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Notebook[{

Cell[CellGroupData[{
Cell["\<\
Examples for Section 7.2
Integration by Parts\
\>", "Subsection"],

Cell["p. 567", "Subsubsection"],

Cell[CellGroupData[{

Cell["5.", "Subsubsection"],

Cell[TextData[{
  "Set ",
  Cell[BoxData[
      \(TraditionalForm\`u\  = \ ln(x)\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`\[DifferentialD]v\  = \ x\ \[DifferentialD]x\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`du\  = \ 1/x\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`v\  = \ \(1\/2\) x\^2\)]],
  ". Then\n\[Integral]x ln(x) \[DifferentialD]x = ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{
            FormBox[\(\(1\/2\) \(x\^2\) \(ln(x)\)\  - \ \[Integral]\),
              "TraditionalForm"], \((\(1\/2\) x\^2)\), \((1\/x)\), " ", 
            \(\[DifferentialD]x\)}], " ", "=", " ", 
          \(\(1\/2\) \(x\^2\) \(ln(x)\)\  - \(1\/4\) x\^4 + \ C\)}], 
        TraditionalForm]]]
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\[Integral]x\ Log[x] \[DifferentialD]x\)], "Input"],

Cell[BoxData[
    \(\(-\(x\^2\/4\)\) + 1\/2\ x\^2\ Log[x]\)], "Output"]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["15.", "Subsubsection"],

Cell[TextData[{
  "Use tabular integration (see p. 567): The derivatives of ",
  Cell[BoxData[
      \(TraditionalForm\`x\^5\)]],
  "are ",
  Cell[BoxData[
      \(TraditionalForm\`5  x\^4\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`20  x\^3\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`60  x\^2\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`120  x\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`120\)]],
  ". The integrals of ",
  Cell[BoxData[
      \(TraditionalForm\`e\^x\)]],
  "are all ",
  Cell[BoxData[
      \(TraditionalForm\`e\^x\)]],
  ". Pair the derivative with the integrals (\"diagonally\") with alternating \
signs to get: ",
  Cell[BoxData[
      \(TraditionalForm
      \`\[Integral]\(x\^5\) \(e\^x\) \[DifferentialD]x\  = \ 
        \(x\^5\) e\^x - \ 5 \( x\^4\) e\^x + \ 20 \( x\^3\) e\^x - 
          60 \( x\^2\) e\^x + \ 120  x\ e\^x - \ 120  e\^x\)]]
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(\ \[Integral]\(x\^5\) \(E\^x\) \[DifferentialD]x\ \)\)], "Input"],

Cell[BoxData[
    \(E\^x\ 
      \((\(-120\) + 120\ x - 60\ x\^2 + 20\ x\^3 - 5\ x\^4 + x\^5)\)\)], 
  "Output"]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["21.", "Subsubsection"],

Cell[TextData[{
  "Let ",
  Cell[BoxData[
      \(TraditionalForm\`u\  = \ e\^\[Theta]\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`\[DifferentialD]v\  = \ sin(\[Theta])\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm
      \`\[DifferentialD]u\  = \ \(e\^\[Theta]\) \[DifferentialD]\[Theta]\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`v\  = \ \(-\(cos(\[Theta])\)\)\)]],
  ". Then ",
  Cell[BoxData[
      \(TraditionalForm
      \`\[Integral]\(e\^\[Theta]\) \(sin(\[Theta])\) 
            \[DifferentialD]\[Theta]\  = \ 
        \(-e\^\[Theta]\) \(cos(\[Theta])\)\  + \ 
          \[Integral]\(e\^\[Theta]\) \(cos(\[Theta])\) 
              \[DifferentialD]\[Theta]\)]],
  ". Repeat with ",
  Cell[BoxData[
      \(TraditionalForm\`u\  = \ e\^\[Theta]\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm
      \`\[DifferentialD]v\  = \ 
        \(cos(\[Theta])\) \[DifferentialD]\[Theta]\)]],
  ", to get",
  Cell[BoxData[
      \(TraditionalForm
      \`\[Integral]\(e\^\[Theta]\) \(sin(\[Theta])\) 
            \[DifferentialD]\[Theta]\  = \ 
        \(\(-e\^\[Theta]\) \(cos(\[Theta])\)\  + \ 
            \[Integral]\(e\^\[Theta]\) \(cos(\[Theta])\) 
                \[DifferentialD]\[Theta]\  = 
          \(-e\^\[Theta]\) \(cos(\[Theta])\)\ \  + \ 
            \(e\^\[Theta]\) \(sin(\[Theta])\)\ \  - 
            \[Integral]\(e\^\[Theta]\) \(sin(\[Theta])\) 
                \[DifferentialD]\[Theta]\)\)]],
  " so ",
  Cell[BoxData[
      \(TraditionalForm
      \`\(2 \(\[Integral]\(e\^\[Theta]\) \(sin(\[Theta])\) 
              \[DifferentialD]\[Theta]\)\  = 
        \(-e\^\[Theta]\) \(cos(\[Theta])\)\ \  + \ 
          \(e\^\[Theta]\) \(sin(\[Theta])\)\  + \ C\_0\ \ \)\)]],
  "and ",
  Cell[BoxData[
      \(TraditionalForm
      \`\[Integral]\(e\^\[Theta]\) \(sin(\[Theta])\) 
            \[DifferentialD]\[Theta]\  = 
        \(1\/2\) \(\(e\^\[Theta]\)(sin(\[Theta]) - cos(\[Theta]))\)\  + \ C
          \)]],
  "."
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\[Integral]\(E\^\[Theta]\) Sin[\[Theta]] \[DifferentialD]\[Theta]\)], 
  "Input"],

Cell[BoxData[
    \(\(-\(1\/2\)\)\ E\^\[Theta]\ \((Cos[\[Theta]] - Sin[\[Theta]])\)\)], 
  "Output"]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["29.", "Subsubsection"],

Cell[TextData[{
  "Substitute ",
  Cell[BoxData[
      \(TraditionalForm\`u = ln(x)\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`\[DifferentialD]u = \(1\/x\) \[DifferentialD]x\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm
      \`\[DifferentialD]x = 
        \(x\ \[DifferentialD]u = \(e\^u\) \[DifferentialD]u\)\)]],
  ", to get ",
  Cell[BoxData[
      \(TraditionalForm
      \`\[Integral]\(sin(ln(x))\) \[DifferentialD]x = 
        \(\[Integral]\(sin(u)\) \(e\^u\) \[DifferentialD]u = 
          \(1\/2\) \(\(e\^u\)(sin(u) - cos(u))\)\  + \ C\)\)]],
  " (see problem 21). Substituting back ",
  Cell[BoxData[
      \(TraditionalForm\`\(u\  = \ ln(x)\ \)\)]],
  "gives ",
  Cell[BoxData[
      \(TraditionalForm
      \`\(\(\[Integral]\(sin(ln(x))\) \[DifferentialD]x = 
                  \(1\/2\) \(\(e\^\(ln(x)\)\)(sin(ln(x)) - cos(ln(x)))\))
                \)\  + \ C\  = \(1\/2\) \(x(sin(ln(x)) - cos(ln(x)))\))\)\  + 
        \ C\)]]
}], "Text"],

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Cell[BoxData[
    \(\(\ \[Integral]Sin[Log[x]] \[DifferentialD]x\)\)], "Input"],

Cell[BoxData[
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}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["42.", "Subsubsection"],

Cell[TextData[{
  "The average value of a function ",
  Cell[BoxData[
      \(TraditionalForm\`f(x)\)]],
  " on an interval ",
  Cell[BoxData[
      \(TraditionalForm\`\([a, b]\)\)]],
  " is given by ",
  Cell[BoxData[
      \(TraditionalForm
      \`avg(f)\  = \ 
        \(1\/\(b - a\)\) 
          \(\[Integral]\_a\%b\( f(x)\) \[DifferentialD]x\)\)]],
  ". So the average value of ",
  Cell[BoxData[
      \(TraditionalForm
      \`f(t)\  = \ 4 \(\( e\^\(-t\)\)(sin(t) - cos(t))\)\)]],
  " on ",
  Cell[BoxData[
      \(TraditionalForm\`\([0, 2  \[Pi]]\)\)]],
  " is ",
  Cell[BoxData[
      \(TraditionalForm
      \`\(1\/\(2  \[Pi]\)\) 
        \(\[Integral]4 \(\( e\^\(-t\)\)(sin(t) - cos(t))\) 
            \[DifferentialD]t\)\)]],
  ". Using integration by parts, we get ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{
            FormBox[
              RowBox[{
                RowBox[{
                  FormBox[\(avg(f)\  = \(-\ \(2\/\[Pi]\)\)\),
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              "TraditionalForm"], \(|\_0\%\(2  \[Pi]\)\)}], "=", " ", "0"}], 
        TraditionalForm]]],
  "."
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(\ \(1\/\(2  \[Pi]\)\) 
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Cell[BoxData[
    \(\(-\(\(2\ E\^\(-t\)\ Sin[t]\)\/\[Pi]\)\)\)], "Output"]
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Cell[CellGroupData[{

Cell[BoxData[
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