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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[ 63136, 1821]*) (*NotebookOutlinePosition[ 63999, 1851]*) (* CellTagsIndexPosition[ 63955, 1847]*) (*WindowFrame->Normal*) Notebook[{ Cell["Calculus I Lab I.6: Visualizing Riemann Sums", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Avantgarde", FontSize->20, FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontColor->RGBColor[1, 0, 1], FontVariations->{"Underline"->True, "Outline"->False, "Shadow"->False}], Cell[TextData[StyleBox["The purpose of this lab is to help you visualize \ Riemann sums, their limit and definite integrals.", 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[TextData[{ "To understand ", StyleBox["instantaneous velocity", FontSlant->"Italic"], " one considers a graph of distance versus time. The ", StyleBox["average ", FontColor->RGBColor[1, 0, 1]], "velocity is represented on the graph by the slope of a secant line. ", StyleBox["Instantaneous", FontColor->RGBColor[1, 0, 1]], " velocity was defined to be the limit of the slopes of the secant lines, \ i.e. the slope of the tangent line. In this way, we constructed the velocity \ function from the distance function. We applied this idea to other functions \ (not just the distance function). \nConclusion: if (a,f(a)) is a point on \ the graph of f, then", StyleBox["\n the instantaneous rate of change of f = the slope\n \ of the tangent line to the graph of f at (a,f(a))", FontColor->RGBColor[0, 0, 1]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[{ StyleBox["Later we started with a velocity function and asked about \ distance. Again, we used the graph to lead us. Here's a quick summary. \nThe \ distance travelled during a small time interval can be well approximated by \ the area of a particular rectangle associated with the graph. The distance \ travelled during any time interval can then be approximated by dividing the \ interval into small subintervals and adding the areas of the corresponding \ rectangles (such a sum is called a ", FontSize->14], StyleBox["Riemann", FontSize->14, FontSlant->"Italic"], StyleBox[" sum). The limit of the Riemann sums (as the subintervals get \ smaller) is the total distance travelled during the time interval. \n\nAs was \ the case for differentiation, we apply the limiting process to other \ functions (not just the velocity function). The ", FontSize->14], StyleBox["definite integral", FontSize->14, FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], StyleBox[" of f from a to b is the ", FontSize->14], StyleBox["limit of", FontSize->14, FontColor->RGBColor[0, 0, 1]], StyleBox[" the corresponding ", FontSize->14], StyleBox["Riemann sums", FontSize->14, FontColor->RGBColor[0, 0, 1]], StyleBox[". In this lab we study the geometric interpretation of these \ ideas.", FontSize->14] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt"], Cell[TextData[{ StyleBox["As usual, enter input cells by putting the cursor anywhere\nin \ the cell and hit \"Enter\", or \"Shift\" and \"Return\" together,\nor \"Shift\ \" and \"Enter\" together. The only exception is the\nfollowing cell, which \ contains some \"hidden\" programs.\n", Evaluatable->False, CellGroupingRules->"OutputGrouping", AspectRatioFixed->True, FontSize->14], StyleBox["If you restart this notebook at a later date, you will need to\n\ reenter the following cell before you can continue successfully!", FontSize->14] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontColor->RGBColor[1, 0, 1]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Click the ", FontWeight->"Bold"], StyleBox["outermost", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], StyleBox[" bracket on the ", FontWeight->"Bold"], StyleBox["far right", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], StyleBox[" of ", FontWeight->"Bold"], StyleBox["THIS", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0], FontVariations->{"Underline"->True}], StyleBox[" cell; it should become highlighted. Hit ", FontWeight->"Bold"], StyleBox["\"", Evaluatable->False, CellGroupingRules->"OutputGrouping", AspectRatioFixed->True, FontSize->14], StyleBox["Enter\", or \"Shift\" and \"Return\" together, or \"Shift\" and \ \"Enter\" together.", Evaluatable->False, CellGroupingRules->"OutputGrouping", AspectRatioFixed->True, FontSize->14, FontWeight->"Bold"], StyleBox[" ", Evaluatable->False, CellGroupingRules->"OutputGrouping", AspectRatioFixed->True, FontSize->14], " ", StyleBox["A horizontal line should appear below this cell.", FontWeight->"Bold"] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[BoxData[{ \(Lhs[f_, {x_, a_, b_, n_}] := Module[{delta, i, pic, plt, r}, delta = N[\(b - a\)\/n]; pic = Table[{{Hue[ .5, .2, 1], Rectangle[{a + i\ delta, 0}, {a + \((i + 1)\)\ delta, f /. {x \[Rule] a + i\ delta}}]}, {RGBColor[0, 0, 1], Line[{{a + i\ delta, 0}, {a + i\ delta, f /. {x \[Rule] a + i\ delta}}, { a + \((i + 1)\)\ delta, f /. {x \[Rule] a + i\ delta}}, { a + \((i + 1)\)\ delta, 0}}]}}, {i, 0, n - 1}]; plt = Plot[f, {x, a, b}, PlotRange \[Rule] All, AxesOrigin \[Rule] {0, 0}, DisplayFunction \[Rule] Identity]; Show[plt, Graphics[pic], FullGraphics[plt], DisplayFunction \[Rule] $DisplayFunction]; N[\((\[Sum]\+\(i = 0\)\%\(n - 1\)\((f /. {x \[Rule] a + i\ delta})\)) \)\ delta]]\), \(Rhs[f_, {x_, a_, b_, n_}] := Module[{delta, i, pic, plt, r}, delta = N[\(b - a\)\/n]; pic = Table[{{Hue[ .5, .2, 1], Rectangle[{a + i\ delta, 0}, {a + \((i + 1)\)\ delta, f /. {x \[Rule] a + \((i + 1)\)\ delta}}]}, { RGBColor[0, 0, 1], Line[{{a + i\ delta, 0}, {a + i\ delta, f /. {x \[Rule] a + \((i + 1)\)\ delta}}, { a + \((i + 1)\)\ delta, f /. {x \[Rule] a + \((i + 1)\)\ delta}}, { a + \((i + 1)\)\ delta, 0}}]}}, {i, 0, n - 1}]; plt = Plot[f, {x, a, b}, PlotRange \[Rule] All, AxesOrigin \[Rule] {0, 0}, DisplayFunction \[Rule] Identity]; Show[plt, Graphics[pic], FullGraphics[plt], DisplayFunction \[Rule] $DisplayFunction]; N[\((\[Sum]\+\(i = 1\)\%n\((f /. {x \[Rule] a + i\ delta})\))\)\ delta]]\), \(Bhs[f_, {x_, a_, b_, n_}] := Module[{delta, i, pic, plt, r}, delta = N[\(b - a\)\/n]; pic1 = Table[{{Hue[ .3, .4, 1], Rectangle[{a + i\ delta, 0}, {a + \((i + 1)\)\ delta, f /. {x \[Rule] a + \((i + 1)\)\ delta}}]}, Line[{{a + i\ delta, 0}, {a + i\ delta, f /. {x \[Rule] a + \((i + 1)\)\ delta}}, { a + \((i + 1)\)\ delta, f /. {x \[Rule] a + \((i + 1)\)\ delta}}, { a + \((i + 1)\)\ delta, 0}}]}, {i, 0, n - 1}]; pic11 = Table[{RGBColor[0, 0, 1], Line[{{a + i\ delta, f /. {x \[Rule] a + \((i + 1)\)\ delta}}, { a + \((i + 1)\)\ delta, f /. {x \[Rule] a + \((i + 1)\)\ delta}}}]}, {i, 0, n - 1}]; pic2 = Table[{{Hue[ .3, .4, 1], Rectangle[{a + i\ delta, 0}, {a + \((i + 1)\)\ delta, f /. {x \[Rule] a + i\ delta}}]}, Line[{{a + i\ delta, 0}, {a + i\ delta, f /. {x \[Rule] a + i\ delta}}, {a + \((i + 1)\)\ delta, f /. {x \[Rule] a + i\ delta}}, {a + \((i + 1)\)\ delta, 0}}]}, {i, 0, n - 1}]; pic22 = Table[{RGBColor[1, 0, 0], Line[{{a + i\ delta, f /. {x \[Rule] a + i\ delta}}, { a + \((i + 1)\)\ delta, f /. {x \[Rule] a + i\ delta}}}]}, {i, 0, n - 1}]; plt = Plot[f, {x, a, b}, PlotRange \[Rule] All, AxesOrigin \[Rule] {0, 0}, DisplayFunction \[Rule] Identity]; Show[plt, Graphics[pic1], Graphics[pic2], Graphics[pic11], Graphics[pic22], FullGraphics[plt], DisplayFunction \[Rule] $DisplayFunction]; N[\((\[Sum]\+\(i = 1\)\%n\((f /. {x \[Rule] a + i\ delta})\) - \[Sum]\+\(i = 0\)\%\(n - 1\)\((f /. {x \[Rule] a + i\ delta}) \))\)\ delta]]\), \(Area[f_, {x_, a_, b_}] := Module[{p, q, r, s}, p = Join[{{a, 0}}, \(Plot[f, {x, a, b}, DisplayFunction \[Rule] Identity] \)\[LeftDoubleBracket]1, 1, 1, 1 \[RightDoubleBracket], {{b, 0}}]; q = Graphics[{Hue[ .5, .2, 1], Polygon[p]}]; r = Plot[f, {x, a, b}, AxesOrigin \[Rule] {0, 0}, PlotRange \[Rule] All, DisplayFunction \[Rule] Identity]; Show[{r, q, FullGraphics[r]}, DisplayFunction \[Rule] $DisplayFunction]; ]\)}], "Input"] }, Closed]], Cell[TextData[{ "You've just entered new commands into ", StyleBox["Mathematica", FontSlant->"Italic"], ": one is called ", StyleBox["Lhs", FontWeight->"Bold"], ", another is called ", StyleBox["Rhs", FontWeight->"Bold"], ". ", StyleBox["Lhs", FontWeight->"Bold"], " is designed to help you visualize a left hand Riemann sum. You input a \ function, a left endpoint, a right endpoint, and the number of subintervals. \ Let's do an example to see how it works." }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[{ StyleBox["Example", FontSize->14, FontWeight->"Plain", FontColor->RGBColor[1, 0, 1], FontVariations->{"Underline"->True}], StyleBox[" ", FontSize->14], StyleBox["(As usual, enter any input cells you encounter)", FontSize->14, FontColor->GrayLevel[0]] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->18, FontWeight->"Bold", FontSlant->"Plain", FontTracking->"Plain", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False}], Cell[BoxData[ \(Lhs[x\^2, {x, 0, 2, 4}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "To find the area of a rectangle, you multiply its height by its width. \ The shaded rectangles above all have width 1/2 (Why can you see only three of \ the four?). Their respective heights are found by evaluating the function at \ the left endpoints (be sure you see why this is true). Adding up the areas of \ the shaded rectangles, we get the following Riemann sum: \n \n \ f(0)*1/2 + f(1/2)*1/2 + f(1)*1/2 + f(1.5)*1/2 \n \n \ = 0^2*1/2+ (1/2)^2*1/2+ (1)^2*1/2 + (1.5)^2* 1/2\n \n = 1.75 \ (", StyleBox[ "this number is recorded below the graph \n to 2 \ decimal places", FontColor->RGBColor[0, 0, 1]], ")" }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[StyleBox["Exercise 1", FontSize->14, FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->18, FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True, "Outline"->False, "Shadow"->False}], Cell["\<\ Is the left hand Riemann sum for 8 subintervals going to be bigger \ or smaller than the one we just computed for 4? Explain your reasoning. Now, \ enter the following cell to see if you were right.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[BoxData[ \(Lhs[x^2, {x, 0, 2, 8}]\)], "Input"], Cell[TextData[StyleBox["The Limit of the Riemann Sums", FontSize->14, FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->18, FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True, "Outline"->False, "Shadow"->False}], Cell["\<\ What happens if we use more subintervals? Keep in mind that we are \ interested in the limit of these Riemann sums as the number of subintervals \ increases. Let's try to visualize this limiting process. Enter the following \ cells.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[BoxData[ \(Lhs[x\^2, {x, 0, 2, 20}]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Lhs[x\^2, {x, 0, 2, 40}]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Lhs[x\^2, {x, 0, 2, 70}]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Lhs[x\^2, {x, 0, 2, 100}]\)], "Input", AspectRatioFixed->True], Cell["\<\ The sum of the areas of the shaded rectangles is moving toward the \ area under the curve. More generally, if the graph of f is above the x-axis, \ then the integral of f from x=a to x=b is the area under the curve, above \ the x-axis, between x=a and x=b. \ \>", "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[BoxData[ \(Area[x\^2, {x, 0, 2}]\)], "Input", AspectRatioFixed->True], Cell[TextData[StyleBox["Exercise 2", FontSize->14, FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->18, FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True, "Outline"->False, "Shadow"->False}], Cell[TextData[{ "Repeat the above process using ", StyleBox["Rhs", FontWeight->"Bold"], " instead of", StyleBox[" Lhs", FontWeight->"Bold"], ". Are successive Riemann sums going to get bigger? Why? Is the limit going \ to be the same?" }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[StyleBox["Area", FontSize->14, FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->18, FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True, "Outline"->False, "Shadow"->False}], Cell[TextData[{ "We already noted that if the graph of f is above the x-axis, the integral \ is the area under the curve. If the graph of f is below the x-axis, the \ function values are negative, so each piece of the Riemann sum is negative. \ (Recall that each piece of the Riemann sum is a function value times the \ length of the subinterval). The integral is therefore the negative of the \ area trapped between the curve and x-axis. If the graph of a function g on an \ interval [a,b] has parts above the x-axis and parts below the x-axis, the \ integral of g from a to b is the area above the x-axis ", StyleBox["minus", FontColor->RGBColor[0, 0, 1]], " the area ", StyleBox["below", FontColor->RGBColor[0, 0, 1]], " the x-axis." }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell[TextData[StyleBox["Example", FontSize->14, FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->18, FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True, "Outline"->False, "Shadow"->False}], Cell[TextData[ "Using geometric reasoning, find the integral of f[x]=Sin[x] from x=0 to x=2\ \[Pi]."], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Swiss721bt", FontSize->14], Cell["Solution", "Text", Evaluatable->False, AspectRatioFixed->True, 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