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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[ 67835, 2423]*) (*NotebookOutlinePosition[ 113259, 3997]*) (* CellTagsIndexPosition[ 113097, 3988]*) (*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[{ "2.06 More Tools and Measurements:\nTechniques for Calculating 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[CellGroupData[{ Cell["B.2) Integration by parts", "Subsection", CellTags->"2.06.B2"], Cell["B.2.a) The integration by parts formula", "Subsubsection"], Cell[CellGroupData[{ Cell[TextData[{ "Use the product rule for differentiation to explain why\n ", Cell[BoxData[ RowBox[{\(\[Integral]\_a\%b u[x] \(v'\)[x] \[DifferentialD]x\), "=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "a", "b"], \(-\(\[Integral]\_a\%b v[x] \(u'\)[x] \[DifferentialD]x\)\)}]}]]], ".\nThis is called the integration by parts formula.\nThis formula can be \ especially handy for hand calculation of integrals." }], "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "The product rule says\n ", Cell[BoxData[ RowBox[{ StyleBox[\(\[DifferentialD]\((u[x] v[x])\)\/\[DifferentialD]x\), FontSize->16], "=", \(u[x] \(v'\)[x] + \(u'\)[x] v[x]\)}]]], ".\nIntegrating this from ", Cell[BoxData[ \(x = a\)]], " to ", Cell[BoxData[ \(x = b\)]], " gives:\n ", Cell[BoxData[ RowBox[{\(\[Integral]\_a\%b\), RowBox[{ RowBox[{"(", StyleBox[\(\[DifferentialD]\((u[x] v[x])\)\/\[DifferentialD]x\), FontSize->16], ")"}], \(\[DifferentialD]x\)}]}]]], "\n ", Cell[BoxData[ \(\( = \[Integral]\_a\%b u[x] \(v'\)[x] \[DifferentialD]x + \[Integral]\_a\%b\( u'\)[x] v[x] \[DifferentialD]x\)\)]], ".\nThis is the same as:\n ", Cell[BoxData[ RowBox[{ RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "a", "b"]}], "=", \(\[Integral]\_a\%b u[x] \(v'\)[x] \[DifferentialD]x + \[Integral]\_a\%b\( u'\)[x] v[x] \[DifferentialD]x\)}]]], ".\nRearranging gives\n ", Cell[BoxData[ RowBox[{\(\[Integral]\_a\%b u[x] \(v'\)[x] \[DifferentialD]x\), "=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "a", "b"], \(-\(\[Integral]\_a\%b v[x] \(u'\)[x] \[DifferentialD]x\)\)}]}]]], ",\nand this explains why the advertised formula holds up." }], "SmallText"] }, Closed]], Cell["B.2.b.i)", "Subsubsection"], Cell[CellGroupData[{ Cell[TextData[{ "Use the integration by parts formula to calculate\n ", Cell[BoxData[ \(\[Integral]\_0\%1 x\ \(E\^x\) \[DifferentialD]x\)]], "." }], "Text"], Cell["Answer:", "Subsubsection"], Cell[TextData[{ "The integration by parts formula is\n ", Cell[BoxData[ RowBox[{\(\[Integral]\_a\%b u[x] \(v'\)[x] \[DifferentialD]x\), "=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "a", "b"], \(-\(\[Integral]\_a\%b v[x] \(u'\)[x] \[DifferentialD]x\)\)}]}]]], ".\nMake the assignments:\n ", Cell[BoxData[ \(u[x] = x\)]], " and ", Cell[BoxData[ \(\(v'\)[x] = E\^x\)]], ".\nThis gives\n ", Cell[BoxData[ \(\(u'\)[x] = 1\)]], " and ", Cell[BoxData[ \(v[x] = E\^x\)]], ".\nWith these assignments,\n ", Cell[BoxData[ \(\[Integral]\_0\%1 x\ \(E\^x\) \[DifferentialD]x\)]], "\n ", Cell[BoxData[ \(\( = \[Integral]\_0\%1 u[x] \(v'\)[x] \[DifferentialD]x\)\)]], " \n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "1"], \(-\(\[Integral]\_0\%1 v[x] \(u'\)[x] \[DifferentialD]x\)\)}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(x\ E\^x\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "1"], \(-\(\[Integral]\_0\%1\( E\^x\) 1 \[DifferentialD]x\)\)}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(x\ E\^x\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "1"], \(-\(\[Integral]\_0\%1\( E\^x\) \[DifferentialD]x\)\)}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(x\ E\^x\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "1"], \(-E\^x\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "1"]}]}]]], "\n ", Cell[BoxData[ \(\( = \((E - 0)\) - \((E - 1)\)\)\)]], "\n ", Cell[BoxData[ \(\( = 1\)\)]], ".\nCheck:" }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[ \(NIntegrate[x\ E\^x, {x, 0, 1}]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(1.`\)], "Output"] }, Open ]], Cell["Got it.", "SmallText"] }, Closed]], Cell["B.2.b.ii)", "Subsubsection"], Cell[CellGroupData[{ Cell[TextData[{ "The integration by parts formula is\n ", Cell[BoxData[ RowBox[{\(\[Integral]\_a\%b u[x] \(v'\)[x] \[DifferentialD]x\), "=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "a", "b"], \(-\(\[Integral]\_a\%b v[x] \(u'\)[x] \[DifferentialD]x\)\)}]}]]], ".\nWhen this formula was used to help calculate\n ", Cell[BoxData[ \(\[Integral]\_0\%1 x\ \(E\^x\) \[DifferentialD]x\)]], " \nabove, the assignments\n ", Cell[BoxData[ \(u[x] = x\)]], " and ", Cell[BoxData[ \(\(v'\)[x] = E\^x\)]], "\nwere made. \nDo you get into deep do-do if you make the assignments\n \ ", Cell[BoxData[ \(u[x] = E\^x\)]], " and ", Cell[BoxData[ \(\(v'\)[x] = x\)]], "? " }], "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "Try it and see.\nThe integration by parts formula is\n ", Cell[BoxData[ RowBox[{\(\[Integral]\_a\%b u[x] \(v'\)[x] \[DifferentialD]x\), "=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "a", "b"], \(-\(\[Integral]\_a\%b v[x] \(u'\)[x] \[DifferentialD]x\)\)}]}]]], ".\nMake the assignments:\n ", Cell[BoxData[ \(u[x] = E\^x\)]], " and ", Cell[BoxData[ \(\(v'\)[x] = x\)]], ".\nThis gives\n ", Cell[BoxData[ \(\(u'\)[x] = E\^x\)]], " and ", Cell[BoxData[ RowBox[{\(v[x]\), "=", StyleBox[\(x\^2\/2\), FontSize->16]}]]], ".\nWith these assignments,\n ", Cell[BoxData[ \(\[Integral]\_0\%1 x\ \(E\^x\) \[DifferentialD]x\)]], "\n ", Cell[BoxData[ \(\( = \[Integral]\_0\%1 u[x] \(v'\)[x] \[DifferentialD]x\)\)]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "1"], \(-\(\[Integral]\_0\%1 v[x] \(u'\)[x] \[DifferentialD]x\)\)}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{ RowBox[{\(E\^x\), RowBox[{"(", StyleBox[\(x\^2\/2\), FontSize->16], ")"}]}], SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "1"], RowBox[{"-", RowBox[{\(\[Integral]\_0\%1\), RowBox[{ RowBox[{"(", StyleBox[\(x\^2\/2\), FontSize->16], ")"}], \(E\^x\), \(\[DifferentialD]x\)}]}]}]}]}]]], ".\nThis puts you in deep do-do because now you have to calculate\n", Cell[BoxData[ RowBox[{\(\[Integral]\_0\%1\), RowBox[{ RowBox[{"(", StyleBox[\(x\^2\/2\), FontSize->16], ")"}], \(E\^x\), \(\[DifferentialD]x\)}]}]]], ", and this integral is harder to calculate by hand than the original \ integral ", Cell[BoxData[ \(\[Integral]\_0\%1 x\ \(E\^x\) \[DifferentialD]x\)]], "." }], "SmallText"] }, Closed]], Cell["B.2.b.iii)", "Subsubsection"], Cell[CellGroupData[{ Cell["What moral is suggested by part ii)?", "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "When you use the integration by parts formula\n ", Cell[BoxData[ RowBox[{\(\[Integral]\_a\%b u[x] \(v'\)[x] \[DifferentialD]x\), "=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "a", "b"], \(-\(\[Integral]\_a\%b v[x] \(u'\)[x] \[DifferentialD]x\)\)}]}]]], ",\nyou are free to assign ", Cell[BoxData[ \(u[x]\)]], " and ", Cell[BoxData[ \(\(v'\)[x]\)]], ". \nGenerally, you will want to make the assignments so that the \ integral on the right-hand side is easier to calculate than the integral on \ the left." }], "SmallText"] }, Closed]], Cell["B.2.c.i) Making the assignments", "Subsubsection"], Cell[CellGroupData[{ Cell[TextData[{ "Use the integration by parts formula to help to calculate\n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi] x\ Cos[x] \[DifferentialD]x\)]], "." }], "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "The integration by parts formula is\n ", Cell[BoxData[ RowBox[{\(\[Integral]\_a\%b u[x] \(v'\)[x] \[DifferentialD]x\), "=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "a", "b"], \(-\(\[Integral]\_a\%b v[x] \(u'\)[x] \[DifferentialD]x\)\)}]}]]], ".\nYou have your choice of the assignments:\n\n", StyleBox["\[RightArrow] Choice 1:", FontWeight->"Bold"], "\n ", Cell[BoxData[ \(u[x] = Cos[x]\)]], " and ", Cell[BoxData[ \(\(v'\)[x] = x\)]], ".\nThis gives\n ", Cell[BoxData[ \(\(u'\)[x] = \(-Sin[x]\)\)]], " and ", Cell[BoxData[ RowBox[{\(v[x]\), "=", StyleBox[\(x\^2\/2\), FontSize->16]}]]], ",\nand\n ", Cell[BoxData[ RowBox[{\(v[x] \(u'\)[x]\), "=", " ", RowBox[{ RowBox[{"-", RowBox[{"(", StyleBox[\(x\^2\/2\), FontSize->16], ")"}]}], \(Sin[x]\)}]}]]], ".\n\n", StyleBox["\[RightArrow] Choice 2:", FontWeight->"Bold"], "\n ", Cell[BoxData[ \(u[x] = x\)]], " and ", Cell[BoxData[ \(\(v'\)[x] = Cos[x]\)]], ".\nThis gives \n ", Cell[BoxData[ \(\(u'\)[x] = 1\)]], " and ", Cell[BoxData[ \(v[x] = Sin[x]\)]], "\nand\n ", Cell[BoxData[ \(v[x] \(u'\)[x] = Sin[x]\)]], ".\n\nBecause ", Cell[BoxData[ \(v[x] \(u'\)[x]\)]], " is simpler with Choice 2 than it is with Choice 1, Choice 2 is the way to \ go.\nHere is Choice 2 again:\n ", Cell[BoxData[ \(u[x] = x\)]], " and ", Cell[BoxData[ \(\(v'\)[x] = Cos[x]\)]], ".\nThis gives\n ", Cell[BoxData[ \(\(u'\)[x] = 1\)]], " and ", Cell[BoxData[ \(v[x] = Sin[x]\)]], ".\nPlug into the integration by parts formula to get \n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi] x\ Cos[x] \[DifferentialD]x\)]], "\n ", Cell[BoxData[ \(\( = \[Integral]\_0\%\[Pi] u[x] \(v'\)[x] \[DifferentialD]x\)\)]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "\[Pi]"], \(-\(\[Integral]\_0\%\[Pi] v[x] \(u'\)[x] \[DifferentialD]x\)\)}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(x\ Sin[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "\[Pi]"], \(-\(\[Integral]\_0\%\[Pi] Sin[x] \[DifferentialD]x\)\)}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(x\ Sin[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "\[Pi]"], \(-\((\(-Cos[x]\))\)\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "\[Pi]"]}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(x\ Sin[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "\[Pi]"], \(+\((Cos[x])\)\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "\[Pi]"]}]}]]], " \n ", Cell[BoxData[ \(\( = \((0 - 0)\) + \((\(-1\) - 1)\)\) = \(-2\)\)]], ".\nCheck:" }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[ \(NIntegrate[x\ Cos[x], {x, 0, \[Pi]}]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\(-1.9999999999999998`\)\)], "Output"] }, Open ]], Cell["Nailed it.", "SmallText"] }, Closed]], Cell["B.2.c.ii) Making the assignments", "Subsubsection"], Cell[CellGroupData[{ Cell[TextData[{ "Use the integration by parts formula to help calculate\n ", Cell[BoxData[ \(\[Integral]\_1\%t Log[x] \[DifferentialD]x\)]], "." }], "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "The integration by parts formula is\n ", Cell[BoxData[ RowBox[{\(\[Integral]\_a\%b u[x] \(v'\)[x] \[DifferentialD]x\), "=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "a", "b"], \(-\(\[Integral]\_a\%b v[x] \(u'\)[x] \[DifferentialD]x\)\)}]}]]], ".\nRewrite the integral as\n ", Cell[BoxData[ \(\[Integral]\_1\%t Log[x] \[DifferentialD]x = \[Integral]\_1\%t Log[x] 1 \[DifferentialD]x\)]], ".\n You have your choice of the assignments:\n\n", StyleBox["\[RightArrow] Choice 1:", FontWeight->"Bold"], "\n ", Cell[BoxData[ \(u[x] = Log[x]\)]], " and ", Cell[BoxData[ \(\(v'\)[x] = 1\)]], ".\nThis gives\n ", Cell[BoxData[ RowBox[{\(\(u'\)[x]\), "=", StyleBox[\(1\/x\), FontSize->16]}]]], " and ", Cell[BoxData[ \(v[x] = x\)]], "\nand\n ", Cell[BoxData[ RowBox[{\(v[x] \(u'\)[x]\), "=", RowBox[{ RowBox[{"x", RowBox[{"(", StyleBox[\(1\/x\), FontSize->16], ")"}]}], "=", "1"}]}]]], ".\n\n", StyleBox["\[RightArrow] Choice 2:", FontWeight->"Bold"], "\n ", Cell[BoxData[ \(u[x] = 1\)]], " and ", Cell[BoxData[ \(\(v'\)[x] = Log[x]\)]], ".\nThis gives\n ", Cell[BoxData[ \(\(u'\)[x] = 0\)]], " and ", Cell[BoxData[ \(\(v[x] = \)\)]], " (choke and punt).\n\nChoice 2 stinks because it's hard to come up with ", Cell[BoxData[ \(v[x]\)]], ". In fact, finding ", Cell[BoxData[ \(v[x]\)]], " is where this thing started.\nSo Choice 1 is the way to go.\nHere is \ Choice 1 again:\n ", Cell[BoxData[ \(u[x] = Log[x]\)]], " and ", Cell[BoxData[ \(\(v'\)[x] = 1\)]], ".\nThis gives\n ", Cell[BoxData[ RowBox[{\(\(u'\)[x]\), "=", StyleBox[\(1\/x\), FontSize->16]}]]], " and ", Cell[BoxData[ \(v[x] = x\)]], ".\nPlug into the integration by parts formula to get\n ", Cell[BoxData[ \(\[Integral]\_1\%t Log[x] 1 \[DifferentialD]x\)]], "\n ", Cell[BoxData[ \(\( = \[Integral]\_1\%t u[x] \(v'\)[x] \[DifferentialD]x\)\)]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "1", "t"], \(-\(\[Integral]\_1\%t v[x] \(u'\)[x] \[DifferentialD]x\)\)}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(Log[x] x\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "1", "t"], RowBox[{"-", RowBox[{\(\[Integral]\_1\%t\), RowBox[{"x", RowBox[{"(", StyleBox[\(1\/x\), FontSize->16], ")"}], \(\[DifferentialD]x\)}]}]}]}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(Log[x] x\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "1", "t"], \(-\(\[Integral]\_1\%t 1 \[DifferentialD]x\)\)}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(Log[x] x\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "1", "t"], \(-x\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "1", "t"]}]}]]], "\n ", Cell[BoxData[ \(\( = \((t\ Log[t] - 0)\) - \((t - 1)\)\)\)]], "\n ", Cell[BoxData[ \(\( = t\ Log[t]\) - t + 1\)]], ".\nCheck:" }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x, t];\)\), "\n", \(integral = \[Integral]\_1\%t Log[x] \[DifferentialD]x\)}], "Input", AspectRatioFixed->True], Cell[BoxData[ \(1 - t + t\ Log[t]\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Expand[integral]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(1 - t + t\ Log[t]\)], "Output"] }, Open ]], Cell["Nailed it cold.", "SmallText"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "B.5) The technique of calculating integrals by taking derivatives: \n \ Calculating ", Cell[BoxData[ \(\[Integral]\_0\%t\( x\^n\ E\^\(s\ x\)\) \[DifferentialD]x\)]], " and ", Cell[BoxData[ \(\[Integral]\_0\%t\( x\^n\) Sin[s\ x]\ \[DifferentialD]x\)]], " with minimum hand labor" }], "Subsection", CellTags->"2.06.B5"], Cell["\<\ The technique you will learn here is used heavily by physicists and \ some engineers. For some reason, it appears in very few standard calculus books.\ \>", "Special5"], Cell[TextData[{ "Put\n ", Cell[BoxData[ \(f[s] = \[Integral]\_0\%t\( E\^\(s\ x\)\) \[DifferentialD]x\)]], " \n and look at these:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[f, s];\)\), "\n", \(f[s_] = \[Integral]\_0\%t\( E\^\(s\ x\)\) \[DifferentialD]x\)}], "Input"], Cell[BoxData[ \(\(-\(1\/s\)\) + \[ExponentialE]\^\(s\ t\)\/s\)], "Output"] }, Open ]], Cell[TextData[{ "Compare the first derivative ", Cell[BoxData[ RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "[", "s", "]"}]]], " and ", Cell[BoxData[ \(\[Integral]\_0\%t\( x\ E\^\(s\ x\)\) \[DifferentialD]x\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({Together[\(f'\)[s]], Together[\[Integral]\_0\%t\( x\ E\^\(s\ x\)\) \[DifferentialD]x]}\)], "Input"], Cell[BoxData[ \({\(1 - \[ExponentialE]\^\(s\ t\) + \[ExponentialE]\^\(s\ t\)\ s\ t\)\/s\ \^2, \(1 - \[ExponentialE]\^\(s\ t\) + \[ExponentialE]\^\(s\ t\)\ s\ \ t\)\/s\^2}\)], "Output"] }, Open ]], Cell[TextData[{ "Compare the second derivative ", Cell[BoxData[ RowBox[{ SuperscriptBox["f", "\[DoublePrime]", MultilineFunction->None], "[", "s", "]"}]]], " and ", Cell[BoxData[ \(\[Integral]\_0\%t\( x\^2\ E\^\(s\ x\)\) \[DifferentialD]x\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({Together[\(\(f'\)'\)[s]], Together[\[Integral]\_0\%t\( x\^2\ E\^\(s\ x\)\) \ \[DifferentialD]x]}\)], "Input"], Cell[BoxData[ \({\(\(-2\) + 2\ \[ExponentialE]\^\(s\ t\) - 2\ \[ExponentialE]\^\(s\ t\)\ \ s\ t + \[ExponentialE]\^\(s\ t\)\ s\^2\ t\^2\)\/s\^3, \(\(-2\) + 2\ \ \[ExponentialE]\^\(s\ t\) - 2\ \[ExponentialE]\^\(s\ t\)\ s\ t + \ \[ExponentialE]\^\(s\ t\)\ s\^2\ t\^2\)\/s\^3}\)], "Output"] }, Open ]], Cell[TextData[{ "Compare the third derivative ", Cell[BoxData[ RowBox[{ SuperscriptBox["f", TagBox[\((3)\), Derivative], MultilineFunction->None], "[", "s", "]"}]]], " and ", Cell[BoxData[ \(\[Integral]\_0\%t\( x\^3\ E\^\(s\ x\)\) \[DifferentialD]x\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({Together[\(\(\(f'\)'\)'\)[s]], Together[\[Integral]\_0\%t\( x\^3\ E\^\(s\ x\)\) \ \[DifferentialD]x]}\)], "Input"], Cell[BoxData[ \({\(6 - 6\ \[ExponentialE]\^\(s\ t\) + 6\ \[ExponentialE]\^\(s\ t\)\ s\ \ t - 3\ \[ExponentialE]\^\(s\ t\)\ s\^2\ t\^2 + \[ExponentialE]\^\(s\ t\)\ \ s\^3\ t\^3\)\/s\^4, \(6 - 6\ \[ExponentialE]\^\(s\ t\) + 6\ \[ExponentialE]\^\ \(s\ t\)\ s\ t - 3\ \[ExponentialE]\^\(s\ t\)\ s\^2\ t\^2 + \[ExponentialE]\^\ \(s\ t\)\ s\^3\ t\^3\)\/s\^4}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Evidently you can calculate\n ", Cell[BoxData[ \(\[Integral]\_0\%t\( x\^k\ E\^\(s\ x\)\) \[DifferentialD]x\)]], " \nby going with\n ", Cell[BoxData[ \(f[s] = \[Integral]\_0\%t\ \(E\^\(s\ x\)\) \[DifferentialD]x\)]], "\nand calculating the kth derivative ", Cell[BoxData[ RowBox[{ SuperscriptBox["f", TagBox[\((k)\), Derivative], MultilineFunction->None], "[", "s", "]"}]]], "\nwith the full assurance that\n ", Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["f", TagBox[\((k)\), Derivative], MultilineFunction->None], "[", "s", "]"}], " ", "=", " ", \(\[Integral]\_0\%t\( x\^k\ E\^\(k\ x\)\) \(\[DifferentialD]x . \)\)}], " "}]]], "\nExplain what's going on.." }], "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "When you go with\n ", Cell[BoxData[ \(f[s] = \[Integral]\_0\%t\ \(E\^\(s\ x\)\) \[DifferentialD]x\)]], ",\nand take the derivative with respect to ", Cell[BoxData[ \(s\)]], ", you get\n ", Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "[", "s", "]"}], "=", \(\[Integral]\_0\%t\ x\ \(E\^\(s\ x\)\) \[DifferentialD]x\)}]]], "." }], "SmallText"], Cell[CellGroupData[{ Cell["\<\ If you want to see the reason this works, click on the right.\ \>", "Special5"], Cell[TextData[{ "Reason: The derivative of ", Cell[BoxData[ \(E\^\(s\ x\)\)]], " with respect to ", Cell[BoxData[ \(u\)]], " is ", Cell[BoxData[ \(x\ E\^\(s\ x\)\)]], ". " }], "Text"] }, Closed]], Cell[TextData[{ "When you take the second derivative, you get\n ", Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox["f", "\[DoublePrime]", MultilineFunction->None], "[", "s", "]"}], "=", \(\[Integral]\_0\%t\( x\^2\ E\^\(s\ x\)\) \[DifferentialD]x\)}]]] }], "SmallText"], Cell[CellGroupData[{ Cell["\<\ If you want to see the reason this works, click on the right.\ \>", "Special5"], Cell[TextData[{ "Reason: The derivative of ", Cell[BoxData[ \(x\ E\^\(s\ x\)\)]], " with respect to ", Cell[BoxData[ \(u\)]], " is ", Cell[BoxData[ \(x\^2\ E\^\(s\ x\)\)]], ". " }], "Text"] }, Closed]], Cell[TextData[{ "When you take the ", Cell[BoxData[ \(k\^th\)]], " derivative, you get\n ", Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox["f", TagBox[\((k)\), Derivative], MultilineFunction->None], "[", "s", "]"}], "=", \(\[Integral]\_0\%t\( x\^k\ E\^\(s\ x\)\) \[DifferentialD]x\)}]]] }], "SmallText"], Cell[CellGroupData[{ Cell["\<\ If you want to see the reason this works, click on the right.\ \>", "Special5"], Cell[TextData[{ "Reason: The kth derivative of ", Cell[BoxData[ \(E\^\(s\ x\)\)]], " with respect to ", Cell[BoxData[ \(u\)]], " is ", Cell[BoxData[ \(x\^k\ E\^\(s\ \ x\)\)]], ". " }], "Text"] }, Closed]], Cell["That's it.", "SmallText"] }, Closed]], Cell["B.5.a.ii)", "Subsubsection"], Cell[CellGroupData[{ Cell[TextData[{ "Use the technique of calculating integrals by taking derivatives to \ calculate\n ", Cell[BoxData[ \(\[Integral]\_\(-8\)\%8\ x\ \(E\^\(-x\)\) \[DifferentialD]x\)]], "." }], "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "Insert an extra variable ", Cell[BoxData[ \(s\)]], " to make this new function of ", Cell[BoxData[ \(s\)]], ":\n ", Cell[BoxData[ \(\(\ \ \ \ f[s] = \[Integral]\_\(-8\)\%8\( E\^\(s\ x\)\) \[DifferentialD]x\)\)]], ":" }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[f, s, x];\)\), "\n", \(f[s_] = \[Integral]\_\(-8\)\%8\( E\^\(s\ x\)\) \[DifferentialD]x\)}], "Input"], Cell[BoxData[ \(\(-\(\[ExponentialE]\^\(\(-8\)\ s\)\/s\)\) + \[ExponentialE]\^\(8\ \ s\)\/s\)], "Output"] }, Open ]], Cell[TextData[{ Cell[BoxData[ \(f[s]\)]], " is easily evalauted by hand." }], "Special5"], Cell[TextData[{ "Notice that\n ", Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "[", "s", "]"}], "=", \(\[Integral]\_\(-8\)\%8\( x\ E\^\(s\ x\)\) \[DifferentialD]x\)}]]], ".\nSo\n ", Cell[BoxData[ \(\[Integral]\_\(-8\)\%8\( x\ E\^\(-x\)\) \[DifferentialD]x\)]], " \n is given by: " }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[ \(\(f'\)[\(-1\)]\)], "Input"], Cell[BoxData[ \(\(-\(9\/\[ExponentialE]\^8\)\) - 7\ \[ExponentialE]\^8\)], "Output"] }, Open ]], Cell["You can do this by hand very easily.", "Special5"], Cell["\<\ Sweet and simple. Check:\ \>", "SmallText"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_\(-8\)\%8\( x\ E\^\(-x\)\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(\(-\(9\/\[ExponentialE]\^8\)\) - 7\ \[ExponentialE]\^8\)], "Output"] }, Open ]], Cell["Nailed it.", "SmallText"] }, Closed]], Cell["B.5.a.iii)", "Subsubsection"], Cell[CellGroupData[{ Cell[TextData[{ "Use the technique of calculating integrals by taking derivatives to \ calculate\n ", Cell[BoxData[ \(\[Integral]\_0\%\(12\ \)\ x\^2\ \(E\^\(\(-3\)\ x\)\) \[DifferentialD]x\)]], "." }], "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "Insert an extra variable ", Cell[BoxData[ \(s\)]], " to make this new function of ", Cell[BoxData[ \(s\)]], ":\n ", Cell[BoxData[ \(f[s] = \[Integral]\_0\%\(12\ \)\(E\^\(s\ x\)\) \[DifferentialD]x\)]], ":" }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[f, s, x];\)\), "\n", \(f[s_] = \[Integral]\_0\%12\( E\^\(s\ x\)\) \[DifferentialD]x\)}], "Input"], Cell[BoxData[ \(\(-\(1\/s\)\) + \[ExponentialE]\^\(12\ s\)\/s\)], "Output"] }, Open ]], Cell[TextData[{ "Notice that\n ", Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "[", "s", "]"}], "=", \(\[Integral]\_0\%12\( x\ E\^\(s\ x\)\) \[DifferentialD]x\)}]]], "\nand\n ", Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox["f", "\[DoublePrime]", MultilineFunction->None], "[", "s", "]"}], "=", \(\[Integral]\_0\%12\( x\^2\ E\^\(s\ x\)\) \[DifferentialD]x\)}]]], "\nSo\n ", Cell[BoxData[ \(\[Integral]\_0\%\(12\ \)x\^2\ \(E\^\(\(-3\)\ x\)\) \[DifferentialD]x\)]], " \n is given by: " }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[ \(\(\(f'\)'\)[\(-3\)]\)], "Input"], Cell[BoxData[ \(2\/27 - 1370\/\(27\ \[ExponentialE]\^36\)\)], "Output"] }, Open ]], Cell["\<\ Simple enough to be done by hand. Check:\ \>", "SmallText"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%12\( x\^2\ E\^\(\(-3\)\ x\)\) \[DifferentialD]x\)], \ "Input"], Cell[BoxData[ \(2\/27 - 1370\/\(27\ \[ExponentialE]\^36\)\)], "Output"] }, Open ]], Cell["Was there ever a doubt?", "SmallText"] }, Closed]], Cell["B.5.b)", "Subsubsection"], Cell[CellGroupData[{ Cell[TextData[{ "Use the technique of calculating integrals by taking derivatives to \ calculate\n ", Cell[BoxData[ \(\[Integral]\_\(0\ \)\%\(5\ \[Pi]\)\(x\^3\ Sin[4\ x]\) \[DifferentialD]x\)]], "." }], "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "Insert an extra variable ", Cell[BoxData[ \(s\)]], " to make this new function of ", Cell[BoxData[ \(s\)]], ":\n ", Cell[BoxData[ \(f[s] = \[Integral]\_0\%\(5\ \[Pi]\)\ Cos[s\ x] \[DifferentialD]x\)]], ":" }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[f, s, x];\)\), "\n", \(f[s_] = \[Integral]\_0\%\(5\ \[Pi]\)Cos[ s\ x] \[DifferentialD]x\)}], "Input"], Cell[BoxData[ \(Sin[5\ \[Pi]\ s]\/s\)], "Output"] }, Open ]], Cell["This is easily calculated by hand.", "Special5"], Cell[TextData[{ "Notice that\n ", Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "[", "s", "]"}], "=", " ", \(-\(\[Integral]\_0\%\(5\ \[Pi]\)\ x\ Sin[s\ x] \[DifferentialD]x\)\)}], ","}]]], "\nand\n ", Cell[BoxData[ RowBox[{ SuperscriptBox["f", "\[DoublePrime]", MultilineFunction->None], "[", "s", "]"}]]], Cell[BoxData[ \(\( = \ \(-\ \((\[Integral]\_0\%\(5\ \[Pi]\)\ x\^2\ Cos[s\ x] \[DifferentialD]x)\)\)\)\)]], ",\nSo the third derivative\n ", Cell[BoxData[ RowBox[{ SuperscriptBox["f", TagBox[\((3)\), Derivative], MultilineFunction->None], "[", "s", "]"}]]], " ", Cell[BoxData[ \(\( = \ \[Integral]\_\(0\ \)\%\(5\ \[Pi]\)\ x\^3\ Sin[s\ x] \[DifferentialD]x\)\)]], "." }], "SmallText"], Cell["Watch those minus signs.", "Special5"], Cell[TextData[{ "\n So \n ", Cell[BoxData[ \(\[Integral]\_0\%\(5\ \[Pi]\)\ x\^3\ Sin[4\ x] \[DifferentialD]x\)]], "\n is given by:" }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[f, s, x];\)\), "\n", \(\(f[ s_] = \[Integral]\_0\%\(5\ \[Pi]\)Cos[ s\ x] \[DifferentialD]x;\)\), "\n", \(\(\(\(f'\)'\)'\)[4]\)}], "Input"], Cell[BoxData[ \(\(15\ \[Pi]\)\/32 - \(125\ \[Pi]\^3\)\/4\)], "Output"] }, Open ]], Cell["\<\ Nearly simple enough to be done by hand. Check:\ \>", "SmallText"], Cell[CellGroupData[{ Cell[BoxData[ \(Expand[\[Integral]\_0\%\(5\ \[Pi]\)\(x\^3\ Sin[ 4\ x]\) \[DifferentialD]x]\)], "Input"], Cell[BoxData[ \(\(15\ \[Pi]\)\/32 - \(125\ \[Pi]\^3\)\/4\)], "Output"] }, Open ]], Cell["Had it all the way.", "SmallText"] }, Closed]], Cell["B.5.c)", "Subsubsection"], Cell[CellGroupData[{ Cell["\<\ Who likes this technique of calculating integrals by taking \ derivatives? Tell a Richard Feynman story about this method.\ \>", "Text"], Cell["Answer:", "Special1"], Cell["\<\ For some reason, math instructors prefer other techniques. In fact \ this technique is not represented in any other calculus course the authors \ know of. But physics instructors and some engiuneering instructors regard it \ as their technique of choice. They say it is simple and fun. Legendary physicist Richard Feynman loved to find new applications for this \ technique. In fact, when Feynman's first wife sued him for divorce, she said \ that each morning he got up and instead of talking with her, he did integrals \ all through breakfast and all the way to the office .\ \>", "SmallText"] }, Closed]] }, Closed]], Cell["Tutorial Problem", "Subsubsection"], Cell[CellGroupData[{ Cell["T.4) Which technique to go with", "Subsection", CellTags->"2.06.T4"], Cell["T.4.a) Integration by parts", "Subsubsection"], Cell[TextData[{ "Look at this calculation of ", Cell[BoxData[ \(\[Integral]\_0\%t ArcTan[x] \[DifferentialD]x\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x, t];\)\), "\n", \(\[Integral]\_0\%t ArcTan[x] \[DifferentialD]x\)}], "Input"], Cell[BoxData[ \(\(-\(1\/2\)\)\ \[ImaginaryI]\ \((t\ Log[\[ImaginaryI]\ \((\(-\ \[ImaginaryI]\) + t)\)] - t\ Log[\(-\[ImaginaryI]\)\ \((\[ImaginaryI] + t)\)] - \[ImaginaryI]\ Log[1 + t\^2])\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "What technique would most C&", StyleBox["M", FontSlant->"Italic"], " folks use to explain where this output comes from?\nUse that technique to \ explain where this output comes from." }], "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "Integration by parts. \nReason: Nothing else comes to mind.\nHere it \ goes:", "\n", "The integration by parts formula is\n ", Cell[BoxData[ RowBox[{\(\[Integral]\_a\%b u[x] \(v'\)[x] \[DifferentialD]x\), "=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "a", "b"], \(-\(\[Integral]\_a\%b v[x] \(u'\)[ x] \[DifferentialD]x\)\)}]}]]], ".\nRewrite the integral as\n ", Cell[BoxData[ \(\[Integral]\_0\%t ArcTan[x] \[DifferentialD]x = \[Integral]\_0\%t ArcTan[x] 1\ \[DifferentialD]x\)]], ".\nGo with:\n ", Cell[BoxData[ \(u[x] = ArcTan[x]\)]], " and ", Cell[BoxData[ \(\(v'\)[x] = 1\)]], ".\nThis gives\n ", Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["u", "\[Prime]", MultilineFunction->None], "[", "x", "]"}], "=", \(1\/\(1 + x\^2\)\)}], " "}]]], " and ", Cell[BoxData[ \(v[x] = \(\(x\)\(.\)\)\)]], "\nPlug into the integration by parts formula to get\n ", Cell[BoxData[ \(\[Integral]\_1\%t ArcTan[x] 1 \[DifferentialD]x\)]], "\n ", Cell[BoxData[ \(\(\(=\)\(\[Integral]\_1\%t u[x] \(v'\)[x] \[DifferentialD]x\)\)\)]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "t"], " ", \(-\ \(\[Integral]\_0\%t v[x] \(u'\)[ x] \[DifferentialD]x\)\)}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(x\ ArcTan[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "t"], " ", \(-\ \(\[Integral]\_0\%t\( x\/\(1 + x\^2\)\) \[DifferentialD]x\)\)}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(x\ ArcTan[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "t"], " ", \(\(-\(1\/2\)\)\ \(\[Integral]\_0\%t\(\( 2\ x\)\/\(1 + x\^2\)\) \[DifferentialD]x\)\)}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(x\ ArcTan[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "t"], " ", \(\(-\ \(1\/2\)\)\ Log[1 + x\^2]\)}]}]]], Cell[BoxData[ RowBox[{ SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "t"], " "}]]], "\n ", Cell[BoxData[ \(\(\(=\)\(\((t\ ArcTan[t] - \ 0\ ArcTan[0])\) - \ 1\/2\ \((Log[1 + t\^2]\ - \ Log[1\ + \ 0])\)\)\)\)]], "\n ", Cell[BoxData[ \(\(\(=\)\(\ \)\(t\ ArcTan[t]\)\) - \ 1\/2\ \((Log[1 + t\^2])\)\)]], ".\n Check:" }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x, t];\)\), "\n", \(\[Integral]\_0\%t ArcTan[x] \[DifferentialD]x\)}], "Input"], Cell[BoxData[ \(\(-\(1\/2\)\)\ \[ImaginaryI]\ \((t\ Log[\[ImaginaryI]\ \((\(-\ \[ImaginaryI]\) + t)\)] - t\ Log[\(-\[ImaginaryI]\)\ \((\[ImaginaryI] + t)\)] - \[ImaginaryI]\ Log[1 + t\^2])\)\)], "Output"] }, Open ]], Cell["Nailed it with integration by parts.", "SmallText"] }, Closed]], Cell["\<\ T.4.b.i) The technique of calculating integrals by taking \ derivatives\ \>", "Subsubsection"], Cell[TextData[{ "Look at this calculation of ", Cell[BoxData[ \(\[Integral]\_0\%t\( x\^3\ Sin[\[Pi]\ x]\) \[DifferentialD]x\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x, t];\)\), "\n", \(\[Integral]\_0\%t\( x\^3\ Sin[\[Pi]\ x]\) \[DifferentialD]x\)}], "Input"], Cell[BoxData[ \(\(-\(\(t\ \((\(-6\) + \[Pi]\^2\ t\^2)\)\ Cos[\[Pi]\ t]\)\/\[Pi]\^3\)\) \ + \(3\ \((\(-2\) + \[Pi]\^2\ t\^2)\)\ Sin[\[Pi]\ t]\)\/\[Pi]\^4\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "What technique would most C&", StyleBox["M", FontSlant->"Italic"], " folks use to explain where this output comes from?\nUse that technique \ reproduce the ", StyleBox["Mathematica", FontSlant->"Italic"], " calculation." }], "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "The technique of calculating integrals by taking derivatives.\nReason: \ It's fairly easy.\nInsert an extra variable ", Cell[BoxData[ \(s\)]], " to make this new function of ", Cell[BoxData[ \(s\)]], ":\n ", Cell[BoxData[ \(f[s] = \[Integral]\_0\%t\ Cos[s\ x] \[DifferentialD]x\)]], ".\n Take the derivative with respect to ", Cell[BoxData[ \(s\)]], " to get\n ", Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "[", "s", "]"}], "=", " ", \(-\(\[Integral]\_0\%t\ x\ Sin[s\ x] \[DifferentialD]x\)\)}]]], ".\nTake the second derivative with respect to ", Cell[BoxData[ \(s\)]], " to get\n ", Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox["f", "\[DoublePrime]", MultilineFunction->None], "[", "s", "]"}], "=", \(-\(\[Integral]\_0\%t x\^2\ Cos[ s\ x] \[DifferentialD]x\)\)}]]], ".\n Take the third derivative with respect to ", Cell[BoxData[ \(s\)]], " to get\n ", Cell[BoxData[ RowBox[{ SuperscriptBox["f", TagBox[\((3)\), Derivative], MultilineFunction->None], "[", "s"}]]], Cell[BoxData[ \(\(\(=\)\(\[Integral]\_0\%t x\^3\ Sin[s\ x] \[DifferentialD]x\)\)\)]], ".\n This tells you that\n ", Cell[BoxData[ RowBox[{ SuperscriptBox["f", TagBox[\((3)\), Derivative], MultilineFunction->None], "[", "\[Pi]", "]"}]]], Cell[BoxData[ \(\(\(=\)\(\[Integral]\_0\%t x\^3\ Sin[\[Pi]\ x] \ \[DifferentialD]x\)\)\)]], "\nSo the steps toward getting ", Cell[BoxData[ \(\[Integral]\_0\%t x\^3\ Sin[\[Pi]\ x] \(\(\[DifferentialD]\)\(x\)\(\ \ \)\)\)]], "are:" }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[f, s, x];\)\), "\n", \(f[s_] = \[Integral]\_0\%t Cos[s\ x] \[DifferentialD]x\)}], "Input"], Cell[BoxData[ \(Sin[s\ t]\/s\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\(f'\)[s]\)], "Input"], Cell[BoxData[ \(\(t\ Cos[s\ t]\)\/s - Sin[s\ t]\/s\^2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\(\(f'\)'\)[s]\)], "Input"], Cell[BoxData[ \(\(-\(\(2\ t\ Cos[ s\ t]\)\/s\^2\)\) + \(2\ Sin[s\ t]\)\/s\^3 - \(t\^2\ Sin[s\ \ t]\)\/s\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\(\(\(f'\)'\)'\)[s]\)], "Input"], Cell[BoxData[ \(\(6\ t\ Cos[s\ t]\)\/s\^3 - \(t\^3\ Cos[s\ t]\)\/s - \(6\ Sin[s\ \ t]\)\/s\^4 + \(3\ t\^2\ Sin[s\ t]\)\/s\^2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\(\(\(f'\)'\)'\)[Pi]\)], "Input"], Cell[BoxData[ \(\(6\ t\ Cos[\[Pi]\ t]\)\/\[Pi]\^3 - \(t\^3\ Cos[\[Pi]\ t]\)\/\[Pi] - \ \(6\ Sin[\[Pi]\ t]\)\/\[Pi]\^4 + \(3\ t\^2\ Sin[\[Pi]\ t]\)\/\[Pi]\^2\)], \ "Output"] }, Open ]], Cell["All are fairly easily calculated by hand.", "Special5"], Cell["Check: ", "SmallText"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x, t];\)\), "\n", \(mathematicacalc = \[Integral]\_0\%t\( x\^3\ Sin[\[Pi]\ x]\) \ \[DifferentialD]x\)}], "Input"], Cell[BoxData[ \(\(-\(\(t\ \((\(-6\) + \[Pi]\^2\ t\^2)\)\ Cos[\[Pi]\ t]\)\/\[Pi]\^3\)\) \ + \(3\ \((\(-2\) + \[Pi]\^2\ t\^2)\)\ Sin[\[Pi]\ t]\)\/\[Pi]\^4\)], "Output"] }, Open ]], Cell["\<\ They are the same. Done.\ \>", "SmallText"] }, Closed]], Cell["T.4.b.ii)", "Subsubsection"], Cell[TextData[{ "Stay with the calculation of ", Cell[BoxData[ \(\[Integral]\_0\%t\( x\^3\ Sin[\[Pi]\ x]\) \[DifferentialD]x\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x, t];\)\), "\n", \(\[Integral]\_0\%t\( x\^3\ Sin[\[Pi]\ x]\) \[DifferentialD]x\)}], "Input"], Cell[BoxData[ \(\(-\(\(t\ \((\(-6\) + \[Pi]\^2\ t\^2)\)\ Cos[\[Pi]\ t]\)\/\[Pi]\^3\)\) \ + \(3\ \((\(-2\) + \[Pi]\^2\ t\^2)\)\ Sin[\[Pi]\ t]\)\/\[Pi]\^4\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Can you explain where this integral comes from by using integration \ by parts?\ \>", "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "Yes, you can, but it will be tedious. \nYou will have to integrate by \ parts three (YES THREE) times. \nHere's how it goes:\nThe integral under \ study is:\n ", Cell[BoxData[ \(\[Integral]\_0\%t\( x\^3\ Sin[\[Pi]\ x]\) \[DifferentialD]x\)]], ".\nThe integration by parts formula is\n ", Cell[BoxData[ RowBox[{\(\[Integral]\_0\%t u[x] \(v'\)[x] \[DifferentialD]x\), "=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "t"], \(-\ \(\[Integral]\_0\%t v[x] \(u'\)[ x] \[DifferentialD]x\)\)}]}]]], ".\nMake the assignments:\n ", Cell[BoxData[ \(u[x] = x\^3\)]], " and ", Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox["v", "\[Prime]", MultilineFunction->None], "[", "x", "]"}], "=", \(Sin[\[Pi]\ x]\)}]]], ".\nThis gives\n ", Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox["u", "\[Prime]", MultilineFunction->None], "[", "x", "]"}], "=", \(3\ x\^2\)}]]], " and ", Cell[BoxData[ \(v[x] = \(-Cos[\[Pi]\ x]\)\)]], ".\nWith these assignments,\n ", Cell[BoxData[ \(\[Integral]\_0\%t\( x\^3\ Sin[\[Pi]\ x]\) \[DifferentialD]x\)]], "\n ", Cell[BoxData[ \(\(\(=\)\(\[Integral]\_0\%t u[x] \(v'\)[x] \[DifferentialD]x\)\)\)]], " \n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(u[x] v[x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "t"], \(-\(\[Integral]\_0\%t v[x] \(u'\)[ x] \[DifferentialD]x\)\)}]}]]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(\(-x\^3\)\ Cos[\[Pi]\ x]\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "t"], \(\(+\ 3\)\ \(\[Integral]\_0\%t\( x\^2\ Cos[\[Pi]\ x]\) \(\(\ \[DifferentialD]x\)\(.\)\)\)\)}]}]]], "\nAt this point, you are only ", Cell[BoxData[ \(1\/3\)]], " of the way home.\nNext you will have to go to integration by parts to \ calculate\n ", Cell[BoxData[ \(\[Integral]\_0\%t\( x\^2\ Cos[\[Pi]\ x]\) \[DifferentialD]x\)]], " \nand this will lead to having to use integration by parts on\n \ ", Cell[BoxData[ \(\(\(\ \)\(\[Integral]\_0\%t x\ Sin[\[Pi]\ x] \[DifferentialD]x\)\)\)]], ".\n Not fun.\nThe technique of calculating integrals by taking derivatives \ beats the heck out of integration by parts on this one.\n " }], "SmallText"] }, Closed]], Cell["T.4.c.i) The complex exponential", "Subsubsection"], Cell[TextData[{ "Look at this calculation of ", Cell[BoxData[ \(\[Integral]\_0\%\(2\ \[Pi]\)Sin[3\ x]\ Sin[ 7\ x] \[DifferentialD]x\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x, t];\)\), "\n", \(\[Integral]\_0\%\(2\ \[Pi]\)\(Sin[3\ x]\ Sin[ 7\ x]\) \[DifferentialD]x\)}], "Input"], Cell[BoxData[ \(0\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "What technique would most C&", StyleBox["M", FontSlant->"Italic"], " folks use to explain where this output comes from?\nUse that technique \ reproduce the ", StyleBox["Mathematica", FontSlant->"Italic"], " calculation." }], "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "The complex exponential.\n\nThe integral under study here is \n \ ", Cell[BoxData[ \(\[Integral]\_0\%\(2\ \[Pi]\)\ Sin[m\ x]\ Sin[n\ x] \[DifferentialD]x\)]], "\nwith ", Cell[BoxData[ \(m = 3\)]], " and ", Cell[BoxData[ \(n = 7\)]], "." }], "SmallText"], Cell[TextData[{ "Remember \n ", Cell[BoxData[ \(E\^\(\(\ \)\(Imx\)\(\ \)\) = Cos[m\ x] + I\ Sin[m\ x]\)]], ".\n So \n ", Cell[BoxData[ \(E\^\(\(-I\)\ m\ x\) = \(Cos[m\ x] + I\ Sin[m\ x] = Cos[m\ x] - I\ Sin[m\ x]\)\)]], ".\n \n Subtract to get \n ", Cell[BoxData[ \(2\ I\ Sin[m\ x] = E\^\(\(\ \)\(I\ m\ x\)\) - E\^\(\(-Im\)\ x\)\)]], " \nMultiply both sides by ", Cell[BoxData[ \(\(\(\ \)\(\(-I\)\/2\)\)\)]], " to get\n ", Cell[BoxData[ \(Sin[m\ x] = 1\/2\ \((\(-I\))\)\ \((E\^\(I\ m\ x\) - E\^\(\(\ \)\(I\ m\ x\)\))\)\)]], "." }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x, m];\)\), "\n", \(sinmx = 1\/2\ \((\(-I\))\)\ \((E\^\(I\ m\ x\) - E\^\(\(-I\)\ m\ x\))\)\)}], "Input"], Cell[BoxData[ \(\(-\(1\/2\)\)\ \[ImaginaryI]\ \((\(-\[ExponentialE]\^\(\(-\[ImaginaryI]\ \)\ m\ x\)\) + \[ExponentialE]\^\(\[ImaginaryI]\ m\ x\))\)\)], "Output"] }, Open ]], Cell[TextData[{ "Repeating the same steps gives this expression for ", Cell[BoxData[ \(Sin[n\ x]\)]], ":" }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x, n];\)\), "\n", \(sinnx = 1\/2\ \((\(-I\))\)\ \((E\^\(I\ n\ x\) - E\^\(\(-I\)\ n\ x\))\)\)}], "Input"], Cell[BoxData[ \(\(-\(1\/2\)\)\ \[ImaginaryI]\ \((\(-\[ExponentialE]\^\(\(-\[ImaginaryI]\ \)\ n\ x\)\) + \[ExponentialE]\^\(\[ImaginaryI]\ n\ x\))\)\)], "Output"] }, Open ]], Cell[TextData[{ "So ", Cell[BoxData[ \(Sin[m\ x]\ Sin[n\ x]\)]], " is:" }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[ \(complexform = \ sinmx\ sinnx\)], "Input"], Cell[BoxData[ \(\(-\(1\/4\)\)\ \((\(-\[ExponentialE]\^\(\(-\[ImaginaryI]\)\ m\ x\)\) + \ \[ExponentialE]\^\(\[ImaginaryI]\ m\ x\))\)\ \((\(-\[ExponentialE]\^\(\(-\ \[ImaginaryI]\)\ n\ x\)\) + \[ExponentialE]\^\(\[ImaginaryI]\ n\ x\))\)\)], \ "Output"] }, Open ]], Cell["Check it:", "SmallText"], Cell[CellGroupData[{ Cell[BoxData[ \(ComplexExpand[complexform]\)], "Input"], Cell[BoxData[ \(Sin[m\ x]\ Sin[n\ x]\)], "Output"] }, Open ]], Cell["\<\ Good. Multiply it out:\ \>", "SmallText"], Cell[CellGroupData[{ Cell[BoxData[ \(Expand[complexform]\)], "Input"], Cell[BoxData[ \(\(-\(1\/4\)\)\ \[ExponentialE]\^\(\(-\[ImaginaryI]\)\ m\ x - \ \[ImaginaryI]\ n\ x\) + 1\/4\ \[ExponentialE]\^\(\[ImaginaryI]\ m\ x - \[ImaginaryI]\ n\ x\) + 1\/4\ \[ExponentialE]\^\(\(-\[ImaginaryI]\)\ m\ x + \[ImaginaryI]\ n\ x\ \) - 1\/4\ \[ExponentialE]\^\(\[ImaginaryI]\ m\ x + \[ImaginaryI]\ n\ x\)\)], \ "Output"] }, Open ]], Cell[TextData[{ "This is ", Cell[BoxData[ \(Sin[m\ x]\ Sin[n\ x]\)]], ".\nTo calculate\n ", Cell[BoxData[ \(\[Integral]\_0\%t Sin[m\ x]\ Sin[n\ x] \[DifferentialD]x\)]], ", \nyou just integrate these complex exponentials from ", Cell[BoxData[ \(0\)]], " to ", Cell[BoxData[ \(t\)]], ". This results in:\n\n ", Cell[BoxData[ \(\(-\(E\^\(\(-I\)\ m\ x - I\ n\ x\)\/\(4\ \((\(-I\)\ m - I\ n)\)\)\)\) + E\^\(I\ m\ x - I\ n\ x\)\/\(4\ \((I\ m - I\ n)\)\) + E\^\(\(-I\)\ m\ x + I\ n\ x\)\/\(4\ \((\(-I\)\ m + I\ n)\)\) - E\^\(I\ m\ x + I\ n\ x\)\/\(4\ \((I\ m + I\ n)\)\)\)]], ".\n Now remember that ", Cell[BoxData[ \(m = 3\)]], " and ", Cell[BoxData[ \(n = 7\)]], " and calculate the integral.\n " }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(m\ = \ 3;\)\), "\n", \(\(n\ = \ 7;\)\n\), "\n", \(\(top\ = \(-\(E\^\(\(-I\)\ m\ x - I\ n\ x\)\/\(4\ \((\(-I\)\ m - I\ n)\)\)\)\) + E\^\(I\ m\ x - I\ n\ x\)\/\(4\ \((I\ m - I\ n)\)\) + E\^\(\(-I\)\ m\ x + I\ n\ x\)\/\(4\ \((\(-I\)\ m + I\ n)\)\) - E\^\(I\ m\ x + I\ n\ x\)\/\(4\ \((I\ m + I\ n)\)\) /. x -> 2\ Pi;\)\), "\n", \(\(bottom\ = \ \(-\(E\^\(\(-I\)\ m\ x - I\ n\ x\)\/\(4\ \((\(-I\)\ m - I\ n)\)\)\)\) + E\^\(I\ m\ x - I\ n\ x\)\/\(4\ \((I\ m - I\ n)\)\) + E\^\(\(-I\)\ m\ x + I\ n\ x\)\/\(4\ \((\(-I\)\ m + I\ n)\)\) - E\^\(I\ m\ x + I\ n\ x\)\/\(4\ \((I\ m + I\ n)\)\) /. x -> 0;\)\n\n\), "\n", \(ComplexExpand[top\ - \ bottom]\)}], "Input"], Cell[BoxData[ \(0\)], "Output"] }, Open ]], Cell[TextData[{ "This explains the output from the ", StyleBox["Mathematica", FontSlant->"Italic"], " calculation:" }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%\(2\ \[Pi]\)\(Sin[3\ x]\ Sin[ 7\ x]\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(0\)], "Output"] }, Open ]] }, Closed]], Cell["T.4.c.ii) The complex exponential", "Subsubsection"], Cell[TextData[{ "Look at this calculation of ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi]\ E\^\(\(-0.5\)\ t\)\ Cos[2.4\ t] \[DifferentialD]t\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x, m, n, t];\)\), "\n", \(\[Integral]\_0\%\[Pi]\( E\^\(\(-0.5\)\ t\)\ Cos[ 2.4\ t]\) \[DifferentialD]t\)}], "Input"], Cell[BoxData[ \(\(\(0.15680089529541402`\)\(\[InvisibleSpace]\)\) - 6.882380089345104`*^-19\ \[ImaginaryI]\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "What technique would most C&", StyleBox["M", FontSlant->"Italic"], " folks use to explain where this output comes from?\nUse that technique \ reproduce the ", StyleBox["Mathematica", FontSlant->"Italic"], " calculation." }], "Text"], Cell["Answer:", "Special1"], Cell[TextData[{ "Complex exponential.\nReason: The integral under study here is ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi]\ E\^\(\(-0.5\)\ t\)\ Cos[2.4\ t] \[DifferentialD]t\)]], ".\nWhen you see\n ", Cell[BoxData[ \(E\^\(\(-0.5\)\ t\)\ Cos[2.4\ t]\)]], " ,\n you think of\n ", Cell[BoxData[ \(E\^\(\((\(-0.5\)\ \ + \ \ 2.4\ I\ )\)\ t\)\ = \ E\^\(\(-0.5\)\ t\)\ Cos[2.4\ t] + I\ E\^\(\(-0.5\)\ t\)\ Sin[2.4\ t]\)]], ".\nThis very good because this tells you that\n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi]\( E\^\(\((\(-0.5\)\ \ + \ \ 2.4\ I\ )\)\ t\)\) \ \[DifferentialD]t = \ \[Integral]\_0\%\[Pi] E\^\(\(-0.5\)\ t\)\ Cos[ 2.4\ t]\ \[DifferentialD]t + I\ \(\[Integral]\_0\%\[Pi] E\^\(\(-0.5\)\ x\)\ Sin[ 2.4\ x] \[DifferentialD]t\)\)]], " . \nThe integrals on the right are hard to do, but the integral on the \ left is really easy to do. \nIt's just \n", Cell[BoxData[ \(\(\(\ \ \ \ \ \ \ \ \)\(\[Integral]\_0\%\[Pi]\( E\^\(\((\(-0.5\)\ \ + \ \ 2.4\ I\ )\)\ t\)\) \(\(\ \[DifferentialD]\)\(t\)\(\ \)\)\)\)\)]], Cell[BoxData[ \( = \)]], " ", Cell[BoxData[ RowBox[{\(E\^\(\((\(-0.5\)\ + \ 2.4\ I)\)\ t\)\/\(\(-0.5\)\ + \ 2.4\ I\)\), SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "\[Pi]"]}]]], ".\nThis is the same as\n ", Cell[BoxData[ \(\(\(-0.5\)\ - \ 2.4\ I\)\/\(\(-0.5\)\ - \ 2.4\ I\)\)]], " ", Cell[BoxData[ \(E\^\(\((\(-0.5\)\ + \ 2.4\ I)\)\ t\)\/\(\(-0.5\)\ + \ 2.4\ I\)\)]], Cell[BoxData[ SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "\[Pi]"]]], ".\nAnd this is the same as\n \ ", Cell[BoxData[ \(\(\(\ \)\(\((\(-0.5\)\ - \ 2.4\ I)\)\ E\^\(\((\(-0.5\)\ + \ 2.4\ I)\ \)\ t\)\)\)\/\(\((\(-0.5\))\)\^2\ \ + \ \ 2.4\^2\)\)]], Cell[BoxData[ SubsuperscriptBox[ StyleBox["|", FontSize->24], "0", "\[Pi]"]]], ":\n " }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(top\ = \ \(\((\(-0.5\)\ - \ 2.4\ I)\)\ E\^\(\((\(-0.5\)\ + \ 2.4\ \ I)\)\ t\)\)\/\(\((\(-0.5\))\)\^2\ \ + \ \ 2.4\^2\) /. t -> Pi;\)\), "\n", \(\(bottom\ = \ \(\((\(-0.5\)\ - \ 2.4\ I)\)\ E\^\(\((\(-0.5\)\ + \ \ 2.4\ I)\)\ t\)\)\/\(\((\(-0.5\))\)\^2\ \ + \ \ 2.4\^2\) /. t -> 0;\)\), "\n", \(top\ - \ bottom\)}], "Input"], Cell[BoxData[ \(\(\(0.15680089529541402`\)\(\[InvisibleSpace]\)\) + 0.3572338460318513`\ \[ImaginaryI]\)], "Output"] }, Open ]], Cell[TextData[{ "This is ", Cell[BoxData[ \(\(\(\n\)\(\[Integral]\_0\%\[Pi]\( E\^\(\((\(-0.5\)\ \ + \ \ 2.4\ I\ )\)\ t\)\) \ \[DifferentialD]t = \ \[Integral]\_0\%\[Pi] E\^\(\(-0.5\)\ t\)\ Cos[ 2.4\ t]\ \[DifferentialD]t + I\ \(\[Integral]\_0\%\[Pi] E\^\(\(-0.5\)\ x\)\ Sin[ 2.4\ x] \[DifferentialD]t\)\)\)\)]], ".\nRead off\n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi]\( E\^\(\(-0.5\)\ t\)\ Cos[ 2.4\ t]\) \[DifferentialD]t = 0.156801\)]], "\nand as a fringe benefit you can throw in\n ", Cell[BoxData[ \(\[Integral]\_0\%\[Pi]\( E\^\(\(-0.5\)\ x\)\ Sin[ 2.4\ x]\) \[DifferentialD]t = 0.357234\)]], ".\nConfirm:" }], "SmallText"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%\[Pi]\( E\^\(\(-0.5\)\ t\)\ Cos[ 2.4\ t]\) \[DifferentialD]t\)], "Input"], Cell[BoxData[ \(\(\(0.15680089529541402`\)\(\[InvisibleSpace]\)\) - 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Be careful when modifying, renaming, or removing these \ styles, because the front end associates special meanings with these style \ names.\ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["Input"], CellFrame->{{3, 0}, {0, 0}}, CellMargins->{{45, Inherited}, {Inherited, Inherited}}, Evaluatable->True, CellGroupingRules->"InputGrouping", CellHorizontalScrolling->True, GroupPageBreakWithin->False, CellLabelMargins->{{23, Inherited}, {Inherited, Inherited}}, DefaultFormatType->DefaultInputFormatType, AutoItalicWords->{}, FormatType->InputForm, ShowStringCharacters->True, NumberMarks->True, CounterIncrements->"Input", FontWeight->"Bold"], Cell[StyleData["Input", "Presentation"], CellFrame->{{3, 0}, {0, 0}}, CellMargins->{{45, Inherited}, {20, Inherited}}, LineSpacing->{1, 0}], Cell[StyleData["Input", "Printout"], CellFrame->{{3, 0}, {0, 0}}, CellMargins->{{30, Inherited}, {2, 2}}, PageBreakWithin->True, GroupPageBreakWithin->True, FontSize->9], Cell[StyleData["Input", "TwoColumn"], 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"TwoColumn"], CellMargins->{{30, Inherited}, {2, 2}}, PageBreakWithin->True, GroupPageBreakWithin->True, LineSpacing->{1, 0}, FontSize->12, FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Message"], CellMargins->{{45, Inherited}, {Inherited, Inherited}}, CellGroupingRules->"OutputGrouping", PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, CellLabelMargins->{{23, Inherited}, {Inherited, Inherited}}, DefaultFormatType->DefaultOutputFormatType, FormatType->InputForm, StyleMenuListing->None, FontColor->RGBColor[0, 0, 1]], Cell[StyleData["Message", "Presentation"], CellMargins->{{45, Inherited}, {Inherited, Inherited}}, LineSpacing->{1, 0}], Cell[StyleData["Message", "Printout"], CellMargins->{{30, Inherited}, {2, 2}}, FontSize->9, FontColor->GrayLevel[0]], Cell[StyleData["Message", "TwoColumn"], CellMargins->{{30, Inherited}, {2, 2}}, FontSize->9, FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Print"], CellMargins->{{45, Inherited}, {Inherited, Inherited}}, CellGroupingRules->"OutputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, CellLabelMargins->{{23, Inherited}, {Inherited, Inherited}}, DefaultFormatType->DefaultOutputFormatType, FormatType->InputForm, StyleMenuListing->None], Cell[StyleData["Print", "Presentation"], CellMargins->{{45, Inherited}, {Inherited, Inherited}}, LineSpacing->{1, 0}], Cell[StyleData["Print", "Printout"], CellMargins->{{30, Inherited}, {2, 2}}, FontSize->9], Cell[StyleData["Print", "TwoColumn"], CellMargins->{{30, Inherited}, {2, 2}}, FontSize->16] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Info"], CellMargins->{{45, Inherited}, {Inherited, Inherited}}, CellGroupingRules->"OutputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, CellLabelMargins->{{23, Inherited}, {Inherited, Inherited}}, DefaultFormatType->DefaultOutputFormatType, FormatType->InputForm, StyleMenuListing->None], Cell[StyleData["Info", "Presentation"], CellMargins->{{45, Inherited}, {Inherited, Inherited}}, LineSpacing->{1, 0}], Cell[StyleData["Info", "Printout"], CellMargins->{{30, Inherited}, {Inherited, Inherited}}, FontSize->10], Cell[StyleData["Info", "TwoColumn"], CellMargins->{{30, Inherited}, {Inherited, Inherited}}, FontSize->12] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Graphics"], CellMargins->{{15, Inherited}, {Inherited, Inherited}}, CellGroupingRules->"GraphicsGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, DefaultFormatType->DefaultOutputFormatType, FormatType->InputForm, ImageMargins->{{35, Inherited}, {Inherited, 0}}, AnimationDisplayTime->0.2, StyleMenuListing->None, FontSize->14], Cell[StyleData["Graphics", "Presentation"], CellMargins->{{45, Inherited}, {Inherited, Inherited}}, LineSpacing->{1, 0}], Cell[StyleData["Graphics", "Printout"], CellMargins->{{30, Inherited}, {0, 0}}, CellFrameMargins->False, ImageSize->{Inherited, 150}, ImageMargins->{{45, 0}, {0, 0}}, ImageRegion->{{0, 1}, {0, 1}}, FontSize->9], Cell[StyleData["Graphics", "TwoColumn"], CellMargins->{{20, Inherited}, {0, 0}}, CellFrameMargins->False, ImageSize->{Inherited, 150}, ImageMargins->{{0, 0}, {0, 0}}, ImageRegion->{{0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["CellLabel"], StyleMenuListing->None, FontFamily->"Times", FontSize->9, FontColor->RGBColor[0, 0, 1]], Cell[StyleData["CellLabel", "Presentation"], FontSize->14], Cell[StyleData["CellLabel", "Printout"], FontColor->GrayLevel[1]], Cell[StyleData["CellLabel", "TwoColumn"], FontColor->GrayLevel[1]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Special Headings", "Section"], Cell[CellGroupData[{ Cell[StyleData["PrefaceTitle"], CellFrame->{{1, 1}, {1, 5}}, ShowCellBracket->False, CellMargins->{{24, 24}, {0, 10}}, CellGroupingRules->{"SectionGrouping", 30}, PageBreakBelow->False, CellFrameMargins->{{15, Inherited}, {Inherited, Inherited}}, DefaultInlineFormatType->DefaultInputInlineFormatType, TextAlignment->Center, LineSpacing->{1.4, 1}, FontFamily->"Times", FontSize->24, FontWeight->"Bold", FontColor->GrayLevel[1], Background->RGBColor[0, 0.392187, 0]], Cell[StyleData["PrefaceTitle", "Presentation"], CellMargins->{{24, Inherited}, {60, Inherited}}, TextAlignment->Center, FontSize->38, FontColor->GrayLevel[1], Background->RGBColor[0.596078, 0.65098, 0.0196078]], Cell[StyleData["PrefaceTitle", "Printout"], CellMargins->{{0, Inherited}, {0, Inherited}}, FontSize->16, FontColor->GrayLevel[0], Background->GrayLevel[1]], Cell[StyleData["PrefaceTitle", "TwoColumn"], CellFrame->{{1, 1}, {0, 5}}, CellMargins->{{0, Inherited}, {0, Inherited}}, FontSize->16, FontColor->GrayLevel[0], Background->GrayLevel[1]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Special Body Text and Index", "Section"], Cell[CellGroupData[{ Cell[StyleData["Accident"], CellFrame->3, ShowCellBracket->False, CellMargins->{{24, 24}, {0, 10}}, CellFrameMargins->{{15, Inherited}, {Inherited, Inherited}}, TextAlignment->Center, LineSpacing->{1.4, 1}, FontFamily->"Times", FontSize->24, FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], Cell[StyleData["Accident", "Presentation"], CellMargins->{{24, Inherited}, {60, Inherited}}, TextAlignment->Center, FontSize->36], Cell[StyleData["Accident", "Printout"], CellFrame->2, CellMargins->{{0, Inherited}, {0, Inherited}}, FontSize->16, FontColor->GrayLevel[0], Background->GrayLevel[1]], Cell[StyleData["Accident", "TwoColumn"], CellFrame->2, CellMargins->{{0, Inherited}, {0, Inherited}}, FontSize->16, FontColor->GrayLevel[0], Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["ContentsText"], CellMargins->{{50, 10}, {5, 5}}, FontFamily->"Times", FontSize->16], Cell[StyleData["ContentsText", "Presentation"]], Cell[StyleData["ContentsText", "Printout"], FontColor->GrayLevel[0], Background->None], Cell[StyleData["ContentsText", "TwoColumn"], FontColor->GrayLevel[0], Background->None] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Special1"], CellDingbat->"\[EmptySquare]", ShowClosedCellArea->True, DefaultFormatType->DefaultTextFormatType, DefaultInlineFormatType->DefaultInputInlineFormatType, FontFamily->"Times", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.689998, 0.0899977, 0.119997]], Cell[StyleData["Special1", "Presentation"], FontSize->16], Cell[StyleData["Special1", "Printout"], FontSize->12, FontColor->GrayLevel[0]], Cell[StyleData["Special1", "TwoColumn"], FontSize->12, FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Special2"], CellMargins->{{6, 0}, {0, 0}}, CellGroupingRules->{"SectionGrouping", 40}, DefaultFormatType->DefaultTextFormatType, DefaultInlineFormatType->DefaultInputInlineFormatType, TextAlignment->Center, FontFamily->"Courier", FontSize->10, FontColor->GrayLevel[0.333333]], Cell[StyleData["Special2", "Presentation"], FontSize->12], Cell[StyleData["Special2", "Printout"], FontSize->10, FontColor->GrayLevel[0]], Cell[StyleData["Special2", "TwoColumn"], FontSize->10, FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Special3"], CellDingbat->"\[GraySquare]", ShowClosedCellArea->True, DefaultFormatType->DefaultTextFormatType, DefaultInlineFormatType->DefaultInputInlineFormatType, TextAlignment->Left, FontFamily->"Courier", FontSize->10, FontColor->GrayLevel[0.333333]], Cell[StyleData["Special3", "Presentation"]], Cell[StyleData["Special3", "Printout"], FontColor->GrayLevel[0]], Cell[StyleData["Special3", "TwoColumn"], FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Special4"], ShowClosedCellArea->True, DefaultFormatType->DefaultTextFormatType, DefaultInlineFormatType->DefaultInputInlineFormatType, TextAlignment->Center, FontFamily->"Courier", FontSize->10, FontColor->GrayLevel[0.333333]], Cell[StyleData["Special4", "Presentation"], FontSize->12], Cell[StyleData["Special4", "Printout"], FontColor->GrayLevel[0]], Cell[StyleData["Special4", "TwoColumn"], FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Special5"], ShowClosedCellArea->True, DefaultFormatType->DefaultTextFormatType, DefaultInlineFormatType->DefaultInputInlineFormatType, TextAlignment->Center, FontFamily->"Courier", FontSize->10, FontColor->GrayLevel[0.333333]], Cell[StyleData["Special5", "Presentation"], FontSize->12], Cell[StyleData["Special5", "Printout"]], Cell[StyleData["Special5", "TwoColumn"]] }, Closed]], Cell[StyleData["IndexEntry"], ShowCellBracket->False, CellMargins->{{15, 5}, {0, 5}}, PageBreakBelow->False, DefaultFormatType->DefaultTextFormatType, DefaultInlineFormatType->DefaultInputInlineFormatType, ParagraphIndent->-40, StyleMenuListing->None, FontSize->16], Cell[StyleData["IndexSubEntry"], ShowCellBracket->False, CellMargins->{{45, 5}, {0, 0}}, DefaultFormatType->DefaultTextFormatType, DefaultInlineFormatType->DefaultInputInlineFormatType, ParagraphIndent->-40, StyleMenuListing->None, FontSize->16] }, Closed]], Cell[CellGroupData[{ Cell["Styles for License Agreement", "Section"], Cell[CellGroupData[{ Cell[StyleData["LicenseHeading"], ShowCellBracket->True, ShowGroupOpenCloseIcon->True, CellMargins->{{24, 24}, {-1, 2}}, CellGroupingRules->{"SectionGrouping", 40}, PageBreakBelow->False, CounterIncrements->"Subsection", CounterAssignments->{{"Subsubsection", 0}}, StyleMenuListing->None, FontFamily->"Helvetica", FontSize->12, FontWeight->"Bold", FontColor->RGBColor[0.4, 0.300008, 0.6]], Cell[StyleData["LicenseHeading", "Presentation"], FontSize->12], Cell[StyleData["LicenseHeading", "Printout"], FontSize->10, FontColor->GrayLevel[0], Background->GrayLevel[1]], Cell[StyleData["LicenseHeading", "TwoColumn"], FontSize->10, FontColor->GrayLevel[0], Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["LicenseText"], CellFrame->True, ShowCellBracket->False, CellMargins->{{24, 24}, {5, -1}}, StyleMenuListing->None, FontFamily->"Helvetica", FontSize->12, Background->RGBColor[1, 0.537743, 0.509071]], Cell[StyleData["LicenseText", "Presentation"], FontSize->18], Cell[StyleData["LicenseText", "Printout"], FontSize->10, FontColor->GrayLevel[0], Background->GrayLevel[1]], Cell[StyleData["LicenseText", "TwoColumn"], FontSize->10, FontColor->GrayLevel[0], Background->GrayLevel[1]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Automatic Numbering", "Section"], Cell["\<\ The following styles are useful for numbered equations, figures, \ etc. They automatically give the cell a FrameLabel containing a reference to \ a particular counter, and also increment that counter.\ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["NumberedEquation"], CounterIncrements->"NumberedEquation"], Cell[StyleData["NumberedEquation", "Presentation"]], Cell[StyleData["NumberedEquation", "Printout"]], Cell[StyleData["NumberedEquation", "TwoColumn"]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["NumberedFigure"], CellFrameLabels->{{None, None}, {Cell[ TextData[ {"Figure ", CounterBox[ "NumberedFigure"]}]], None}}, CounterIncrements->"NumberedFigure", FormatTypeAutoConvert->False, FontFamily->"Times"], Cell[StyleData["NumberedFigure", "Presentation"], CellMargins->{{24, Inherited}, {20, Inherited}}, LineSpacing->{1, 0}], Cell[StyleData["NumberedFigure", "Printout"], CellMargins->{{14, Inherited}, {Inherited, Inherited}}, FontSize->10], Cell[StyleData["NumberedFigure", "TwoColumn"], CellMargins->{{14, Inherited}, {Inherited, Inherited}}, FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["NumberedTable"], CellMargins->{{6, Inherited}, {Inherited, Inherited}}, CellFrameLabels->{{None, None}, {Cell[ TextData[ {"Table ", CounterBox[ "NumberedTable"]}]], None}}, CounterIncrements->"NumberedTable", FormatTypeAutoConvert->False, FontFamily->"Times"], Cell[StyleData["NumberedTable", "Presentation"], CellMargins->{{24, Inherited}, {20, Inherited}}, LineSpacing->{1, 0}], Cell[StyleData["NumberedTable", "Printout"], CellMargins->{{14, Inherited}, {Inherited, Inherited}}, FontSize->10], Cell[StyleData["NumberedTable", "TwoColumn"], CellMargins->{{14, Inherited}, {Inherited, Inherited}}, FontSize->10] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Headers and Footers", "Section"], Cell[StyleData["Header"], CellMargins->{{0, 0}, {4, 1}}, StyleMenuListing->None, FontSize->10], Cell[StyleData["Footer"], CellMargins->{{0, 0}, {0, 4}}, StyleMenuListing->None, FontSize->9], Cell[StyleData["PageNumber"], CellMargins->{{0, 0}, {4, 1}}, StyleMenuListing->None, FontFamily->"Times", FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell["Palette Styles", "Section"], Cell["\<\ The cells below define styles that define standard \ ButtonFunctions, for use in palette buttons.\ \>", "Text"], Cell[StyleData["Paste"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, After]}]&)}], Cell[StyleData["Evaluate"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], SelectionEvaluate[ FrontEnd`InputNotebook[ ], All]}]&)}], Cell[StyleData["EvaluateCell"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], FrontEnd`SelectionMove[ FrontEnd`InputNotebook[ ], All, Cell, 1], FrontEnd`SelectionEvaluateCreateCell[ FrontEnd`InputNotebook[ ], All]}]&)}], Cell[StyleData["CopyEvaluate"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionCreateCell[ FrontEnd`InputNotebook[ ], All], FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], FrontEnd`SelectionEvaluate[ FrontEnd`InputNotebook[ ], All]}]&)}], Cell[StyleData["CopyEvaluateCell"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionCreateCell[ FrontEnd`InputNotebook[ ], All], FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], FrontEnd`SelectionEvaluateCreateCell[ FrontEnd`InputNotebook[ ], All]}]&)}] }, Closed]], Cell[CellGroupData[{ Cell["Hyperlink Styles", "Section"], Cell["\<\ The cells below define styles useful for making hypertext \ ButtonBoxes. The \"Hyperlink\" style is for links within the same Notebook, \ or between Notebooks.\ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["Hyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontSize->14, FontColor->RGBColor[0, 0.392187, 0], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["Hyperlink", "Printout"], FontColor->GrayLevel[0], Background->GrayLevel[1], FontVariations->{"Underline"->False}], Cell[StyleData["Hyperlink", "TwoColumn"], FontColor->GrayLevel[0], Background->GrayLevel[1], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["MEIHyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->None, FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.650004, 0.680003, 0.0800031], Background->RGBColor[0, 0.392187, 0], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2]}]&), Active->True, ButtonFrame->"None", ButtonNote->"Come visit us!"}], Cell[StyleData["MEIHyperlink", "Printout"], FontColor->GrayLevel[0], Background->GrayLevel[1], FontVariations->{"Underline"->False}], Cell[StyleData["MEIHyperlink", "TwoColumn"], FontColor->GrayLevel[0], Background->GrayLevel[1], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["BasicsHyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->None, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.848096, 0.171878, 0.228321], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2]}]&), Active->True, ButtonFrame->"None", ButtonNote->"Go to Basics"}], Cell[StyleData["BasicsHyperlink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["BasicsHyperlink", "TwoColumn"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["BasicsIndexHyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->None, FontWeight->"Bold", FontColor->RGBColor[0.848096, 0.171878, 0.228321], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2], FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], Next, CellGroup], FrontEndToken[ "SelectionCloseAllGroups"], FrontEndToken[ "OpenCloseGroup"]}]&), Active->True, ButtonFrame->"None", ButtonNote->"Link into Basics"}], Cell[StyleData["BasicsIndexHyperlink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["BasicsIndexHyperlink", "TwoColumn"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["TutorialsHyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->None, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.0199588, 0.346716, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2]}]&), Active->True, ButtonFrame->"None", ButtonNote->"Go to Tutorials"}], Cell[StyleData["TutorialsHyperlink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["TutorialsHyperlink", "TwoColumn"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["TutorialsIndexHyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->None, FontWeight->"Bold", FontColor->RGBColor[0.0199588, 0.346716, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2], FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], Next, CellGroup], FrontEndToken[ "SelectionCloseAllGroups"], FrontEndToken[ "OpenCloseGroup"]}]&), Active->True, ButtonFrame->"None", ButtonNote->"Link into Tutorials"}], Cell[StyleData["TutorialsIndexHyperlink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["TutorialsIndexHyperlink", "TwoColumn"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["GiveItaTryHyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->None, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.497459, 0.196094, 0.543877], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2]}]&), Active->True, ButtonFrame->"None", ButtonNote->"Go to GiveItaTry"}], Cell[StyleData["GiveItaTryHyperlink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["GiveItaTryHyperlink", "TwoColumn"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["GiveItaTryIndexHyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->None, FontWeight->"Bold", FontColor->RGBColor[0.497459, 0.196094, 0.543877], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2], FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], Next, CellGroup], FrontEndToken[ "SelectionCloseAllGroups"], FrontEndToken[ "OpenCloseGroup"]}]&), Active->True, ButtonFrame->"None", ButtonNote->"Link into GiveItaTry"}], Cell[StyleData["GiveItaTryIndexHyperlink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["GiveItaTryIndexHyperlink", "TwoColumn"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["LiteracyHyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->None, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[1, 0.433326, 0], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2]}]&), Active->True, ButtonFrame->"None", ButtonNote->"Go to Literacy"}], Cell[StyleData["LiteracyHyperlink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["LiteracyHyperlink", "TwoColumn"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["LiteracyIndexHyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->None, FontWeight->"Bold", FontColor->RGBColor[1, 0.433326, 0], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2], FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], Next, CellGroup], FrontEndToken[ "SelectionCloseAllGroups"], FrontEndToken[ "OpenCloseGroup"]}]&), Active->True, ButtonFrame->"None", ButtonNote->"Link into Literacy"}], Cell[StyleData["LiteracyIndexHyperlink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["LiteracyIndexHyperlink", "TwoColumn"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["PreviewHyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->None, FontWeight->"Bold", FontSlant->"Italic", FontColor->GrayLevel[0.250004], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2], FrontEndToken[ "OpenCloseGroup"]}]&), Active->True, ButtonFrame->"None", ButtonNote->"Preview of Lesson"}], Cell[StyleData["PreviewHyperlink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["PreviewHyperlink", "TwoColumn"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["RefGuideLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Courier", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "RefGuideLink", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["RefGuideLink", "Presentation"]], Cell[StyleData["RefGuideLink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["RefGuideLink", "TwoColumn"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["GettingStartedLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "GettingStarted", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["GettingStartedLink", "Presentation"]], Cell[StyleData["GettingStartedLink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["GettingStartedLink", 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