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In particular, we will study how the \ Fundamental Theorem relates a function to its derivative.", Evaluatable->False, CellGroupingRules->"OutputGrouping", AspectRatioFixed->True, FontSize->14]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell[TextData[StyleBox[ "Before continuing, you might wish to alter the magnification by using the \ `format' menu at the top and selecting ??% under ``magnification''. You \ should probably also alter the size of the window to one you prefer.You might \ wish to take notes as you work through this file. Saving this file to your \ disk every few minutes is highly recommended. Your instructor should tell \ you if any part of this lab is to be submitted for grading and what the \ due-date is.", Evaluatable->False, CellGroupingRules->"OutputGrouping", AspectRatioFixed->True, FontSize->14, FontColor->RGBColor[0, 0, 1]]], "Text", Evaluatable->False, CellGroupingRules->"OutputGrouping", AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell["\<\ We have seen that the integral of the velocity function from a to b \ is the total distance travelled during that time interval. Using the same \ reasoning, we deduced the Fundamental Theorem of Calculus (FTC): the \ definite integral of a rate is the total change. More specifically, if f is \ a continuous function on [a,b] and f(t)=F'(t) for some function F, then the \ integral of f from t=a to t=b is F(b)-F(a). \ \>", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[StyleBox[ "As usual, enter input cells by putting the cursor anywhere\nin the cell and \ hit \"Enter\", or \"Shift\" and \"Return\" together, or \"Shift\" and \"Enter\ \" together. ", Evaluatable->False, CellGroupingRules->"OutputGrouping", AspectRatioFixed->True, FontSize->14]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontColor->RGBColor[1, 0, 1]], Cell["Example 1", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->18, FontColor->RGBColor[0, 0, 1]], Cell["Let's plot a piece of the graph of the cosine function.", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData["Plot[Cos[t], {t,0,\[Pi]}, AxesLabel->{t,y}];"], "Input", AspectRatioFixed->True], Cell[TextData[ "Notice that the area trapped above the t-axis is equal to the area trapped \ below the t-axis, so the integral from 0 to \[Pi] must be zero. But we also \ know that (Sin[t])'=Cos[t], so we can apply the FTC with f[t]=Cos[t] and \n\ F[t]=Sin[t]. The integral of Cos[t] from 0 to \[Pi] is \n \ Sin[\[Pi]]-Sin[0] = 0 - 0 = 0.\nNow consider the integral of Cos[t] from 0 to \ \[Pi]/2. The graph of Cos[t] is above the t-axis on [0, \[Pi]/2], so the \ integral is positive, but the exact answer is not clear simply by looking at \ the graph. The FTC makes it clear: the integral of Cos[t] from 0 to \[Pi]/2 \ is Sin[\[Pi]/2] - Sin[0] = 1 - 0 = 1."], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[{ StyleBox["Mathematica", FontSize->14, FontSlant->"Italic"], StyleBox[" also knows the FTC.", FontSize->14] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell[TextData["Integrate[Cos[t],{t,0,\[Pi]/2}]"], "Input", AspectRatioFixed->True], Cell[TextData[{ "Exercise 1 \n", StyleBox["Compute the definite integrals of the following functions.", FontSize->14] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->18, FontColor->RGBColor[0, 0, 1]], Cell[" g[t]=1/t from t=1 to t=4", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Times", FontSize->14], Cell[" h[t]=1/(1+t^2) from t=0 to t=3", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Times", FontSize->14], Cell[" k[t]=E^t from t=-2 to t=2", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Times", FontSize->14], Cell["Example 2 ", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->18, FontColor->RGBColor[0, 0, 1]], Cell[TextData[{ StyleBox[ "After t hours, a population of bacteria is growing at a rate of 2", FontColor->GrayLevel[0]], StyleBox["t", FontColor->GrayLevel[0], FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[ " million bacteria per hour. Estimate the total increase in the bacteria \ population during the first hour.", FontColor->GrayLevel[0]] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14, FontColor->RGBColor[0, 0, 1]], Cell[TextData[StyleBox["Solution", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->13], Cell[TextData[{ "The", StyleBox[" ", FontSlant->"Italic"], "growth ", StyleBox["rate ", FontSlant->"Italic"], " is F'[t]=", Cell[BoxData[ FormBox[ SuperscriptBox["2", StyleBox[\(\ t\), FontSize->15]], TraditionalForm]]], ". We are looking for the total change, i.e. F[1]-F[0]. The FTC says that \ this total change is equal to the integral of F'[t] from 0 to 1. Let's use ", StyleBox["Mathematica", FontSlant->"Italic"], " to compute this integral. " }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[{ "Integrate[2^t, {t,0,1}] \n", StyleBox["(*recall that ", FontColor->RGBColor[0, 0, 1]], StyleBox["Mathematica", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], StyleBox[" represents ln(x) as Log[x]*)\n", FontColor->RGBColor[0, 0, 1]] }], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Note 1", FontSize->17, FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontSlant->"Italic"], "In the above example, ", StyleBox["Mathematica", FontSlant->"Italic"], " returned an exact answer. Here's what happened: it found an \"elementary\ \" formula for F then plugged in the numbers. Sometimes an \"elementary\" \ formula for F doesn't exist. What does ", StyleBox["Mathematica", FontSlant->"Italic"], " do in that case?" }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell["Integrate[2^Sin[t], {t,0,2}]", "Input", AspectRatioFixed->True], Cell["Integrate[Sin[t^2], {t,0,2}]", "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " returns the original problem or writes the answer in terms of a function \ which is beyond the scope of this class! But we still have some options. \ After all, the graph of 2^Sin[t] is decent on [0, 2]:" }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell["\<\ Plot[2^Sin[t], {t,0,2}, AxesOrigin->{0,0}, AxesLabel->{x,y}];\ \>", "Input", AspectRatioFixed->True], Cell[TextData[{ "We can estimate an integral by using Riemann sums - so can ", StyleBox["Mathematica.", FontSlant->"Italic"], " " }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[{ "NIntegrate[2^Sin[t], {t,0,1}] ", StyleBox["(*Notice the \"N\"*)", FontColor->RGBColor[0, 0, 1]] }], "Input", AspectRatioFixed->True], Cell[TextData[{ "The ", StyleBox["N", FontWeight->"Bold"], " tells Mathematica to do the problem numerically (without trying to find \ a formula for F). Now return to the integral in Example 2 and make ", StyleBox["Mathematica", FontSlant->"Italic"], " do the integral numerically - what is it?" }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[{ StyleBox["Exercise", FontSize->18, FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontSize->14, FontColor->RGBColor[0, 0, 1]], StyleBox["2", FontSize->18, FontColor->RGBColor[0, 0, 1]], StyleBox[" \nNumerically integrate the functions in Exercise 1.", FontSize->14] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell["Example 3", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->18, FontColor->RGBColor[0, 0, 1]], Cell[TextData[StyleBox[ "Referring to Example 2, suppose that when t=0, there are 2 million bacteria. \ Use the FTC to determine the number of bacteria after 1 second? After 2 \ seconds? 3?", FontColor->GrayLevel[0]]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14, FontColor->RGBColor[0, 0, 1]], Cell["Solution ", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->13, FontColor->RGBColor[1, 0, 1]], Cell[TextData[{ "We know that F[1]-F[0]=", Cell[BoxData[ FormBox[ RowBox[{\(\[Integral]\_0\%1\), RowBox[{ SuperscriptBox["2", RowBox[{" ", StyleBox["t", FontSize->15]}]], \(\[DifferentialD]t\)}]}], TraditionalForm]]], " (by the FTC), and we're assuming that F[0]=2, so \nF[1]=", Cell[BoxData[ FormBox[ RowBox[{\(\[Integral]\_0\%1\), RowBox[{ SuperscriptBox["2", RowBox[{" ", StyleBox["t", FontSize->15]}]], \(\[DifferentialD]t\)}]}], TraditionalForm]]], " + F[0] = 1/ln(2) + 2 = 1/ln(2) +2.\n\nSimilarly, F[2]-F[0]=", Cell[BoxData[ FormBox[ RowBox[{\(\[Integral]\_0\%2\), RowBox[{ SuperscriptBox["2", RowBox[{" ", StyleBox["t", FontSize->15]}]], \(\[DifferentialD]t\)}]}], TraditionalForm]]], ", so \nF[2]=", Cell[BoxData[ FormBox[ RowBox[{\(\[Integral]\_0\%2\), RowBox[{ SuperscriptBox["2", RowBox[{" ", StyleBox["t", FontSize->15]}]], \(\[DifferentialD]t\)}]}], TraditionalForm]]], " + F[0]=", Cell[BoxData[ FormBox[ RowBox[{\(\[Integral]\_0\%2\), RowBox[{ SuperscriptBox["2", RowBox[{" ", StyleBox["t", FontSize->15]}]], \(\[DifferentialD]t\)}]}], TraditionalForm]]], " + 2 = 3/ln(2)+2. \n \nYou find the answer when t=3." }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[{ StyleBox["Note 2", FontSize->17, FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontSize->14, FontSlant->"Italic"], StyleBox[ "In Example 3 we actually described how to find the function values for \ F[t]:\n F[0]=2\n \t\t\t\t\t\t\t \n F[", FontSize->14], StyleBox["1", FontSize->14, FontColor->RGBColor[0, 0, 1]], StyleBox["]=", FontSize->14], Cell[BoxData[ FormBox[ RowBox[{ SubsuperscriptBox["\[Integral]", "0", StyleBox["1", FontColor->RGBColor[0, 0, 1]]], RowBox[{ SuperscriptBox["2", RowBox[{" ", StyleBox["t", FontSize->15]}]], \(\[DifferentialD]t\)}]}], TraditionalForm]], FontSize->14], StyleBox["+ F[0] = 1/ln(2) + 2\n \t\t\t\t\t\t\t \n F[", FontSize->14], StyleBox["2", FontSize->14, FontColor->RGBColor[0, 0, 1]], StyleBox["]=", FontSize->14], Cell[BoxData[ FormBox[ RowBox[{ SubsuperscriptBox["\[Integral]", "0", StyleBox["2", FontColor->RGBColor[0, 0, 1]]], RowBox[{ SuperscriptBox["2", RowBox[{" ", StyleBox["t", FontSize->15]}]], \(\[DifferentialD]t\)}]}], TraditionalForm]], FontSize->14], " ", StyleBox["+ F[0] = 3/ln(2) + 2\n \t\t\t\t\t\t\t \n F[", FontSize->14], StyleBox["3", FontSize->14, FontColor->RGBColor[0, 0, 1]], StyleBox["]=", FontSize->14], Cell[BoxData[ FormBox[ StyleBox[ RowBox[{ SubsuperscriptBox["\[Integral]", "0", StyleBox["3", FontColor->RGBColor[0, 0, 1]]], RowBox[{ SuperscriptBox["2", RowBox[{" ", StyleBox["t", FontSize->15]}]], \(\[DifferentialD]t\)}]}], FontSize->14], TraditionalForm]]], " ", StyleBox["+ F[0] = 7/ln(2) + 2\n \n ", FontSize->14], Cell[BoxData[ \(TraditionalForm\`\( : \+ : \% : \)\)]], StyleBox["\n\t\t \nF[", FontSize->14], StyleBox["x", FontSize->14, FontColor->RGBColor[0, 0, 1]], StyleBox["]=", FontSize->14], Cell[BoxData[ FormBox[ StyleBox[ RowBox[{ SubsuperscriptBox["\[Integral]", "0", StyleBox["x", FontColor->RGBColor[0, 0, 1]]], RowBox[{ SuperscriptBox["2", RowBox[{" ", StyleBox["t", FontSize->15]}]], \(\[DifferentialD]t\)}]}], FontSize->14], TraditionalForm]]], " ", StyleBox["+ F[0] =", FontSize->14], Cell[BoxData[ FormBox[ RowBox[{" ", StyleBox[ RowBox[{ SubsuperscriptBox["\[Integral]", "0", StyleBox["x", FontColor->RGBColor[0, 0, 1]]], RowBox[{ SuperscriptBox["2", RowBox[{" ", StyleBox["t", FontSize->15]}]], \(\[DifferentialD]t\)}]}], FontSize->14]}], TraditionalForm]]], " ", StyleBox["+ 2\n = Integrate[2^t, {t, 0, ", FontSize->14], StyleBox["x", FontSize->14, FontColor->RGBColor[0, 0, 1]], StyleBox["}] + 2\n \nLet's make a table of function values for F. ", FontSize->14] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell[TextData[{ "Clear[F,x,mytable]\nF[x_]:=Integrate[2^t, {t,0,x}]+2 ", StyleBox["(*first we define F, as above*)", FontColor->RGBColor[0, 0, 1]], "\nmytable=Table[F[x], {x,0,8}] ", StyleBox["(*we call the Table 'mytable'*)", FontColor->RGBColor[0, 0, 1]] }], "Input", AspectRatioFixed->True], Cell[TextData[StyleBox[ "The function values in the table are listed in order, i.e., \nF[0], F[1], \ F[2],...,F[8]. Now we plot the points. ", FontSize->14]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell["\<\ ListPlot[mytable, Prolog->AbsolutePointSize[4], AxesLabel->{x,\"F(x)\"}, PlotStyle->Hue[0.8]];\ \>", "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Do you recognize the function?", FontColor->GrayLevel[0]], "\nBear in mind what F[x] represents geometrically. It is 2 +\nthe area \ under the graph of ", Cell[BoxData[ FormBox[ SuperscriptBox["2", RowBox[{" ", StyleBox["t", FontSize->15], StyleBox[" ", FontSize->15]}]], TraditionalForm]]], "from 0 to the number x. " }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[{ StyleBox["Note 3", FontSize->17, FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontSize->17], StyleBox[" \n", FontSize->13], StyleBox[ "In Note 2 we investigated the function F. Now we try to find its \ derivative", FontSize->14], ".", StyleBox[" Recall that \n F'(a)= lim", FontSize->14], StyleBox["h\[Rule]0", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSize->14], Cell[BoxData[ FormBox[ StyleBox[\(\(F(a + h) - F(a)\)\/h\), FontSize->18, FontWeight->"Plain"], TraditionalForm]], FontSize->14, FontWeight->"Bold"], ".", StyleBox["\nLet's try to find F'(0). ", FontSize->14] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell["Limit[(F[0+h]-F[0])/h, h->0]", "Input", AspectRatioFixed->True], Cell["\<\ We can use this format to compute F' at any x. So let's make a table of values for F'.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[{ "dF[x_]:=Limit[(F[x+h]-F[x])/h, h->0] ", StyleBox["(*defining derivative*)", FontColor->RGBColor[0, 0, 1]], "\nmytable2=Table[dF[x], {x,0,8}]" }], "Input", AspectRatioFixed->True], Cell["\<\ The function values in the above table should look familiar! Can \ you guess a formula for F'? Let's plot the points.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell["\<\ ListPlot[mytable2, Prolog->AbsolutePointSize[4], AxesLabel->{x,\"F'(x)\"}, PlotStyle->Hue[0.8]];\ \>", "Input", AspectRatioFixed->True], Cell["", "Input", AspectRatioFixed->True], Cell["Exercises", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->18, FontColor->RGBColor[0, 0, 1]], Cell[TextData[StyleBox[ "3. Find formulas for F and F' discussed in Notes 2 and 3.", FontSize->14]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell[TextData[{ "4. Consider the integral of ", Cell[BoxData[ \(TraditionalForm\`t\^2\)]], " from 0 to x. This defines \n a function G[x]. \n (a) What \ does G represent geometrically. (Give more\n than a one-word answer!) " }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[" (b) Find a formula for G that has an integral in it.", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[{ " (c) Define a ", StyleBox["Mathematica ", FontSlant->"Italic"], " function which represents\n the function you found in part (b). \ \n (", StyleBox["Hint", FontColor->RGBColor[0, 0, 1]], ": Look back at Note 2.) " }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[StyleBox[ " (d) Make a table of the function values for G. Does \n the \ function look familiar?", FontSize->14]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell[TextData[StyleBox[" (e) Plot the points from the table. ", FontSize->14]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell[TextData[StyleBox[ " (f) Find a formula for G that does not involve an \n integral.", FontSize->14]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell[TextData[StyleBox[ "5. In this exercise we investigate G' where G is given in \n the \ previous exercise.", FontSize->14]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell[TextData[{ StyleBox[ " (a) Use the limit definition of derivative to compute \n \ G'[0] (", FontSize->14], StyleBox["Hint", FontSize->14, FontColor->RGBColor[0, 0, 1]], StyleBox[": Look back at Note 3.)", FontSize->14] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell[TextData[StyleBox[ " (b) Make a table of values for G'. Do you recognize \n the \ function values?", FontSize->14]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell[TextData[StyleBox[" (c) Plot the points from the table. ", FontSize->14]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell[TextData[StyleBox[" (d) Find a formula for G'.", FontSize->14]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell[TextData[{ "\n", StyleBox[ "End Lab I.7. If you wish, save your work. Remember - it's easier to save \ if you FIRST delete the pictures. 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