(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.2' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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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[ 23811, 928]*) (*NotebookOutlinePosition[ 24739, 959]*) (* CellTagsIndexPosition[ 24695, 955]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Integral Calculus", "Title", Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True], Cell[TextData[{ "Sean Mauch\nsean@caltech.edu\n", ButtonBox["http://www.its.caltech.edu/~sean", ButtonData:>{ URL[ "http://www.its.caltech.edu/~sean"], None}, ButtonStyle->"Hyperlink"], "\n", "This work is distributed under the GNU FDL. See ", ButtonBox["license.nb ", ButtonData:>{"license.nb", None}, ButtonStyle->"Hyperlink"], "for details." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Indefinite Integrals" }], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "While ", StyleBox["Mathematica", FontSlant->"Italic"], " can calculate the derivative of any expression, it cannot find the \ integral of an arbitrary expression. In a sense, most functions do not have \ an integral that can be expressed in terms of elementary functions. You can \ find indefinite integrals with the ", StyleBox["Integrate[]", "Input"], " function. This function takes two arguments, the function to integrate, \ and the integration variable." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Integrate[x\^2 + Sin[x], x]\)], "Input"], Cell["\<\ To get the typist version: \t\[EscapeKey]int\[EscapeKey] (x \ \[ControlKey]\[LeftModified]6\[RightModified] 2 \[ControlKey]\[LeftModified]\ \[SpaceKey]\[RightModified] +Sin[x]) \[EscapeKey]dd\[EscapeKey] x\ \>", "Text",\ TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\[Integral]\((x\^2 + Sin[x])\) \[DifferentialD]x\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The indefinite integral of a function is only determined up to a constant. \ This constant is not shown in the output. If ", StyleBox["Mathematica", FontSlant->"Italic"], " is unable to find an expression for an integral in terms of the functions \ that it knows, it returns the input unchanged." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]Sin[Sin[x]] \[DifferentialD]x\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " knows a lot of obscure functions. (Many functions are defined in terms \ of an integral or in terms of the solution of a differential equation.) The \ result of an integration may involve such functions." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\(Sin[x]\/x\) \[DifferentialD]x\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\(?SinIntegral\)\)], "Input"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " can also integrate expressions involving formal functions." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ RowBox[{"\[Integral]", RowBox[{ RowBox[{\(f[x]\), " ", RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "[", "x", "]"}]}], \(\[DifferentialD]x\)}]}]], "Input", AspectRatioFixed->True], Cell[TextData[{ "You must use some caution when integrating expressions that involve \ parameters. When you ask ", StyleBox["Mathematica", FontSlant->"Italic"], " to integrate ", Cell[BoxData[ \(TraditionalForm\`x\^n\)]], ", it gives the ``general'' result and assumes that ", Cell[BoxData[ \(TraditionalForm\`n\ \[NotEqual] \(-1\)\)]], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\(x\^n\) \[DifferentialD]x\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "If you later substitute the value ", Cell[BoxData[ \(TraditionalForm\`n\ = \ \(-1\)\)]], " into the answer you do not get the correct value of the integral." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(% /. n \[Rule] \(-1\)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can get the correct value of the integral only by substituting ", Cell[BoxData[ \(TraditionalForm\`n\ = \ \(-1\)\)]], " before you integrate." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\((x\^n /. n \[Rule] \(-1\))\) \[DifferentialD]x\)], "Input",\ AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[TextData[{ "Evaluate ", Cell[BoxData[ \(TraditionalForm \`\[Integral]\(\((x + 2)\)\^10\) \[DifferentialD]x\)]], " by hand and using ", StyleBox["Mathematica", FontSlant->"Italic"], ". 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Now try ", Cell[BoxData[ \(TraditionalForm\`\[Integral]\(\((x + 2)\)\^n\) \[DifferentialD]x\)]], "." }], "Text", TextJustification->1] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["Solve the problem:", "Text"], Cell[BoxData[ \(TraditionalForm\`\(y'\)' - 3 y' + 2 y = sin(x), \ \ \ \ y(0) = 1, \ \ \ \ y' \((0)\) = 2. \)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "Recall that the general solution of an inhomogeneous, second order, linear \ differential equation, ", Cell[BoxData[ \(TraditionalForm\`L[y] = f(x)\)]], ", is:" }], "Text"], Cell[BoxData[ \(TraditionalForm \`y = y\_1 + y\_2 - \(\(y\_1\)(x)\) \(\[Integral]\(\(\(\(y\_2\)(x)\) \(f(x)\)\)\/\(W(x)\)\) \[DifferentialD]x\) + \(\(y\_2\)(x)\) \(\[Integral]\(\(\(\(y\_1\)(x)\) \(f(x)\)\)\/\(W(x)\)\) \[DifferentialD]x\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`y\_1\)]], " and ", Cell[BoxData[ \(TraditionalForm\`y\_2\)]], " are homogeneous solutions and ", Cell[BoxData[ \(TraditionalForm\`W(x)\)]], " is the Wronskian of ", Cell[BoxData[ \(TraditionalForm\`y\_1\)]], " and ", Cell[BoxData[ \(TraditionalForm\`y\_2\)]], "." }], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Definite Integrals" }], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "For definite integrals, you specify the integration variable and the \ limits of integration in the second argument of ", StyleBox["Integrate[]", "Input"], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Integrate[Sin[x], {x, 0, \[Pi]}]\)], "Input"], Cell["\<\ For the typeset version, \t\[EscapeKey]int\[EscapeKey] \[ControlKey]\[LeftModified]-\[RightModified] 0 \ \[ControlKey]\[LeftModified]5\[RightModified] \[EscapeKey]p\[EscapeKey] \ \[ControlKey]\[LeftModified]\[SpaceKey]\[RightModified] Sin[x] \[EscapeKey]dd\ \[EscapeKey] x or \t\[EscapeKey]int\[EscapeKey] \[ControlKey]\[LeftModified]-\[RightModified] 0 \ \[ControlKey]\[LeftModified]\[SpaceKey]\[RightModified] \[ControlKey]\ \[LeftModified]6\[RightModified] \[EscapeKey]p\[EscapeKey] \[ControlKey]\ \[LeftModified]\[SpaceKey]\[RightModified] Sin[x] \[EscapeKey]dd\[EscapeKey] \ x\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\[Integral]\_0\%\[Pi] Sin[x] \[DifferentialD]x\)], "Input"], Cell[TextData[{ "Again, if ", StyleBox["Mathematica", FontSlant->"Italic"], " is unable to evaluate the integral exactly in terms of the functions it \ knows, it will return the input unchanged." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\_0\%\[Pi] Sin[Sin[x]] \[DifferentialD]x\)], "Input", AspectRatioFixed->True], Cell["\<\ However, you can get a numerical approximation to such a definite \ integral with the numerical function.\ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(N[%]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "If you want a numerical approximation in the first place, use the ", StyleBox["NIntegrate[]", "Input"], " function." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(NIntegrate[Sin[Sin[x]], {x, 0, \[Pi]}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " will notify you if an integral does not converge." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\_\(-1\)\%1\( 1\/x\) \[DifferentialD]x\)], "Input", AspectRatioFixed->True], Cell["\<\ Though the integral diverges, the principal value of the integral \ exists.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Integrate[1\/x, {x, \(-1\), 1}, PrincipalValue -> True]\)], "Input"], Cell[TextData[{ "You must be careful when you do definite integrals where the limits of \ integration are parameters. Below ", StyleBox["Mathematica", FontSlant->"Italic"], " warns you that the integral may not converge for all values of the \ parameters." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\_a\%b\( 1\/x\^2\) \[DifferentialD]x\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Indeed, if you substitute the limits ", Cell[BoxData[ \(TraditionalForm\`a\ = \ \(-1\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`b\ = \ 1\)]], ", you get an incorrect result." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(% /. {a \[Rule] \(-1\), b \[Rule] 1}\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "If ", StyleBox["Mathematica", FontSlant->"Italic"], " had known the limits of integration, it would have warned that the \ integral diverges." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\_\(-1\)\%1\( 1\/x\^2\) \[DifferentialD]x\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "If you were to do the integral ", Cell[BoxData[ \(TraditionalForm \`\[Integral]\_0\%\[Infinity]\ \(\[ExponentialE]\^\(\(-c\)\ x\)\) \[DifferentialD]x\)]], ", you would probably assume that ", Cell[BoxData[ \(TraditionalForm\`Re(c) > 0\)]], ", however ", StyleBox["Mathematica ", FontSlant->"Italic"], "will not make this assumption, it returns a result that is valid for all \ values of ", Cell[BoxData[ \(TraditionalForm\`c\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\[Integral]\_0\%\[Infinity] Exp[\(-c\)\ x] \[DifferentialD]x\)], "Input"], Cell[TextData[{ "You can tell ", StyleBox["Mathematica", FontSlant->"Italic"], " to make assumptions about the parameters in an integral. Here we \ indicate that the parameter, ", Cell[BoxData[ \(TraditionalForm\`c\)]], ", has positive real part." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Integrate[Exp[\(-c\)\ x], {x, 0, \[Infinity]}, Assumptions -> Re[c] > 0] \)], "Input"], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[TextData[{ "Find the area of the region below the curve ", Cell[BoxData[ \(TraditionalForm\`y = \(-x\^2\) - 2 x + 10\)]], " and above the curve ", Cell[BoxData[ \(TraditionalForm\`y = 2 x - 1\)]], "." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[TextData[{ "For what values of \[Omega] does the integral of the Fourier transform of \ ", Cell[BoxData[ \(TraditionalForm\`1/\((x\^2 + 1)\)\)]], " converge? What is the value of the Fourier transform in the domain of \ convergence? 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Now try ", Cell[BoxData[ \(TraditionalForm\`\[Integral]\(\((x + 2)\)\^n\) \[DifferentialD]x\)]], "." }], "Text", TextJustification->1], Cell["By inspection we see that", "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm \`\[Integral]\(\((x + 2)\)\^10\) \[DifferentialD]x = \((x + 2)\)\^11\/11 + \(c . \)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "When ", StyleBox["Mathematica", FontSlant->"Italic"], " does the integral it expand the integrand first. " }], "Text", TextJustification->1], Cell[BoxData[ \(\[Integral]\(\((x + 2)\)\^10\) \[DifferentialD]x\)], "Input"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], "'s answer differs from ours by a constant." }], "Text", TextJustification->1], Cell[BoxData[ \(% - \((x + 2)\)\^11\/11 // Expand\)], "Input"], Cell[TextData[{ "If you asked ", StyleBox["Mathematica", FontSlant->"Italic"], " to integrate ", Cell[BoxData[ \(TraditionalForm\`\((x + 2)\)\^1000000000\)]], " it would expand the integrand into 1000000001 terms before doing the \ integration. If your computer has the system resources for this computation \ it would take quite a while. Whether or not your computer can handle it, ", StyleBox["Mathematica", FontSlant->"Italic"], " would give it a try. Sometimes it is a real glutton for punishment. Now \ you can tell all your friends that you can integrate faster than ", StyleBox["Mathematica", FontSlant->"Italic"], ". Note that ", StyleBox["Mathematica", FontSlant->"Italic"], " is aware of the power rule." }], "Text", TextJustification->1], Cell[BoxData[ \(\[Integral]\(\((x + 2)\)\^n\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(% // Simplify\)], "Input"], Cell["\<\ It's just that it's rule for integration tell it to \"simplify\" \ the integrand before doing the integration.\ \>", "Text", TextJustification->1] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["We consider the problem", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(TraditionalForm\`\(y'\)' - 3 y' + 2 y = sin(x), \ \ \ \ y(0) = 1, \ \ \ \ y' \((0)\) = 2. \)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "First we substitute ", Cell[BoxData[ \(TraditionalForm\`y = \[ExponentialE]\^\(\[Alpha]\ x\)\)]], " into the differential equation to obtain a quadratic equation for \ \[Alpha]." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(y[x_] := Exp[\[Alpha]\ x]; \n \(\(y'\)'\)[x] - 3 \( y'\)[x] + 2 y[x]\)], "Input"], Cell["The roots of the quadratic are", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Solve[% == 0, \[Alpha]]\)], "Input"], Cell["\<\ A set of homogeneous solutions of the differential equation \ are\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(y[x] /. %\)], "Input"], Cell[TextData[{ "We denote the homogeneous set of solutions ", Cell[BoxData[ \(TraditionalForm\`{y\_1, y\_2}\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(y1[x_] = %\[LeftDoubleBracket]1\[RightDoubleBracket]; \n y2[x_] = %\[LeftDoubleBracket]2\[RightDoubleBracket]; \)], "Input"], Cell["The Wronskian of the homogeneous solutions is", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ RowBox[{\(w[x_]\), "=", RowBox[{"Det", "[", RowBox[{"(", GridBox[{ {\(y1[x]\), \(y2[x]\)}, {\(\(y1'\)[x]\), \(\(y2'\)[x]\)} }], ")"}], "]"}]}]], "Input"], Cell["\<\ Now we use variation of parameters to find a particular solution of \ the differential equation.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(-y1[x]\) \(\[Integral]\(\(y2[x] Sin[x]\)\/w[x]\) \[DifferentialD]x\) + y2[x] \(\[Integral]\(\(y1[x] Sin[x]\)\/w[x]\) \[DifferentialD]x\) // Simplify\)], "Input"], Cell["\<\ The general solution of the differential equation is a linear \ combination of the two homogeneous solutions plus the particular \ solution.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(y[x_] = c1\ y1[x] + c2\ y2[x] + %\)], "Input"], Cell["\<\ We apply the initial conditions to find the constants in the linear \ combination.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Solve[{y[0] == 1, \(y'\)[0] == 2}, {c1, c2}]\)], "Input"], Cell["\<\ The solution of the differential equation subject to the initial \ conditions is\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(y[x_] = y[x] /. First[%]\)], "Input"], Cell[TextData[{ "The solution is oscillatory for negative ", Cell[BoxData[ \(TraditionalForm\`x\)]], " and exponentially growing for positive ", Cell[BoxData[ \(TraditionalForm\`x\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Plot[y[x], {x, \(-15\), 1}, PlotRange -> All]; \)\)], "Input"], Cell[BoxData[ \(Clear[y, y1, y2]\)], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Definite Integrals" }], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[TextData["First we find where the curves intersect."], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(endPts = x /. Solve[\(-x\^2\) - 2\ x + 10 \[Equal] 2\ x - 1]\)], "Input",\ AspectRatioFixed->True], Cell[TextData["The approximate value of the intersections are:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(TableForm[N[endPts]]\)], "Input", AspectRatioFixed->True], Cell[TextData["Here is a plot of the region."], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(Plot[{\(-x\^2\) - 2\ x + 10, 2\ x - 1}, {x, \(-6\), 2}]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData["Now we find the area of the region."], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\_\(endPts[\([1]\)]\)\%\(endPts[\([2]\)]\)\((\(-x\^2\) - 2\ x + 10 - \((2\ x - 1)\))\) \[DifferentialD]x\)], "Input", AspectRatioFixed->True], Cell[TextData[ "Mathematica is able to simplify the above result considerably."], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Simplify[%]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Clear[endPts]\)], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["First we evaluate the Fourier transform integral.", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\[Integral]\_\(-\[Infinity]\)\%\[Infinity]\(( \(1\/\(2 \[Pi]\)\) Exp[\(-I\)\ \[Omega]\ x]\/\(x\^2 + 1\))\) \[DifferentialD]x\)], "Input"], Cell[TextData[{ "We see that it converges for ", Cell[BoxData[ \(TraditionalForm\`Im(\[Omega]) = 0\)]], ", that is, when \[Omega] is real. 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Notebook file. *******************************************************************)