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Cell["\<\
Summing It up with Riemann, Definite Integrals, and the Fundamental Theorem \
of Calculus\
\>", "Title",
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Cell["Introduction", "Section",
  PageWidth->PaperWidth],

Cell["\<\
OBJECTIVE: Visualize the relationship among Riemann sums, the definite \
integral, and the Fundamental Theorem of the Calculus. Use each of the \
concepts to compute or approximate the signed area under the graph of a \
function representing various applications.\
\>", "Text",
  PageWidth->PaperWidth],

Cell[TextData[{
  "In practical applications, the signed area between the graph of a function \
and the ",
  StyleBox["x ",
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  "or ",
  StyleBox["t",
    FontSlant->"Italic"],
  "-axis",
  StyleBox[" ",
    FontSlant->"Italic"],
  "can represent a wide variety of different things. For example, if a \
function of time represents the rate of water flowing into a storage tank, \
the total signed area between the graph of the flow rate function and the ",
  StyleBox["t",
    FontSlant->"Italic"],
  "-axis is equal to the total amount of water that accumulates or drains \
from the tank during the time interval. To see that this is the case, you \
should recall something that you have probably known for a long time, that \
is, that change in a quantity is equal to its rate of change times time. Some \
familiar examples of this are \"distance is equal to speed times time\" and \
\"amount is equal to rate times time.\" If water flows into a tank at a \
constant rate of 30 liters per second for 100 seconds, then it easy to \
determine that the total water added to the tank during this time is 300\
\[Times]100 = 30000 liters. These simple rules are based, of course, on the \
assumption that the rate of change is constant during the time interval. Now \
that we are more sophisticated, we know that rates of change do not have to \
be constant and can, in fact, vary continuously over a time interval. So, how \
do we calculate the total change over a time interval when the rate varies \
from one instant to the next? Calculus provides the answer, and we begin by \
looking at a technique for estimating change from rates of change using a \
summation process called Riemann sums."
}], "Text",
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evaluate all the initializations cells in the \tnotebook . . . ,\" respond by \
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Cell["Part I: An Example", "Section",
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Cell[TextData[{
  "The idea of Riemann sums builds upon what you already know, that is that \
change is equal to rate of change times time. Suppose that ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") represents the rate of change of some quantity ",
  StyleBox["Q",
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  Cell[BoxData[
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        RowBox[{
          FractionBox[
            StyleBox[
              RowBox[{"d", 
                StyleBox["Q",
                  FontSlant->"Italic"]}]], 
            StyleBox["dt",
              FontSlant->"Italic"]], "=", \(f(t)\)}], TraditionalForm]]],
  ". The quantity ",
  StyleBox["Q",
    FontSlant->"Italic"],
  " might represent the amount of water in a reservoir, the distance traveled \
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the forces that influence change and thereby determine ",
  StyleBox["f",
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  "(",
  StyleBox["t",
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  StyleBox["t",
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  "), there is a way to estimate the change in ",
  StyleBox["Q ",
    FontSlant->"Italic"],
  "over an interval of time, even though the rate of change varies from one \
instant to the next. Here it is."
}], "Text",
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  "Let's consider a specific example. Suppose that ",
  Cell[BoxData[
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          FractionBox[
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            StyleBox["dt",
              FontSlant->"Italic"]], "=", " "}], TraditionalForm]]],
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    FontSlant->"Italic"],
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          RowBox[{"cos", "(", 
            FractionBox[
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                StyleBox["t",
                  FontSlant->"Italic"]}], "12"], ")"}]}], TraditionalForm]]],
  " for 0\[LessEqual] ",
  StyleBox["t ",
    FontSlant->"Italic"],
  "\[LessEqual] 15  and that we wish to calculate \[CapitalDelta]",
  StyleBox["Q ",
    FontSlant->"Italic"],
  "over this time interval. First, let's graph ",
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  "(",
  StyleBox["t",
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  ").\n\nIf you are running Version 4 of ",
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  ", you will get an error message when the initialization cells are executed \
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    \(\(f[t_] = 1\/4 - Cos[\(\[Pi]*t\)\/12];\)\), "\n", 
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        PlotStyle \[Rule] {RGBColor[0, 1, 0]}];\)\)}], "Input",
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  "We know from the definition of differentials that over a short interval of \
time \[CapitalDelta]",
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  "This suggests a method of attack. If we partition the 15-hour period into \
very short time intervals, called subintervals, and then approximate the \
change in ",
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    FontSlant->"Italic"],
  " over each of these subintervals and add them up, we should get an \
approximation of the total \[CapitalDelta]",
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  "(",
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period into ",
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  "Then, \[CapitalDelta]",
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    FontSlant->"Italic"],
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                    StyleBox["Q",
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  "There is a geometric view that can be attached to the Riemann sum, ",
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  ". The product, ",
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  ", can be thought of as the area of a rectangle that is ",
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  " units high. In each subinterval, we can draw this rectangle that is ",
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  " units wide and ",
  Cell[BoxData[
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  " units high. The base of each rectangle is on the ",
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  " or ",
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    FontSlant->"Italic"],
  " axis, and the opposite side is on the graph of ",
  StyleBox["f",
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  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "). The Riemann sum is the sum of the areas of these rectangles. If we \
admit that ",
  Cell[BoxData[
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  " could be negative, then the area of a rectangle can be negative; hence, \
we add signed areas in a Riemann sum, which itself can be positive, negative, \
or zero. The product can be negative when ",
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  " is positive and ",
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    FontSlant->"Italic"],
  " is negative. This would occur if we were to sum back through the time \
history of ",
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  "(",
  StyleBox["t",
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  "). It makes sense that if ",
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  " increases as time progresses, then it would decrease if we were to go \
backwards through the same time interval. And, if ",
  StyleBox["Q ",
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  "decreases as time progresses, then it would increase if we were to go \
backwards through the same time interval.\n\nTo illustrate the geometric view \
of a Riemann sum, we have written a command called ",
  StyleBox["riemannleft[f_, t_, a_, b_, n_ ,window_]",
    FontWeight->"Bold"],
  ". The arguments are ",
  StyleBox["f",
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  ", the function ",
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    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "), ",
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    FontWeight->"Bold"],
  "the independent variable, ",
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    FontWeight->"Bold"],
  " the left endpoint of the interval, ",
  StyleBox["b",
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  ", the right endpoint of the interval, ",
  StyleBox["n",
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  ", the number of rectangles to be used, and ",
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    FontWeight->"Bold"],
  ", the viewing window for the graph. The ",
  StyleBox["riemannleft[ ] ",
    FontWeight->"Bold"],
  "command uses the left endpoint value in each subinterval for ",
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      \(TraditionalForm\`c\_i\)]],
  ".\n\nLet's use ",
  StyleBox["riemannleft[ ] ",
    FontWeight->"Bold"],
  "on the function we have been considering and use 15 rectangles."
}], "Text",
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Cell[BoxData[
    \(riemannleft[f[t], t, 0, 15, 15, {{0, 16}, {\(-1\), 1.5}}]\)], "Input",
  PageWidth->PaperWidth],

Cell[TextData[{
  "Note that the blue rectangles have negative areas, whereas the red ones \
have positive areas. Try changing the number of rectangles in the ",
  StyleBox["riemannleft[ ]",
    FontWeight->"Bold"],
  " command to see that the total signed area of the rectangles gets closer \
to the signed area under the graph of ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "). You can also change the function and the bounds on the interval to \
explore other possibilities if you wish. See what happens when you reverse \
the bounds on the time interval.\n"
}], "Text",
  PageWidth->PaperWidth]
}, Closed]],

Cell[CellGroupData[{

Cell["You Try It: Part I", "Section",
  PageWidth->PaperWidth],

Cell[TextData[{
  "In addition to ",
  StyleBox["riemannleft[ ]",
    FontWeight->"Bold"],
  ", we have written ",
  StyleBox["riemannright[ ] ",
    FontWeight->"Bold"],
  "and ",
  StyleBox["riemannmiddle[ ]",
    FontWeight->"Bold"],
  " commands so that you can see the geometry of a Riemann sum when we pick \
points other than the left endpoint in each subinterval. The arguments are \
the same as for ",
  StyleBox["riemannleft[ ]",
    FontWeight->"Bold"],
  ". Use these commands to visualize the Riemann sums for the function \
considered in the preceding section and for some other functions that you \
choose. Here is an example. First, run this function, and then try it out \
with one of your own by changing the terms in red. Be careful to use correct \
notation when inserting a function."
}], "Text",
  PageWidth->PaperWidth],

Cell[BoxData[{\(Clear[f, x]\), "\n", 
    RowBox[{
      RowBox[{\(f[x]\), "=", 
        StyleBox[\(\(-3\) x\^3\),
          FontColor->RGBColor[1, 0, 0]]}], ";"}], "\n", 
    RowBox[{
      RowBox[{"a", "=", 
        StyleBox[\(-2\),
          FontColor->RGBColor[1, 0, 0]]}], ";"}], "\n", 
    RowBox[{
      RowBox[{"b", "=", 
        StyleBox["2",
          FontColor->RGBColor[1, 0, 0]]}], ";"}], "\n", 
    RowBox[{
      RowBox[{"n", "=", 
        StyleBox["20",
          FontColor->RGBColor[1, 0, 0]]}], ";"}], "\n", 
    RowBox[{
      RowBox[{"window", "=", 
        StyleBox[\({{\(-2\), 2}, {\(-24\), 24}}\),
          FontColor->RGBColor[1, 0, 0]]}], 
      ";"}], "\n", \(riemannleft[f[x], x, a, b, n, 
        window];\), "\n", \(riemannmiddle[f[x], x, a, b, n, 
        window];\), "\n", \(riemannright[f[x], x, a, b, n, 
        window];\)}], "Input",
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Cell["\<\
Geometrically, which looks as if it would give the best results? Try it with \
another function.\
\>", "Text",
  PageWidth->PaperWidth]
}, Closed]],

Cell[CellGroupData[{

Cell["Part II: Summing It Up with Riemann", "Section",
  PageWidth->PaperWidth],

Cell[TextData[{
  "We need to calculate the total signed area of the rectangles. First we \
calculate ",
  StyleBox["h",
    FontSlant->"Italic"],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`c\_i\)]],
  " for each of the subintervals, and then we add up the signed areas of the \
rectangles. We use the left endpoint in each interval for ",
  Cell[BoxData[
      \(TraditionalForm\`c\_i\)]],
  "."
}], "Text",
  PageWidth->PaperWidth],

Cell[BoxData[{
    \(\(Clear[f, h, a, b, n, t];\)\), "\n", 
    \(\(f[t_] = 1\/4 - Cos[\(\[Pi]*t\)\/12];\)\), "\n", 
    \(\(n = 15;\)\), "\n", 
    \(\(a = 0;\)\), "\n", 
    \(\(b = 15;\)\), "\n", 
    \(\(h = \((b - a)\)/n;\)\), "\n", 
    \(\(c[i_] = a + i*h;\)\), "\n", 
    \(area[n_] = 
      Sum[f[c[i]]*h, {i, 0, 
          n - 1}]; \  (*This\ is\ the\ Riemann\ \(\(sum\)\(.\)\)\
*) \[IndentingNewLine]Print["\<The total signed area of the rectangles is \
\>", area[n] // N]\)}], "Input",
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Cell[TextData[{
  "To see what happens as we increase the number of rectangles, we combine \
the ",
  StyleBox["riemannleft[ ] ",
    FontWeight->"Bold"],
  "command together with the commands in the preceding cell and put them all \
in a single cell, the one that follows. With this we can easily change ",
  StyleBox["f[t]",
    FontWeight->"Bold"],
  ", ",
  StyleBox["n",
    FontWeight->"Bold"],
  ", ",
  StyleBox["a",
    FontWeight->"Bold"],
  ", and/or ",
  StyleBox["b",
    FontWeight->"Bold"],
  "."
}], "Text",
  PageWidth->PaperWidth],

Cell[BoxData[{
    \(\(Clear[f, a, b, n, h];\)\), "\n", 
    \(\(f[t_] = 1\/4 - Cos[\(\[Pi]*t\)\/12];\)\), "\n", 
    \(\(n = 30;\)\), "\n", 
    \(\(a = 0;\)\), "\n", 
    \(\(b = 15;\)\), "\n", 
    \(\(h = \((b - a)\)/n;\)\), "\n", 
    \(\(c[i_] = a + i*h;\)\), "\n", 
    \(\(area[n_] = Sum[f[c[i]]*h, {i, 0, n - 1}];\)\), "\n", 
    \(\(Print["\<The total signed area of the rectangles is \>", 
        area[n] // N];\)\), "\n", 
    \(\(riemannleft[f[t], t, a, b, 
        n, {{0, 16}, {\(-1\), 1.5}}];\)\)}], "Input",
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Cell[TextData[{
  "If we let the number of subintervals and corresponding rectangles approach \
infinity, we might expect the Riemann sum to approach the actual signed area \
under the graph of ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "), provided that ",
  StyleBox["f",
    FontSlant->"Italic"],
  " is continuous on [",
  StyleBox["a",
    FontSlant->"Italic"],
  ", ",
  StyleBox["b",
    FontSlant->"Italic"],
  "]",
  StyleBox[". ",
    FontSlant->"Italic"],
  "The value that the Riemann sum approaches is exactly equal to \
\[CapitalDelta]",
  StyleBox["Q",
    FontSlant->"Italic"],
  " over the interval."
}], "Text",
  PageWidth->PaperWidth]
}, Closed]],

Cell[CellGroupData[{

Cell["You Try It: Part II", "Section",
  PageWidth->PaperWidth],

Cell["\<\
Try out your own function by changing the terms in red. Be careful to use the \
correct notation. Replace any of the terms in red with functions and domains \
of your own choosing.\
\>", "Text",
  PageWidth->PaperWidth],

Cell[BoxData[{\(Clear[f, a, b, n, h, x];\), "\n", 
    RowBox[{
      RowBox[{\(f[x_]\), "=", 
        StyleBox[\(\(1\/\@\(2  \[Pi]\)\) Exp[\(-x\^2\)/2]\),
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    RowBox[{
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        StyleBox["20",
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    RowBox[{
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        StyleBox[\(-2\),
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        StyleBox["2",
          FontColor->RGBColor[1, 0, 0]]}], ";"}], "\n", 
    RowBox[{
      RowBox[{"window", "=", 
        StyleBox[\({{\(-3\), 3}, {0, 0.5}}\),
          FontColor->RGBColor[1, 0, 0]]}], 
      ";"}], "\n", \(h = \((b - a)\)/n;\), "\n", \(c[i_] = 
        a + i*h;\), "\n", \(area[n_] = 
        Sum[f[c[i]]*h, {i, 0, 
            n - 1}];\), "\n", \(Print["\<The total signed area of the \
rectangles is \>", area[n] // N];\), "\n", \(riemannleft[f[x], x, a, b, n, 
        window];\)}], "Input",
  PageWidth->PaperWidth]
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Cell[CellGroupData[{

Cell[TextData[{
  "Part III: Take it to the Limit ",
  Cell[BoxData[
      \(TraditionalForm\`-\)]],
  " The Definite Integral"
}], "Section",
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Cell[TextData[{
  "To evaluate the limit of the Riemann sum, we form the sum without \
assigning a specific value to ",
  StyleBox["n ",
    FontSlant->"Italic"],
  "and then take the limit as ",
  StyleBox["n",
    FontSlant->"Italic"],
  " goes to infinity. In the first cell that follows, we use ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " to find an expression for the Riemann sum in terms of ",
  StyleBox["n",
    FontSlant->"Italic"],
  ",",
  " and in the next cell we take the limit."
}], "Text",
  PageWidth->PaperWidth],

Cell[BoxData[{
    \(\(Clear[f, a, b, n, h];\)\), "\n", 
    \(\(f[t_] = 1\/4 - Cos[\(\[Pi]*t\)\/12];\)\), "\n", 
    \(\(a = 0;\)\), "\n", 
    \(\(b = 15;\)\), "\n", 
    \(\(h = \((b - a)\)/n;\)\), "\n", 
    \(\(c[i_] = a + i*h;\)\), "\n", 
    \(riemannSum[n_] = \(Sum[f[c[i]]*h, {i, 0, n - 1}] // ExpToTrig\) // 
        Simplify\)}], "Input",
  PageWidth->PaperWidth],

Cell["Now we take the limit.", "Text",
  PageWidth->PaperWidth],

Cell[BoxData[
    \(deltaQ = 
      Limit[riemannSum[n], n \[Rule] \[Infinity]] // Simplify\)], "Input",
  PageWidth->PaperWidth],

Cell[TextData[{
  "This is the exact value for the signed area under the graph of ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "), and hence it is also the exact value for \[CapitalDelta]",
  StyleBox["Q",
    FontSlant->"Italic"],
  ". The limit of a Riemann sum is called a definite integral and is \
specified by the following notation, \n\n",
  Cell[BoxData[
      \(TraditionalForm\`\[Integral]\_a\%b\( f(
              t)\) \[DifferentialD]t = \(\(\(lim\)\(\ \)\)\+\(h \[Rule] 
                0\)\) \(\[Sum]\+\(i = 1\)\%n\( f(c\_i)\) h\)\)]],
  ". \n\nWe can use ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " to evaluate definite integrals as in the following cell."
}], "Text",
  PageWidth->PaperWidth],

Cell[BoxData[
    \(deltaQ = \[Integral]\_0\%15 f[t] \[DifferentialD]t\)], "Input",
  PageWidth->PaperWidth],

Cell["\<\
To compare this result with the values of the Riemann sums calculated above, \
we express the preceding result as a decimal approximation.\
\>", "Text",
  PageWidth->PaperWidth],

Cell[BoxData[
    \(deltaQ // N\)], "Input",
  PageWidth->PaperWidth],

Cell[TextData[{
  "It is important to remember that a definite integral is defined as the \
limit of a Riemann sum. We are sure that the definite integral of a function \
will exist (i.e., the limit of the Riemann sum will be a finite number), \
provided that ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") is continuous on the closed interval over which we perform the \
summation."
}], "Text",
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}, Closed]],

Cell[CellGroupData[{

Cell["You Try It: Part III", "Section",
  PageWidth->PaperWidth],

Cell[TextData[{
  StyleBox["Exercise 1",
    FontWeight->"Bold"],
  "\nFor each of the following functions, use the group of commands copied \
into the cell below to estimate the definite integral using a Riemann sum. \
First use the ",
  StyleBox["riemannleft[ ]",
    FontWeight->"Bold"],
  " command to show the graph and some rectangles (use fewer than 50 \
rectangles for a clear image). Then use the commands in the next cell to see \
if you can make an estimate of the definite integral that is accurate to 3 \
digits by varying ",
  StyleBox["n",
    FontSlant->"Italic"],
  ". Make note of how many subintervals you need to use to do this. After you \
make an estimate, use the definite integral command to see ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  "'s decimal estimate of the exact value, and compare it with yours. Here \
are some functions to try. We show (c) as an example and find by trial and \
error that 5853 rectangles are needed to get three-digit accuracy.\n\na)  ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") = ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{"3", 
            FormBox[\(t\^2\),
              "TraditionalForm"]}], "-", " ", "6"}], TraditionalForm]]],
  "  for  ",
  Cell[BoxData[
      \(TraditionalForm\`\(-1\) \[LessEqual] \ t\  \[LessEqual] \ 3\)]],
  "\n\nb)   ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") =2 sin(",
  StyleBox["t",
    FontSlant->"Italic"],
  ")   for  ",
  Cell[BoxData[
      \(TraditionalForm\`\(-\[Pi]\) \[LessEqual] t \[LessEqual] 2  \[Pi]\)]],
  "\n\nc)   ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") =",
  Cell[BoxData[
      \(TraditionalForm\`2\/t\^2\)]],
  "   for   ",
  Cell[BoxData[
      \(TraditionalForm\`0.5 \[LessEqual] t \[LessEqual] 3\)]],
  "\n\nd)   ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") =",
  Cell[BoxData[
      \(TraditionalForm\`\@t\)]],
  "   for   1 \[LessEqual] t \[LessEqual] 4\n\ne)   ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") = ",
  Cell[BoxData[
      \(TraditionalForm\`1\/t\)]],
  "    for   ",
  Cell[BoxData[
      \(TraditionalForm\`1 \[LessEqual] t \[LessEqual] 2\)]]
}], "Text",
  PageWidth->PaperWidth],

Cell["\<\
You can change the terms in red to explore the functions listed above.\
\>", "Text",
  PageWidth->PaperWidth],

Cell[BoxData[{\(Clear[f, a, b, n, h, t];\), "\n", 
    RowBox[{
      RowBox[{\(f[t_]\), "=", 
        StyleBox[\(2/t\^2\),
          FontColor->RGBColor[1, 0, 0]]}], ";"}], "\n", 
    RowBox[{
      RowBox[{"a", "=", 
        StyleBox["0.5",
          FontColor->RGBColor[1, 0, 0]]}], ";"}], "\n", 
    RowBox[{
      RowBox[{"b", "=", 
        StyleBox["3",
          FontColor->RGBColor[1, 0, 0]]}], ";"}], "\n", 
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      RowBox[{"n", "=", 
        StyleBox["15",
          FontColor->RGBColor[1, 0, 0]]}], ";"}], "\n", 
    RowBox[{
      RowBox[{"window", "=", 
        StyleBox[\({{0, 3}, {0, 8}}\),
          FontColor->RGBColor[1, 0, 0]]}], 
      ";"}], "\n", \(riemannleft[f[t], t, a, b, 15, window];\)}], "Input",
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Cell[BoxData[
    RowBox[{
      RowBox[{"n", "=", 
        StyleBox["5853",
          FontColor->RGBColor[1, 0, 0]]}], ";", "\n", \(h = \((b - a)\)/n\), 
      ";", "\n", \(c[i_] = a + i*h\), ";", 
      "\n", \(area[n_] = Sum[f[c[i]]*h, {i, 0, n - 1}]\), ";", 
      "\n", \(Print["\<The total signed area of the rectangles is \>", 
        area[n] // N]\), ";"}]], "Input",
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Cell[BoxData[
    \(\[Integral]\_a\%b f[t] \[DifferentialD]t // N\)], "Input",
  PageWidth->PaperWidth],

Cell[TextData[{
  "Exercise 2",
  StyleBox["\nWrite a series of commands to calculate the Riemann sum using \
the midpoint of each subinterval. Use your series of commands to repeat the \
problems in exercise 1. Compare the number of rectangles that are required to \
get three-digit accuracy using the left Riemann sum, and the midpoint Riemann \
sum and discuss what you find. Use the command ",
    FontWeight->"Plain"],
  "riemannmiddle[ ] ",
  StyleBox["to visualize the midpoint rectangles if you wish.",
    FontWeight->"Plain"]
}], "Text",
  PageWidth->PaperWidth,
  FontWeight->"Bold"]
}, Closed]],

Cell[CellGroupData[{

Cell[TextData[{
  "Part IV: Sometimes There is an Easier Way ",
  Cell[BoxData[
      \(TraditionalForm\`-\)]],
  " The Fundamental Theorem of Calculus"
}], "Section",
  PageWidth->PaperWidth],

Cell[TextData[{
  "Remember that our motivation for calculating the area under the graph of a \
function ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          FractionBox[
            StyleBox[
              RowBox[{"d", 
                StyleBox["Q",
                  FontSlant->"Italic"]}]], 
            StyleBox[
              RowBox[{"d", 
                StyleBox["t",
                  FontSlant->"Italic"]}]]], "=", \(f(t)\)}], 
        TraditionalForm]]],
  " is to determine \[CapitalDelta]",
  StyleBox["Q",
    FontSlant->"Italic"],
  " over a time interval ",
  StyleBox["a",
    FontSlant->"Italic"],
  " \[LessEqual] ",
  StyleBox["t",
    FontSlant->"Italic"],
  " \[LessEqual] ",
  StyleBox["b. ",
    FontSlant->"Italic"],
  "Our approach has been:\n\n1. to first divide the time interval into a \
sequence of subintervals,\n\n2. to then approximate \[CapitalDelta]",
  Cell[BoxData[
      \(TraditionalForm\`Q\_i\)]],
  " over each of these subintervals of time with d",
  Cell[BoxData[
      \(TraditionalForm\`Q\_i\)]],
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \ \)\(\(=\)\(\ \)\(\(f(c\_i)\) h\)\)\)\)]],
  " (i.e., over a short time interval, the change in ",
  StyleBox["Q",
    FontSlant->"Italic"],
  " is approximately equal its rate of change times time),\n\n3. to finally \
add up the d",
  Cell[BoxData[
      \(TraditionalForm\`Q\_i\)]],
  "'s in a Riemann sum and take the limit of the sum as the number of \
subintervals goes to infinity.\n\nAs ",
  "the number of subintervals",
  " increases, the Riemann sum approximation improves, approaching the exact \
value for \[CapitalDelta]",
  StyleBox["Q",
    FontSlant->"Italic"],
  ".  That is, as the number of subintervals goes to infinity, the value of \
the Riemann sum approaches the exact value of \[CapitalDelta]",
  StyleBox["Q",
    FontSlant->"Italic"],
  ", and this we call the definite integral of ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") over the interval [",
  StyleBox["a",
    FontSlant->"Italic"],
  ", ",
  StyleBox["b",
    FontSlant->"Italic"],
  "].\n\nYou are probably wondering, why all the fuss? Wouldn't it be easier \
simply to find ",
  StyleBox["Q = F",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "), an antiderivative of ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "), and calculate \[CapitalDelta]Q = ",
  StyleBox["Q",
    FontSlant->"Italic"],
  Cell[BoxData[
      FormBox[
        RowBox[{\(\( \[VerticalSeparator] \_\(t = b\)\)\(-\ 
              Q\)\( \[VerticalSeparator] \_\(t = a\)\)\), "=", 
          RowBox[{
            RowBox[{
              StyleBox["F",
                FontSlant->"Italic"], "(", 
              StyleBox["b",
                FontSlant->"Italic"], ")"}], "-", 
            RowBox[{
              StyleBox["F",
                FontSlant->"Italic"], "(", 
              StyleBox["a",
                FontSlant->"Italic"], ")"}]}]}], TraditionalForm]]],
  "? Yes, in fact we can do exactly that. This can make the whole business of \
evaluating a definite integral a lot easier, provided there is a simple \
formula for the antiderivative, ",
  StyleBox["which isn't always the case",
    FontSlant->"Italic"],
  ". Let's do it for the integral ",
  Cell[BoxData[
      \(TraditionalForm\`\[Integral]\_0\%15\( f(t)\) \[DifferentialD]t\)]],
  ", where ",
  Cell[BoxData[
      \(TraditionalForm\`f(t) = 1\/4 - cos(\(\[Pi]\ t\)\/12)\)]],
  ". "
}], "Text",
  PageWidth->PaperWidth],

Cell[BoxData[{
    \(\(Clear[f, a, b, n, h];\)\), "\n", 
    \(\(f[t_] = 1\/4 - Cos[\(\[Pi]*t\)\/12];\)\)}], "Input",
  PageWidth->PaperWidth],

Cell[TextData[{
  "First, we find an antiderivative of ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") and then calculate \[CapitalDelta]",
  StyleBox["Q",
    FontSlant->"Italic"],
  "."
}], "Text",
  PageWidth->PaperWidth],

Cell[BoxData[
    \(F[t_] = \[Integral]f[t] \[DifferentialD]t + C\)], "Input",
  PageWidth->PaperWidth],

Cell[BoxData[{
    \(\(Clear[a, b, deltaQ];\)\), "\n", 
    \(deltaQ[a_, b_] = F[b] - F[a]\)}], "Input",
  PageWidth->PaperWidth],

Cell[BoxData[
    \(deltaQ[0, 15]\)], "Input",
  PageWidth->PaperWidth],

Cell[TextData[{
  "This is the same value we obtained by taking the limit of a Riemann sum! \
This result is ",
  StyleBox["fundamental",
    FontSlant->"Italic"],
  ". It tells us that we can evaluate the definite integral of a function, ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "), by either: 1) taking the limit of a Riemann sum or  2) finding an \
antiderivative of ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "), let's call it ",
  StyleBox["F",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "), and taking the difference of this function over the interval of \
summation. We can summarize this as follows: \n\n",
  Cell[BoxData[
      \(TraditionalForm\`\[Integral]\_a\%b\( f(
              t)\) \[DifferentialD]t = \(\(\(\(lim\)\(\ \)\)\+\(h \[Rule] 
                  0\)\) \(\[Sum]\+\(i = 1\)\%n\( f(c\_i)\) h\) = 
          F(b) - F(a)\)\)]],
  " \n\nor, in terms of ",
  StyleBox["Q",
    FontSlant->"Italic"],
  " and ",
  Cell[BoxData[
      FormBox[
        FractionBox[
          StyleBox["dQ",
            FontSlant->"Italic"], 
          StyleBox["dt",
            FontSlant->"Italic"]], TraditionalForm]]],
  ", \n\n",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{\(\[Integral]\_a\%b\), 
            RowBox[{
              StyleBox[\(dQ\/dt\),
                FontSlant->"Italic"], \(\[DifferentialD]t\)}]}], 
          "=", \(Q(b) - Q(a)\)}], TraditionalForm]]],
  ". \n\nThis all makes good sense if you think of it as a generalization of \
our old friend, change is equal to rate times time. This result is so \
important that it is called the Fundamental Theorem of Calculus, Part 2. The \
theorem assures us that we can evaluate a definite integral in this simpler \
way whenever ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") is continuous over the interval of summation and provided a formula \
exists for the antiderivative of ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "). \n\nThe Fundamental Theorem of Calculus, Part 2, helps us in two ways. \
First, as we have discovered, it tells us that instead of taking the limit of \
a Riemann sum, we can evaluate the definite integral of a function ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") by taking the difference of values of an antiderivative ",
  StyleBox["F",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") over the interval of integration, provided there is a formula for ",
  StyleBox["F",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "). But what if, as is often the case, there is no algebraic formula for ",
  StyleBox["F",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ")? Well, now we can turn the problem around. Even when we don't have a \
formula for ",
  StyleBox["F",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "), we can evaluate a change in the value of ",
  StyleBox["F",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") over an interval [",
  StyleBox["a",
    FontSlant->"Italic"],
  ", ",
  StyleBox["b",
    FontSlant->"Italic"],
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  StyleBox["F",
    FontSlant->"Italic"],
  "(",
  StyleBox["b",
    FontSlant->"Italic"],
  ") - ",
  StyleBox["F",
    FontSlant->"Italic"],
  "(",
  StyleBox["a",
    FontSlant->"Italic"],
  ") =",
  Cell[BoxData[
      \(TraditionalForm\`\[Integral]\_a\%b\( f(
              t)\) \[DifferentialD]t = \(\(\(lim\)\(\ \)\)\+\(h \[Rule] 
                0\)\) \(\[Sum]\+\(i = 1\)\%n\( f(c\_i)\) h\)\)]],
  ". \n\nThis is the second way that the Fundamental Theorem of Calculus, \
Part 2, helps us out. \n\nBut in this last case, how do we know that the \
value we calculate is actually a difference of an antiderivative of ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ")? That is, if there is no algebraic formula for ",
  StyleBox["F",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "), how do we know that the value we calculate by taking the limit of a \
Riemann sum with ",
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  "(",
  StyleBox["t",
    FontSlant->"Italic"],
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  Cell[BoxData[
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  ", where ",
  StyleBox["F",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") is an antiderivative of ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ")? \n\nThe Fundamental Theorem of Calculus, Part 1, provides the answer. \
It assures us that if ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") is continuous on [",
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  StyleBox["t",
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  StyleBox["t",
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  " in [",
  StyleBox["a",
    FontSlant->"Italic"],
  ", ",
  StyleBox["b",
    FontSlant->"Italic"],
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sum, to form ",
  StyleBox["F",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  "). Specifically, ",
  StyleBox["F",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") = ",
  Cell[BoxData[
      \(TraditionalForm\`\[Integral]\_a\%t\( f(u)\) \[DifferentialD]u\)]],
  ". As a result, we can say for sure that ",
  Cell[BoxData[
      \(TraditionalForm\`\[Integral]\_a\%b\( f(t)\) \[DifferentialD]t = 
        F(b) - F(a)\)]],
  ", where ",
  StyleBox["F",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
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  StyleBox["f",
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    FontSlant->"Italic"],
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Cell["You Try It: Part IV", "Section",
  PageWidth->PaperWidth],

Cell[TextData[{
  "Now you try repeating the steps in Part IV with the functions in the list \
that follows. First, calculate the limit of a Riemann sum, and then find an \
antiderivative and calculate the difference over the interval of summation. \
Change the limits on the interval (i.e., ",
  StyleBox["a",
    FontSlant->"Italic"],
  " and ",
  StyleBox["b",
    FontSlant->"Italic"],
  ") as well as the function definition for each exercise. We show (d) as an \
example. (Note that because ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " has difficulty evaluating some limits symbolically, we use the",
  StyleBox[" NLimit[ ]",
    FontWeight->"Bold"],
  " command, which gives a numeric estimate of the limit that is precise to \
the number of digits available in your computer's numeric processor. The ",
  StyleBox["Terms->10",
    FontWeight->"Bold"],
  " option is to ensure accurate estimates of the limit.) \n\na)  ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") = 3",
  Cell[BoxData[
      \(TraditionalForm\`t\^2\)]],
  "- 6  for  -1 \[LessEqual] ",
  StyleBox["t",
    FontSlant->"Italic"],
  " \[LessEqual] 3\n\nb)   ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") =2 sin(",
  StyleBox["t",
    FontSlant->"Italic"],
  ")  for  ",
  Cell[BoxData[
      \(TraditionalForm\`\(-\[Pi]\) \[LessEqual] t \[LessEqual] 2  \[Pi]\)]],
  "\n\nc)   ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") =",
  Cell[BoxData[
      \(TraditionalForm\`2\/t\^2\)]],
  "   for   ",
  Cell[BoxData[
      \(TraditionalForm\`0.5 \[LessEqual] t \[LessEqual] 3\)]],
  "\n\nd)   ",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["t",
    FontSlant->"Italic"],
  ") =",
  Cell[BoxData[
      \(TraditionalForm\`\@t\)]],
  "   for   1 \[LessEqual] t \[LessEqual] 4"
}], "Text",
  PageWidth->PaperWidth],

Cell["\<\
Change the terms in red to explore the functions listed above.\
\>", "Text",
  PageWidth->PaperWidth],

Cell[BoxData[
    RowBox[{\(Clear[f, a, b, n, h, t]\), ";", "\n", 
      RowBox[{\(f[t_]\), "=", 
        StyleBox[\(\@t\),
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      RowBox[{"a", "=", 
        StyleBox["1",
          FontColor->RGBColor[1, 0, 0]]}], ";", "\n", 
      RowBox[{"b", "=", 
        StyleBox["4",
          FontColor->RGBColor[1, 0, 0]]}], ";", "\n", \(h = \((b - a)\)/n\), 
      ";", "\n", \(c[i_] = a + i*h\), ";", 
      "\n", \(riemannSum[
          n_] = \(Sum[f[c[i]]*h, {i, 0, n - 1}] // ExpToTrig\) // Simplify\), 
      ";", "\n", \(NLimit[riemannSum[n], n \[Rule] \[Infinity], 
        Terms \[Rule] 10]\)}]], "Input",
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Cell["Now we use the Fundamental Theorem.", "Text",
  PageWidth->PaperWidth],

Cell[BoxData[
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Cell[BoxData[
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Cell[CellGroupData[{

Cell["Part V: Calculus Gives Birth to New Functions", "Section",
  PageWidth->PaperWidth],

Cell[CellGroupData[{

Cell["The Natural Log Function", "Subsection",
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Cell[TextData[{
  "As we indicated before, many functions have antiderivatives that cannot be \
expressed with algebraic formulas. One example is the function ",
  Cell[BoxData[
      \(TraditionalForm\`f(x) = 1\/x\)]],
  ". It is important to realize that while the antiderivatives of this \
function cannot be represented by algebraic formulas, antiderivatives do, \
nonetheless, exist. We can use the Fundamental Theorem of Calculus, Part 1, \
to define its antiderivatives, thus giving birth to a new function that can \
only be defined using calculus. Specifically, an antiderivative of ",
  Cell[BoxData[
      \(TraditionalForm\`f(x) = 1\/x\)]],
  " is defined as ",
  Cell[BoxData[
      \(TraditionalForm\`F(
          x) = \[Integral]\_1\%x\( 1\/u\) \[DifferentialD]u\)]],
  " for ",
  Cell[BoxData[
      \(TraditionalForm\`x > 0\)]],
  ". Since the function ",
  Cell[BoxData[
      \(TraditionalForm\`F(x)\)]],
  " is so important it is given a special name. It is called the natural log \
of ",
  StyleBox["x ",
    FontSlant->"Italic"],
  "and is denoted by",
  StyleBox[" ",
    FontSlant->"Italic"],
  Cell[BoxData[
      \(TraditionalForm\`ln\ x\)]],
  ". There is a very good reason for calling it a logarithm: it exhibits all \
of the characteristics of a log function. All of the properties of logarithms \
hold for ",
  Cell[BoxData[
      \(TraditionalForm\`ln\ x\)]],
  ".\n\nHow do we evaluate this function? While the Fundamental Theorem of \
Calculus, Part 2, still holds true for ",
  Cell[BoxData[
      \(TraditionalForm\`f(x) = 1\/x\)]],
  " and its antiderivative ",
  Cell[BoxData[
      \(TraditionalForm\`ln\ x\)]],
  ", that is, \n\n",
  Cell[BoxData[
      \(TraditionalForm\`\[Integral]\_a\%b\( 1\/x\) \[DifferentialD]x = 
        ln\ b - ln\ a\)]],
  ", \n\nit is not much help in evaluating ",
  Cell[BoxData[
      \(TraditionalForm\`\[Integral]\_a\%b\( 1\/x\) \[DifferentialD]x\)]],
  " because there is no algebraic formula for ",
  Cell[BoxData[
      \(TraditionalForm\`ln\ x\)]],
  ". Consequently, we are forced to resort to its definition to find values \
for ",
  Cell[BoxData[
      \(TraditionalForm\`ln\ x\)]],
  ". Specifically, ",
  Cell[BoxData[
      \(TraditionalForm\`ln\ x\)]],
  " is the limit of a Riemann sum. We know that ",
  Cell[BoxData[
      \(TraditionalForm\`ln\ x = \[Integral]\_1\%x\( 
                1\/u\) \[DifferentialD]u\)]],
  " exists because ",
  Cell[BoxData[
      \(TraditionalForm\`f(x) = 1\/x\)]],
  " is continuous on any interval from 1 to ",
  StyleBox["x",
    FontSlant->"Italic"],
  ", provided ",
  Cell[BoxData[
      \(TraditionalForm\`x > 0\)]],
  ". But how do we represent the values of ",
  Cell[BoxData[
      \(TraditionalForm\`ln\ x\)]],
  " as decimals? Usually we settle for an approximation that is precise to \
some specified number of digits. One approach is to approximate ",
  Cell[BoxData[
      \(TraditionalForm\`ln\ x = \[Integral]\_1\%x\( 
                1\/u\) \[DifferentialD]u\)]],
  " with a Riemann sum using enough rectangles to give the desired precision. \
Let's do it.\n\nWe define our own approximate ",
  Cell[BoxData[
      \(TraditionalForm\`ln\ x\)]],
  " function in the following cell and call it ",
  StyleBox["approxln[x]",
    FontWeight->"Bold"],
  ". The variable ",
  StyleBox["n",
    FontSlant->"Italic"],
  " is the number of rectangles used in the Riemann sum that approximates ",
  Cell[BoxData[
      \(TraditionalForm\`ln\ x\)]],
  "."
}], "Text",
  PageWidth->PaperWidth],

Cell[BoxData[
    \(\(approxln[x_, n_] := 
        Block[{a, b, h, c}, \[IndentingNewLine]a = 1; 
          b = x; \[IndentingNewLine]h = \((b - a)\)/n; \[IndentingNewLine]c[
              i_] = a + 
              i*h; \[IndentingNewLine]Sum[\((1/c[i])\)*h, {i, 0, n - 1}] // 
            N];\)\)], "Input",
  PageWidth->PaperWidth],

Cell[TextData[{
  "As an exercise, vary the value of ",
  StyleBox["n",
    FontSlant->"Italic"],
  " in the next group of cells to determine how many rectangles are needed to \
approximate ",
  Cell[BoxData[
      \(TraditionalForm\`ln\ x\)]],
  " accurate to three decimal places for ",
  Cell[BoxData[
      \(TraditionalForm\`1\/10 \[LessEqual] x \[LessEqual] 10\)]],
  ". We start with ",
  Cell[BoxData[
      \(TraditionalForm\`n = 12\)]],
  " and calculate the error, that is, the difference between our approximate \
value and that given by ",
  StyleBox["Log[x]",
    FontWeight->"Bold"],
  ",",
  " which is ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  "'s command for calculating ",
  Cell[BoxData[
      \(TraditionalForm\`ln\ x\)]],
  "."
}], "Text",
  PageWidth->PaperWidth],

Cell[BoxData[{
    \(\(n = 12;\)\), "\n", 
    \(approxln[10, n] - Log[10]\)}], "Input",
  PageWidth->PaperWidth],

Cell[BoxData[
    \(approxln[1/10, n] - Log[1/10]\)], "Input",
  PageWidth->PaperWidth],

Cell[TextData[{
  "Let's compare our approximating function ",
  StyleBox["approxln[x]",
    FontWeight->"Bold"],
  " with ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  "'s ",
  StyleBox["Log[x] ",
    FontWeight->"Bold"],
  "by graphing the two together. First use ",
  Cell[BoxData[
      \(TraditionalForm\`n = 12\)]],
  ", and then replace it with the value that you found above that ensures \
three decimal places of precision for ",
  Cell[BoxData[
      \(TraditionalForm\`1\/10 \[LessEqual] x \[LessEqual] 10\)]],
  "."
}], "Text",
  PageWidth->PaperWidth],

Cell[BoxData[{
    \(\(n = 12;\)\), "\n", 
    \(\(Plot[{Log[x], approxln[x, n]}, {x, 0.1, 10}, 
        PlotStyle \[Rule] {{RGBColor[1, 0, 0]}, {RGBColor[0, 0, 1]}}, 
        AxesLabel \[Rule] {"\<x\>", "\<y\>"}];\)\), "\[IndentingNewLine]", 
    \(Print["\<The natural log function is plotted in red and the \
approximation to it is plotted in blue.\>"]\)}], "Input",
  PageWidth->PaperWidth],

Cell["What do you observe?", "Text",
  PageWidth->PaperWidth]
}, Closed]]
}, Closed]],

Cell[CellGroupData[{

Cell["\<\
You Try It: The Standard Normal Distribution and the Error Function\
\>", "Section",
  PageWidth->PaperWidth],

Cell[TextData[{
  "When you did the \"You Try It: Part II,\" you may have thought that the \
function ",
  Cell[BoxData[
      \(TraditionalForm\`f(x) = \(1\/\@\(2  \[Pi]\)\) 
          e\^\(-\(x\^2\/2\)\)\)]],
  " seemed a bit unusual, but the bell-shaped graph that resulted probably \
looked very familiar. This function is a classic example of one that does not \
have algebraic forms for its antiderivatives, and it is called the \
standard-normal-distribution function, which is used in probability and \
statistics. Let's examine it in more detail."
}], "Text",
  PageWidth->PaperWidth],

Cell[BoxData[{
    \(\(Clear[f];\)\), "\n", 
    \(\(f[x_] = \(1\/\@\(2  \[Pi]\)\) E\^\(-\(x\^2\/2\)\);\)\)}], "Input",
  PageWidth->PaperWidth],

Cell[BoxData[
    \(\(Plot[f[x], {x, \(-5\), 5}, PlotRange \[Rule] All, 
        PlotStyle \[Rule] {RGBColor[0, 1, 0]}, 
        AxesLabel \[Rule] {"\<x\>", "\<y\>"}];\)\)], "Input",
  PageWidth->PaperWidth],

Cell[TextData[{
  "Now we consider the definite integral, ",
  Cell[BoxData[
      \(TraditionalForm\`F(
          x) = \(\[Integral]\_\(-x\)\%x\( f(
                t)\) \[DifferentialD]t = \(1\/\@\(2  \[Pi]\)\) \
\(\[Integral]\_\(-x\)\%x\( e\^\(\(-t\^2\)/2\)\) \[DifferentialD]t\)\)\)]],
  ", which gives the area under the bell-shaped curve between ",
  Cell[BoxData[
      \(TraditionalForm\`\(-x\)\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  ". Because ",
  Cell[BoxData[
      \(TraditionalForm\`f(x)\)]],
  " is an even function, we can rewrite the integral as ",
  Cell[BoxData[
      \(TraditionalForm\`F(
          x) = \(\@\(2\/\[Pi]\)\) \(\[Integral]\_0\%x\( 
                  e\^\(\(-t\^2\)/2\)\) \[DifferentialD]t\)\)]],
  ".  \n\nA function that is closely related to ",
  Cell[BoxData[
      \(TraditionalForm\`F(x)\)]],
  " is the error function, which is denoted by ",
  Cell[BoxData[
      \(TraditionalForm\`erf\ x\)]],
  " and is defined as follows: ",
  Cell[BoxData[
      \(TraditionalForm\`erf\ x = \(2\/\@\[Pi]\) \(\[Integral]\_0\%x\( 
                  e\^\(-t\^2\)\) \[DifferentialD]t\)\)]],
  ". We leave it as an exercise for you to show that ",
  Cell[BoxData[
      \(TraditionalForm\`F(x) = erf(\ x\/\@2)\)]],
  ". We will use ",
  Cell[BoxData[
      \(TraditionalForm\`F(x)\)]],
  " in the remainder of this module. Since there are no algebraic formulas \
for the antiderivatives of ",
  Cell[BoxData[
      \(TraditionalForm\`f(x)\)]],
  ", we will approximate values of ",
  Cell[BoxData[
      \(TraditionalForm\`F(x)\)]],
  " using a Riemann sum with a sufficient number of rectangles to ensure the \
desired precision. \n\nUsing the method that we used for ",
  Cell[BoxData[
      \(TraditionalForm\`ln\ x\)]],
  " in Part V above, write ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " commands to approximate values of ",
  Cell[BoxData[
      \(TraditionalForm\`F(x)\)]],
  " to three decimal places of accuracy on the interval ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(-4\)\(\[LessEqual]\)\(x\)\(\[LessEqual]\)\(4\)\(\
\ \)\)\)]],
  ". First use a ",
  StyleBox["left-hand",
    FontSlant->"Italic"],
  " Riemann sum, and then use a ",
  StyleBox["midpoint",
    FontSlant->"Italic"],
  " Riemann sum. To test your commands, plot the graph of ",
  Cell[BoxData[
      \(TraditionalForm\`erf(\ x\/\@2)\)]],
  " and the graph of your own ",
  Cell[BoxData[
      \(TraditionalForm\`F(x)\)]],
  " to see if the functions match. (In ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  ", the error function is ",
  StyleBox["Erf[x]",
    FontWeight->"Bold"],
  ".) Comment on what you find. We illustrate the Riemann sum with the ",
  StyleBox["riemannmiddle[ ] ",
    FontWeight->"Bold"],
  "command."
}], "Text",
  PageWidth->PaperWidth],

Cell[BoxData[
    \(\(riemannmiddle[f[x], x, 0, \(-4\), 
        15, {{\(-4\), 0}, {0, 0.4}}];\)\)], "Input",
  PageWidth->PaperWidth],

Cell[TextData[{
  "Note that the signed areas are negative because we are summing from right \
to left, and therefore ",
  Cell[BoxData[
      \(TraditionalForm\`h < 0\)]],
  ", giving ",
  Cell[BoxData[
      \(TraditionalForm\`\(f(c\_i)\) h < 0\)]],
  " in the sum."
}], "Text",
  PageWidth->PaperWidth],

Cell["\<\
If you need help with this exercise, the following cell, which is closed, \
contains our solution. Try doing it on your own before you check out what we \
did.\
\>", "Text",
  PageWidth->PaperWidth],

Cell[CellGroupData[{

Cell["Solution", "Subsubsection",
  PageWidth->PaperWidth],

Cell[BoxData[{
    \(\(Clear[approxF];\)\), "\n", 
    \(\(approxF[x_, n_] := 
        Block[{a, b, h, c}, \[IndentingNewLine]a = 0; 
          b = x; \[IndentingNewLine]h = \((b - a)\)/n; \[IndentingNewLine]c[
              i_] = a + \((i + 1\/2)\)*h; \[IndentingNewLine]\@\(2\/\[Pi]\)*
            Sum[Exp[\(-\((c[i]^2/2)\)\)]*h // N, {i, 0, 
                n - 1}]];\)\)}], "Input",
  PageWidth->PaperWidth],

Cell[BoxData[
    \(approxF[\(-0.1\), 4]\)], "Input",
  PageWidth->PaperWidth],

Cell[BoxData[
    \(Erf[\(-0.1\)/Sqrt[2]] // N\)], "Input",
  PageWidth->PaperWidth],

Cell[BoxData[{
    \(\(n = 2;\)\), "\n", 
    \(\(Plot[{approxF[x, n], Erf[x/Sqrt[2]]}, {x, \(-2\), 2}, 
        PlotStyle \[Rule] {{RGBColor[1, 0, 0]}, {RGBColor[0, 0, 1]}}, 
        AxesLabel \[Rule] {"\<x\>", "\<y\>"}];\)\)}], "Input",
  PageWidth->PaperWidth],

Cell["\<\
There is one final note. Though the use of a Riemann sum to estimate a \
definite integral is a very direct approach, it is not very efficient. More \
efficient methods of estimating definite integrals have been developed, and \
some of these are studied in the module on numerical integration entitled \
\"Riemann, Trapezoids, and Simpson.\"\
\>", "Text",
  PageWidth->PaperWidth]
}, Closed]]
}, Closed]]
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*)




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End of Mathematica Notebook file.
***********************************************************************)