi]\( E\^Sin[a\ x]\) a\ Cos[a\ x] \[DifferentialD]x\)}]]], "." }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "This is a legal step because \n", Cell[BoxData[ \(a\)]], " \nis a constant." }], "Special2", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "Now there is the \n ", Cell[BoxData[ \(a\ Cos[a\ x]\)]], " \nterm right where you want it.\nThis gives the pairings\n ", Cell[BoxData[ \(E\^Sin[a\ x]\)]], " <-------------------------> ", Cell[BoxData[ \(E\^u\)]], "\n ", Cell[BoxData[ \(a\ Cos[a\ x] \[DifferentialD]x = \(u'\)[x] \[DifferentialD]x\)]], " <---> ", Cell[BoxData[ \(\[DifferentialD]u\)]], "\n ", Cell[BoxData[ RowBox[{ RowBox[{"(", StyleBox[\(1\/a\), FontSize->16], ")"}], \(\[Integral]\_0\%\[Pi]\)}]]], " <---------------> ", Cell[BoxData[ RowBox[{ RowBox[{"(", StyleBox[\(1\/a\), FontSize->16], ")"}], \(\[Integral]\_0\%\(Sin[a\ \[Pi]]\)\)}]]], ".\nSo\n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi]\( E\^Sin[a\ x]\) Cos[a\ x] \[DifferentialD]x\)]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{ RowBox[{"(", StyleBox[\(1\/a\), FontSize->16], ")"}], \(\[Integral]\_0\%\[Pi]\( E\^Sin[a\ x]\) a\ Cos[a\ x] \[DifferentialD]x\)}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{ RowBox[{"(", StyleBox[\(1\/a\), FontSize->16], ")"}], \(\[Integral]\_0\%\(Sin[a\ \[Pi]]\)\(E\^u\) \[DifferentialD]u\)}]}]]], ".\nNow look for a function ", Cell[BoxData[ \(f[u]\)]], " whose derivative is ", Cell[BoxData[ \(E\^u\)]], ".\nOne such is:" }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[f, u];\)\), "\n", \(f[u_] = E\^u\)}], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(\[ExponentialE]\^u\)], "Output"] }, Open ]], Cell[TextData[{ "Now you know for sure that\n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi]\( E\^Sin[a\ x]\) Cos[a\ x] \[DifferentialD]x\)]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{ RowBox[{"(", StyleBox[\(1\/a\), FontSize->16], ")"}], \(\[Integral]\_0\%\(Sin[a\ \[Pi]]\)\(E\^u\) \[DifferentialD]u\)}]}]]], " is given by:" }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(\(f[Sin[a\ \[Pi]]] - f[0]\)\/a\)], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(\(\(-1\) + \[ExponentialE]\^Sin[a\ \[Pi]]\)\/a\)], "Output"] }, Open ]], Cell["Compare:", "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(Together[MathematicaCalculation]\)], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(\(\(-1\) + \[ExponentialE]\^Sin[a\ \[Pi]]\)\/a\)], "Output"] }, Open ]], Cell["Got it.", "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}] }, Closed]], Cell["B.1.e)", "Subsubsection", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "Here is ", StyleBox["Mathematica", FontSlant->"Italic"], "'s calculation of \n ", Cell[BoxData[ \(\[Integral]\_0\%1\(\@\( 1 - x\^2\)\) \[DifferentialD]x\)]], ": " }], "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x];\)\), "\n", \(MathematicaCalculation = \[Integral]\_0\%1\(\@\( 1 - x\^2\)\) \[DifferentialD]x\)}], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(\[Pi]\/4\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Use a transformation to explain where this result comes from.\ \>", "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell["Answer:", "Special1", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "The integral under study is \n ", Cell[BoxData[ \(\[Integral]\_0\%1\(\@\( 1 - x\^2\)\) \[DifferentialD]x\)]], ".\nMake the wild card substitution \n ", Cell[BoxData[ \(x = Sin[t]\)]], ". \nThis gives you the pairings\n ", Cell[BoxData[ \(\@\(1 - x\^2\)\)]], " <-------> ", Cell[BoxData[ \(\@\(1 - Sin[t]\^2\) = Cos[t]\)]], ";\n ", Cell[BoxData[ \(\[DifferentialD]x\)]], " <--------> ", Cell[BoxData[ \(Cos[t] \[DifferentialD]t\)]], ";\n ", Cell[BoxData[ \(\[Integral]\_0\%1\)]], " <--------> ", Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]\/2\)\)]], "\nbecause ", Cell[BoxData[ \(Sin[0] = 0\)]], " and ", Cell[BoxData[ RowBox[{ RowBox[{"Sin", "[", StyleBox[\(\[Pi]\/2\), FontSize->16], "]"}], "=", "1"}]]], ".\nNow you know that\n ", Cell[BoxData[ RowBox[{ RowBox[{\(\[Integral]\_0\%1\), RowBox[{\(\@\(1 - x\^2\)\), RowBox[{ StyleBox["\[DifferentialD]", FontColor->GrayLevel[0]], "x"}]}]}], "=", \(\[Integral]\_0\%\(\[Pi]\/2\)Cos[t] Cos[t] \[DifferentialD]t\)}]]], "\n ", Cell[BoxData[ \(\( = \[Integral]\_0\%\(\[Pi]\/2\)\(Cos[t]\^2\) \[DifferentialD]t\)\)]], ". \nTo calculate ", Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]\/2\)\(Cos[t]\^2\) \[DifferentialD]t\)]], ", apply a trig identity to ", Cell[BoxData[ \(Cos[t]\^2\)]], ":" }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[t];\)\), "\n", \(TrigReduce[Cos[t]\^2]\)}], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(1\/2\ \((1 + Cos[2\ t])\)\)], "Output"] }, Open ]], Cell[TextData[{ "Now you know that \n ", Cell[BoxData[ \(\[Integral]\_0\%1\(\@\( 1 - x\^2\)\) \[DifferentialD]x = \[Integral]\_0\%\(\[Pi]\/2\)\(Cos[t]\^2\) \[DifferentialD]t\)]], " \n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(\[Integral]\_0\%\(\[Pi]\/2\)\), RowBox[{ RowBox[{"(", RowBox[{ StyleBox[\(1\/2\), FontSize->16], "+", StyleBox[\(Cos[2 t]\/2\), FontSize->16]}], ")"}], \(\[DifferentialD]t\)}]}]}]]], ". \nTo complete the calculation, look for a function ", Cell[BoxData[ \(f[t]\)]], " with\n ", Cell[BoxData[ RowBox[{\(\(f'\)[t]\), "=", RowBox[{ StyleBox[\(1\/2\), FontSize->16], "+", StyleBox[\(Cos[2 t]\/2\), FontSize->16]}]}]]], ".\nHere is one such:" }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[BoxData[{ \(\(Clear[f, t];\)\), "\n", \(\(f[t_] = t\/2 + 1\/4\ Sin[2\ t];\)\)}], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[TextData[{ "Check whether ", Cell[BoxData[ RowBox[{\(\(f'\)[t]\), "=", RowBox[{ StyleBox[\(1\/2\), FontSize->16], "+", StyleBox[\(Cos[2 t]\/2\), FontSize->16]}]}]]], ":" }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(\(f'\)[t]\)], "Input"], Cell[BoxData[ \(1\/2 + 1\/2\ Cos[2\ t]\)], "Output"] }, Open ]], Cell[TextData[{ "Good; now you can say with confidence and considerable authority that\n \ ", Cell[BoxData[ RowBox[{ \(\[Integral]\_0\%1\(\@\( 1 - x\^2\)\) \[DifferentialD]x\), "=", RowBox[{\(\[Integral]\_0\%\(\[Pi]\/2\)\), RowBox[{ RowBox[{"(", RowBox[{ StyleBox[\(1\/2\), FontSize->16], "+", StyleBox[\(Cos[2 t]\/2\), FontSize->16]}], ")"}], \(\[DifferentialD]t\)}]}]}]]], " \nis given by:" }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(f[\[Pi]\/2] - f[0]\)], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(\[Pi]\/4\)], "Output"] }, Open ]], Cell["Compare:", "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(MathematicaCalculation\)], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(\[Pi]\/4\)], "Output"] }, Open ]], Cell["Got it.", "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}] }, Closed]], Cell["B.1.f)", "Subsubsection", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "Look at ", StyleBox["Mathematica", FontSlant->"Italic"], "'s calculation of \n ", Cell[BoxData[ \(\[Integral]\_3\%\(5\ \)\(E\^\(-\((\(x - 3\)\/2)\)\^2\)\) \[DifferentialD]x\)]], ":" }], "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x];\)\), "\n", \(\[Integral]\_3\%5\( E\^\(-\((\(x - 3\)\/2)\)\^2\)\) \[DifferentialD]x\)}], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(\@\[Pi]\ Erf[1]\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Remembering that ", Cell[BoxData[ \(Erf[x]\)]], " is defined by\n ", Cell[BoxData[ RowBox[{\(Erf[x]\), "=", RowBox[{ RowBox[{"(", StyleBox[\(2\/\@\[Pi]\), FontSize->16], ")"}], \(\[Integral]\_0\%x\( E\^\(-t\^2\)\) \[DifferentialD]t\)}]}]]], ", \nyou see that ", StyleBox["Mathematica", FontSlant->"Italic"], " is telling you that \n ", Cell[BoxData[ \(\[Integral]\_3\%5\( E\^\(-\((\(x - 3\)\/2)\)\^2\)\) \[DifferentialD]x\)]], "\ntransforms into\n ", Cell[BoxData[ RowBox[{\(\@\[Pi]\), RowBox[{"(", StyleBox[\(2\/\@\[Pi]\), FontSize->16], ")"}], \(\[Integral]\_0\%1\( E\^\(-t\^2\)\) \[DifferentialD]t\)}]]], "\n ", Cell[BoxData[ \(\( = 2 \(\[Integral]\_0\%1\( E\^\(-t\^2\)\) \[DifferentialD]t\)\)\)]], ". \nUse a substitution to explain why\n ", Cell[BoxData[ \(\[Integral]\_3\%5\( E\^\(-\((\(x - 3\)\/2)\)\^2\)\) \[DifferentialD]x = 2 \(\[Integral]\_0\%1\( E\^\(-t\^2\)\) \[DifferentialD]t\)\)]], "." }], "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell["Answer:", "Special1", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "The goal is to explain why\n ", Cell[BoxData[ \(\[Integral]\_3\%5\( E\^\(-\((\(x - 3\)\/2)\)\^2\)\) \[DifferentialD]x = 2 \(\[Integral]\_0\%1\( E\^\(-t\^2\)\) \[DifferentialD]t\)\)]], ".\nTo do this, take \n ", Cell[BoxData[ \(\[Integral]\_3\%5\( E\^\(-\((\(x - 3\)\/2)\)\^2\)\) \[DifferentialD]x\)]], "\nand make the substitution\n ", Cell[BoxData[ RowBox[{"t", "=", StyleBox[\(\((x - 3)\)\/2\), FontSize->16]}]]], ". \nThis gives\n ", Cell[BoxData[ RowBox[{\(\[DifferentialD]t\), "=", RowBox[{ RowBox[{"(", StyleBox[\(1\/2\), FontSize->16], ")"}], \(\[DifferentialD]x\)}]}]]], ",\nwhich is the same as\n ", Cell[BoxData[ \(2 \[DifferentialD]t = \[DifferentialD]x\)]], ".\nThis gives the pairings\n ", Cell[BoxData[ \(E\^\(-\((\(x - 3\)\/2)\)\^2\)\)]], " <--------------> ", Cell[BoxData[ \(E\^\(-t\^2\)\)]], ";\n ", Cell[BoxData[ \(\[DifferentialD]x\)]], " <----------------------> ", Cell[BoxData[ \(2 \[DifferentialD]t\)]], ";\n ", Cell[BoxData[ \(\[Integral]\_3\%5\)]], " <------------------> ", Cell[BoxData[ \(\[Integral]\_\(\(3 - 3\)\/2\)\%\(\(5 - 3\)\/2\)\( = \[Integral]\_0\%1\)\)]], ".\nThe upshot is\n ", Cell[BoxData[ \(\[Integral]\_3\%5\( E\^\(-\((\(x - 3\)\/2)\)\^2\)\) \[DifferentialD]x = \(\[Integral]\_0\%1\( E\^\(-t\^2\)\) 2 \[DifferentialD]t = 2 \(\[Integral]\_0\%1\( E\^\(-t\^2\)\) \[DifferentialD]t\)\)\)]], ".\nDone. " }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}] }, Closed]] }, Closed]], Cell["Tutorial Problems", "Subsubsection"], Cell[CellGroupData[{ Cell[TextData[{ "T.2) Transforming integrals to help understand ", StyleBox["Mathematica", FontSlant->"Italic"], " output" }], "Subsection", CellTags->"2.04.T2"], Cell["T.2.a)", "Subsubsection", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "In the course of your scientific endeavors, you find that you need to know \ the value of\n ", Cell[BoxData[ StyleBox[\(\[Integral]\_0\%15\( Sin[x\/3]\/x\) \[DifferentialD]x\), FontSize->16]]], "; \nso you ask ", StyleBox["Mathematica", FontSlant->"Italic"], ":" }], "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x];\)\), "\n", \(result = \[Integral]\_0\%15\( Sin[ x\/3]\/x\) \[DifferentialD]x\)}], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(SinIntegral[5]\)], "Output"] }, Open ]], Cell["You have no trouble getting the decimal result:", "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(N[result]\)], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(1.549931244944675`\)], "Output"] }, Open ]], Cell["And you can get it very, very accurately:", "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(N[result, 13]\)], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(1.549931244944675`\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "But how do you go about explaining to someone else the meaning of the \ original ", StyleBox["Mathematica", FontSlant->"Italic"], " output\n ", Cell[BoxData[ RowBox[{ StyleBox[\(\[Integral]\_0\%15\( Sin[x\/3]\/x\) \[DifferentialD]x\), FontSize->16], "=", \(SinIntegral[5]\)}]]], "?" }], "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell["Answer:", "Special1", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "You look in Stephen Wolfram's book ", StyleBox["Mathematica", FontSlant->"Italic"], ", Addison-Wesley, 1991 to see what the meaning of ", Cell[BoxData[ \(SinIntegral[x]\)]], " is. On page 868, you learn that\n ", Cell[BoxData[ RowBox[{\(SinIntegral[x]\), "=", StyleBox[\(\[Integral]\_0\%x\( Sin[t]\/t\) \[DifferentialD]t\), FontSize->16]}]]], ".\nThe ", StyleBox["Mathematica", FontSlant->"Italic"], " calculation\n ", Cell[BoxData[ RowBox[{ StyleBox[\(\[Integral]\_0\%15\( Sin[x\/3]\/x\) \[DifferentialD]x\), FontSize->16], "=", \(SinIntegral[5]\)}]]], " \nmeans no more or less than\n ", Cell[BoxData[ RowBox[{ StyleBox[\(\[Integral]\_0\%15\( Sin[x\/3]\/x\) \[DifferentialD]x\), FontSize->16], "=", StyleBox[\(\[Integral]\_0\%5\( Sin[t]\/t\) \[DifferentialD]t\), FontSize->16]}]]], ".\nYou can explain this with a transformation.\nStart with\n ", Cell[BoxData[ StyleBox[\(\[Integral]\_0\%15\( Sin[x\/3]\/x\) \[DifferentialD]x\), FontSize->16]]], " .\nMake the substitution\n ", Cell[BoxData[ RowBox[{"t", "=", StyleBox[\(x\/3\), FontSize->16]}]]], " (i.e., ", Cell[BoxData[ \(x = 3 t\)]], ").\nThis gives you the pairings\n ", Cell[BoxData[ StyleBox[\(Sin[x\/3]\/x\), FontSize->16]]], " <--------> ", Cell[BoxData[ StyleBox[\(Sin[t]\/\(3 t\)\), FontSize->16]]], ";\n ", Cell[BoxData[ \(\[DifferentialD]x\)]], " <----------------> ", Cell[BoxData[ \(3 \[DifferentialD]t\)]], ";\n ", Cell[BoxData[ \(\[Integral]\_0\%15\)]], " <---------------> ", Cell[BoxData[ \(\[Integral]\_0\%5\)]], "\nbecause ", Cell[BoxData[ \(t = 0\)]], " when ", Cell[BoxData[ \(x = 0\)]], " and ", Cell[BoxData[ \(t = 5\)]], " when ", Cell[BoxData[ \(x = 15\)]], ". \nSo:\n ", Cell[BoxData[ RowBox[{ StyleBox[\(\[Integral]\_0\%15\( Sin[x\/3]\/x\) \[DifferentialD]x\), FontSize->16], "=", RowBox[{\(\[Integral]\_0\%5\), RowBox[{ StyleBox[\(Sin[t]\/\(3 t\)\), FontSize->16], "3", \(\[DifferentialD]t\)}]}]}]]], " \n ", Cell[BoxData[ RowBox[{"=", StyleBox[\(\[Integral]\_0\%5\( Sin[t]\/t\) \[DifferentialD]t\), FontSize->16]}]]], " \n ", Cell[BoxData[ \(\(\(=\)\(SinIntegral[5]\)\)\)]], ".\nExplanation complete." }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}] }, Open ]], Cell["T.2.b)", "Subsubsection", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "Another time, you find that you need to know the value of\n ", Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]\/2\)\(E\^\(-\((\(Sin[x] - \ 1\/2\)\/6)\)\^2\)\ \) Cos[x] \[DifferentialD]x\)]], "; \nso you ask ", StyleBox["Mathematica", FontSlant->"Italic"], ":" }], "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x];\)\), "\n", \(result = \[Integral]\_0\%\(\[Pi]\/2\)\(E\^\(-\((1\/6\ \((Sin[x] - 1\/2)\ \))\)\^2\)\ Cos[x]\) \[DifferentialD]x\)}], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(6\ \@\[Pi]\ Erf[1\/12]\)], "Output"] }, Open ]], Cell["You have no trouble getting the decimal result:", "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(N[result]\)], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(0.9976899997530428`\)], "Output"] }, Open ]], Cell["And you can get this with great accuracy:", "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(N[result, 25]\)], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(0.99768999975304284874324092849467913618`25\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "But how do you go about explaining to someone else the meaning of the \ original ", StyleBox["Mathematica", FontSlant->"Italic"], " output\n ", Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]\/2\)\(E\^\(-\((\(Sin[x] - \ 1\/2\)\/6)\)\^2\)\ \) Cos[x] \[DifferentialD]x = 6 \(\@ \[Pi]\) Erf[1\/12]\)]], "\nand where it comes from?" }], "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell["Answer:", "Special1", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "You look elsewhere in this lesson or in Stephen Wolfram's book ", StyleBox["Mathematica", FontSlant->"Italic"], ", Addison-Wesley, 1991 to see what the meaning of ", Cell[BoxData[ \(Erf[x]\)]], " is. On page 787 of that book, you learn that\n ", Cell[BoxData[ RowBox[{\(Erf[x]\), "=", RowBox[{ RowBox[{"(", StyleBox[\(2\/\@\[Pi]\), FontSize->16], ")"}], \(\[Integral]\_0\%x\( E\^\(-t\^2\)\) \[DifferentialD]t\)}]}]]], "\nThe ", StyleBox["Mathematica", FontSlant->"Italic"], " calculation\n ", Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]\/2\)\(E\^\(-\((\(Sin[x] - \ 1\/2\)\/6)\)\^2\)\ \) Cos[x] \[DifferentialD]x = 6 \(\@ \[Pi]\) Erf[1\/12]\)]], "\nmeans no more or less than \n ", Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]\/2\)\(E\^\(-\((\(Sin[x] - \ 1\/2\)\/6)\)\^2\)\ \) Cos[x] \[DifferentialD]x\)]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{"6", \(\@\[Pi]\), RowBox[{"(", StyleBox[\(2\/\@\[Pi]\), FontSize->16], ")"}], \(\[Integral]\_0\%\(1\/12\)\(E\^\(-t\^2\)\) \ \[DifferentialD]t\)}]}]]], "\n ", Cell[BoxData[ \(\(\(=\)\(12 \(\[Integral]\_0\%\(1\/12\)\(E\^\(-t\^2\)\) \ \[DifferentialD]t\)\)\)\)]], "\nYou can explain why \n ", Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]\/2\)\(E\^\(-\((\(Sin[x] - \ 1\/2\)\/6)\)\^2\)\ \) Cos[x] \[DifferentialD]x = 12 \(\[Integral]\_0\%\(1\/12\)\(E\^\(-t\^2\)\) \ \[DifferentialD]t\)\)]], "\nwith a transformation.\nStart with \n ", Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]\/2\)\(E\^\(-\((\(Sin[x] - \ 1\/2\)\/6)\)\^2\)\ \) Cos[x] \[DifferentialD]x\)]], "\nMake the substitution\n ", Cell[BoxData[ RowBox[{"t", "=", StyleBox[\(\((Sin[x] - 1\/2)\)\/6\), FontSize->16]}]]], ".\nThis gives you the pairings\n ", Cell[BoxData[ \(E\^\(-\((\(Sin[x] - \ 1\/2\)\/6)\)\^2\)\)]], " <-------> ", Cell[BoxData[ \(E\^\(-t\^2\)\)]], ";\n ", Cell[BoxData[ \(Cos[x] \[DifferentialD]x\)]], " <-----------> ", Cell[BoxData[ \(6 \[DifferentialD]t\)]], ";\n ", Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]\/2\)\)]], " <--------------------> ", Cell[BoxData[ \(\[Integral]\_\(-\(1\/12\)\)\%\(1\/12\)\)]], " \nbecause ", Cell[BoxData[ \(t = \(-\(1\/12\)\)\)]], " when ", Cell[BoxData[ \(x = 0\)]], " and ", Cell[BoxData[ \(t = 1\/12\)]], " when ", Cell[BoxData[ \(x = \[Pi]\/2\)]], ". \nSo\n ", Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]\/2\)\(E\^\(-\((\(Sin[x] - \ 1\/2\)\/6)\)\^2\)\ \) Cos[x] \[DifferentialD]x\)]], "\n ", Cell[BoxData[ \(\(\(=\)\(\[Integral]\_\(-\(1\/12\)\)\%\(1\/12\)\(E\^\(-t\^2\)\) 6 \[DifferentialD]t\)\)\)]], "\n ", Cell[BoxData[ \(\(\(=\)\(6 \(\[Integral]\_\(-\(1\/12\)\)\%\(1\/12\)\(E\^\(-t\^2\)\) \ \[DifferentialD]t\)\)\)\)]], "\n ", Cell[BoxData[ \(\(\(=\)\(12 \(\[Integral]\_0\%\(1\/12\)\(E\^\(-t\^2\)\) \ \[DifferentialD]t\)\)\)\)]], "\n because ", Cell[BoxData[ \(E\^\(-t\^2\)\)]], " is an even function:" }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[t];\)\), "\n", \(\(Plot[E\^\(-t\^2\), {t, \(-\(1\/12\)\), 1\/12}, PlotStyle \[Rule] {{Blue, Thickness[0.01]}}, AspectRatio \[Rule] 1\/GoldenRatio, AxesLabel \[Rule] {"\", "\"}];\)\)}], "Input", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica 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The negative sign comes from the fact that ", Cell[BoxData[ \(y\)]], " is negative on the bottom part of the curve.\nConsequently, when you \ calculate\n ", Cell[BoxData[{ \(\[Integral]\_tlow\%thigh y[t] \(x'\)[t] \[DifferentialD]t\), "\n", \(\ \ \ \ \ \ \ \ \ \ \(\(=\)\(\[Integral]\_tlow\%trev y[t] \(x'\)[ t] \[DifferentialD]t + \[Integral]\_trev\%thigh y[t] \(x'\)[ t] \[DifferentialD]t\)\)\)}]], ",\nyou get the negative of the area enclosed by the entire curve." }], "SmallText", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}] }, Closed]], Cell["T.3.a.iii)", "Subsubsection", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell["\<\ Does this work for other closed curves as well? What do you have to watch out for?\ \>", "Text", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell["Answer:", "Special1", CellMargins->{{Inherited, 111}, {Inherited, Inherited}}], Cell[TextData[{ "Yes, it does work for other closed curves. \nIf you have a closed curve \ given in parametric form by\n ", Cell[BoxData[ \({x[t], y[t]}\)]], " with ", Cell[BoxData[ \(tlow \[LessEqual] t \[LessEqual] thigh\)]], ", \nthen\n ", Cell[BoxData[ \(\[Integral]\_tlow\%thigh y[t] \(x'\)[t] \[DifferentialD]t\)]], "\nmeasures the negative of the area enclosed by the curve provided that \ the curve is traced out in the counterclockwise direction as ", Cell[BoxData[ \(t\)]], " advances from ", Cell[BoxData[ \(tlow\)]], " to ", Cell[BoxData[ \(thigh\)]], ".\nOn the other hand,\n ", Cell[BoxData[ \(\[Integral]\_tlow\%thigh y[t] \(x'\)[t] \[DifferentialD]t\)]], "\nmeasures the actual area enclosed by the curve provided that the curve \ is traced out in the clockwise direction as ", Cell[BoxData[ 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Data collections show that the amounts it actually dispenses are \ normally distributed with ", Cell[BoxData[ \(average = c\)]], " and ", Cell[BoxData[ \(standard\ deviation = 0.35\)]], " ounces.\nIf you set the machine to dispense exactly ", Cell[BoxData[ \(7\)]], " ounces, then you expect the machine to overflow a ", Cell[BoxData[ \(7\)]], "-ounce cup half the time. If you don't believe this, look at\n ", Cell[BoxData[ \(\[Integral]\_7\%\[Infinity] normal[x, 7, 0.35] \[DifferentialD]x\)]], ":" }], "Text", CellMargins->{{Inherited, 112}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x, bell];\)\), "\n", \(\(bell[x_] = E\^\(-\(x\^2\/2\)\)\/\@\(2\ \[Pi]\);\)\n\), "\n", \(\(Clear[normal, mean, dev];\)\), "\n", \(\(normal[x_, mean_, dev_] = bell[\(x - mean\)\/dev]\/dev;\)\n\), "\n", \(\(mean = 7;\)\), "\n", \(\(dev = 0.35;\)\), "\n", \(NIntegrate[normal[x, mean, dev], {x, mean, \[Infinity]}]\)}], "Input", CellMargins->{{Inherited, 112}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[BoxData[ \(0.49999999999998895`\)], "Output"] }, Open ]], Cell[TextData[{ "If you set the machine to dispense ", Cell[BoxData[ \(6.65\)]], " ounces, how often do you expect the machine to overflow a ", Cell[BoxData[ \(7\)]], "-ounce cup?" }], "Text", CellMargins->{{Inherited, 112}, {Inherited, Inherited}}], Cell["G.5.c.ii)", 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