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1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Plot" -1 13 1 {CSTYLE "" -1 -1 "Time s" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 24 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 258 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 1 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Heading 1" -1 259 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 260 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 261 1 {CSTYLE "" -1 -1 "T imes" 1 12 0 0 0 1 1 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Heading 1" -1 262 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }} {SECT 0 {PARA 256 "" 0 "" {TEXT -1 25 "The Accumulation Function" }} {PARA 257 "" 0 "" {TEXT -1 0 "" }{TEXT 258 32 "Fundamental Theorem of \+ Calculus " }}{PARA 257 "" 0 "" {TEXT -1 0 "" }{TEXT 349 34 "An Explora tory Introduction-Part 2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 17 "Notes to Student:" }}{PARA 0 "" 0 "" {TEXT -1 207 "Now that we have explored the Riemann Rectangle Approxim ation Method we are ready to investigate the area question further. Pa y particular attention to the notation used in the lab and the stated \+ objectives." }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 13 "Introduction:" }} {PARA 0 "" 0 "" {TEXT -1 81 "Once again, we will focus on seeking the \+ answer to the area question posed below." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 199 "\"Given a function f which is c ontinuous and non-negative over a closed interval [a,b], how do we fin d the area enclosed beneath the curve and above the horizontal axis ov er the designated interval ?\"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 173 "In this second module we will discover a n alternative method to the Riemann Rectangle Approximation Method tha t we examined in our first module. Proceed to the next section." }}} {SECT 1 {PARA 3 "" 0 "" {TEXT -1 19 "Objectives Defined:" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }{TEXT 359 0 "" }{TEXT 360 5 "Time:" }{TEXT 361 11 " 50 minutes" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT -1 214 "The purpose of this module is to explore and define a ne w method of determining the area beneath a curve and further extension s other than the Riemann Rectangle Approximation Method. This goal wil l be achieved by: " }}{PARA 260 "" 0 "" {TEXT -1 74 "1. Defining a con tinuous function called the Accumulation Function, A(x). " }}{PARA 260 "" 0 "" {TEXT -1 124 "2. Exploring animated graphs of a rectangle, triangle, trapezoid, and parabola with their respective Accumulation \+ Functions." }}{PARA 260 "" 0 "" {TEXT -1 72 "3. Formulating generaliza tions drawn upon observations and computations." }{TEXT 354 0 "" }} {PARA 0 "" 0 "" {TEXT -1 28 "Proceed to the next section." }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 6 "Focus:" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "Suppose we wish to " }{TEXT 355 21 "find the area beneath" }{TEXT -1 12 " the curve " }{XPPEDIT 257 1 "f(x) = 3*x^2;" "6#/-%\"fG6#%\"xG *&\"\"$\"\"\"*$F'\"\"#F*" }{TEXT -1 55 " over the interval [2,4]. Pla ce the cursor in the red " }{TEXT 256 7 "restart" }{TEXT 265 1 " " } {TEXT -1 32 "Maple command and press \"Enter\"." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 279 "restart:\nwith(plots):\nf(x)=3*x^2:\na1:=plot (3*x^2,x=2..4,title=\"What is the Area\\n Beneath the Curve?\",filled= true,view=[0..4,0..60],font=[TIMES,BOLD,14],titlefont=[TIMES,BOLD,14], color=aquamarine):\na2:=plot(3*x^2,x=2..4,view=[0..4,0..60],thickness= 3, color=red):\ndisplay(a1,a2);\n" }}}{PARA 0 "" 0 "" {TEXT -1 330 "Up to this point the only approach to this problem has been the Riemann \+ Rectangle Approximation Method. This method as you recall was very lab or intinsive and notationally cumbersome to do by hand. It is our goal to devise another approach to determining the area beneath a curve by constructing a new function which we will call " }{TEXT 356 28 "\" Th e Accumulation Function\"" }{TEXT -1 30 ". Proceed to the next section ." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 " " 0 "" {TEXT -1 35 "Defining the Accumulation Function:" }}{PARA 0 "" 0 "" {TEXT -1 17 "We will define a " }{TEXT 263 8 "function" }{TEXT -1 7 " which " }{TEXT 264 13 "\"accumulates\"" }{TEXT -1 49 " area ove r an interval. We'll call this function " }{TEXT 260 50 "the Accumulat ion Function and designate it as A(x)" }{TEXT 261 1 "." }{TEXT -1 1 " \+ " }{TEXT 262 241 "Let's see how this function might behave for a recta ngle, a triangle, a trapezoid, and a parabola. Work through each figur e and the questions which accompany them. Examine the graphs carefully taking time to view and analyze the animations. " }{TEXT -1 28 "Proce ed to the next section." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "The Rectangle:(Animated)" }}{PARA 0 "" 0 "" {TEXT -1 114 "Consider a very simple rectangle lying in the first q uadrant over the interval [0,2]. Place the cursor in the red " }{TEXT 259 8 "restart " }{TEXT -1 32 "Maple command and press \"Enter\"." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 197 "restart:\nwith(plots):\na:=0:\nb:=2:\nc:=0:\nd:=10:\nn:=20:\nf:=x ->1:\nplot(f,x=a..b,filled=true,title=\"Simple Rectangle\\nArea = base x height\",tickmarks=[2,2], titlefont=[TIMES,BOLD,14],color=aquamarin e);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 " " }{TEXT 279 25 "/*Press Enter to Continue" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 " We can see that the " }{TEXT 269 9 "area is 2" }{TEXT -1 6 ". Our " } {TEXT 266 21 "Accumulation Function" }{TEXT -1 2 ", " }{TEXT 267 4 "A( x)" }{TEXT -1 29 ", needs to return a value of " }{TEXT 268 1 "2" } {TEXT -1 165 " when our inputs for x vary across the same interval [0, 2]. The code below will graph the Accumulation Function, A(x), on the \+ same graph as the rectangle. The graph " }{TEXT 288 8 "animates" } {TEXT -1 137 " the \"sweeping out\" of the area beneath the rectangle \+ as x varies from 0 to 2 along with the plot of A(x). In other words, \+ it shows the " }{TEXT 280 73 "connection between the swept out area an d the Accumulation Function, A(x)" }{TEXT -1 28 ". Press \"Enter\" to \+ continue." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 768 "a:=0:\nb:=2: \nc:=0:\nd:=3:\nn:=20:\nDigits:=4:\nf:x->1:g:=value(Int(f(t),t=a..x)): \nan1:=plot(f(x),x=a..b,view=[a..b,c..d],color=red,thickness=3):\nfor \+ i from 1 to n do \nk:=a+(b-a)*i/n:\nh:=evalf(value(Int(f(x),x=a..k))): \nan2[i]:=plot(f(x),x=a..k,filled=true,view =[a..b,c..d],color=AQUAMA RINE):\nan3[i]:=plot(g(x),x=a..k,view=[a..b,c..d],color=blue,thickness =3):\nan4[i]:=display(textplot([1,2,convert(h,string)],font=[TIMES,BOL D,14],color=blue)):\nend do:\np:=plots[display]([seq(an2[i],i=1..n)],i nsequence=true):q:=plots[display]([seq(an3[i],i=1..n)],insequence=true ):\nr:=plots[display]([seq(an4[i],i=1..n)],insequence=true):\ndisplay( r,an1,p,q,title=\"Rectangle and A(x)\",titlefont=[TIMES,BOLD,14],tickm arks=[2,4],TEXT([0.6,2],'`A(x) = `',FONT(TIMES,BOLD,14),COLOR(RGB,0,0, 1)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "1. \"Click\" on the figu re above and use the " }{TEXT 357 18 "Animation Toolbar " }{TEXT -1 25 "which will appear at the " }{TEXT 358 15 "top of the page" }{TEXT -1 113 ". Run the animation several times.When done, place the cursor \+ back at the prompt > and press \"Enter\" to continue." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 7 "2. The " }{TEXT 272 4 "blue" }{TEXT -1 16 " function is the" } {TEXT 271 1 " " }{TEXT 270 28 "Accumulation Function, A(x)," }{TEXT -1 11 " while the " }{TEXT 273 3 "red" }{TEXT -1 10 " function " } {TEXT 281 21 "defines the rectangle" }{TEXT -1 30 " and is the constan t function " }{TEXT 274 8 "f(x) = 1" }{TEXT -1 89 " over [0,2]. As x i ncreases, the area increases, and hence, the height of A(x) increases. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 239 "Q1. As the rectangle's area \+ filled, in what manner did the area function A(x) grow? Explain your a nswer (You may view the animation as often as you like by clicking on \+ it. Place the cursor at the prompt > and press \"Enter\" when finished )." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 117 "Q2. From the graph determine A(x) for the inputs \+ x = 0, 0.2, 0.4, 0.6, 0.8, 1, and 2. Create a t-table for your work." }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{TEXT 275 0 "" }{TEXT 276 20 "x-in put | A(x) value" }}{PARA 257 "" 0 "" {TEXT -1 5 "| " }}{PARA 257 " " 0 "" {TEXT -1 5 "| " }}{PARA 257 "" 0 "" {TEXT -1 5 "| " }} {PARA 257 "" 0 "" {TEXT -1 7 "... " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "Q3. If our in terval is designated as [a,b], then what are the values of A(a) & A(b) ? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "Q4. Write out the equation A(b) - A(a) = . How does this compare to the area?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "Q5. Using the graph and your t-tab le from Q2, write an equation for A(x). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 12 "Q6. Compute " }{TEXT 277 5 "A'(x)" }{TEXT -1 54 " \+ and compare it to the rectangle's defining function, " }{TEXT 278 4 "f (x)" }{TEXT -1 22 ". What do you notice? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 23 "The Triangl e:(Animated)" }}{PARA 0 "" 0 "" {TEXT -1 113 "Our second familiar shap e is the triangle. Execute the following Maple commands by placing the cursor in the red " }{TEXT 362 8 "restart " }{TEXT -1 82 "command and pressing \"Enter\". The code displays a triangle over the interval [0 ,2]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 206 "restart;\nwith(plots):\na:=0:\nb:=2:\nc:=0:\nd:=10: \nn:=20:\nf:=x->2*x:\nplot(f(x),x=a..b,filled=true,title=\"Simple Tria ngle\\nArea = (base x height)/2\",tickmarks=[2,2], titlefont=[TIMES,BO LD,14],color=aquamarine);;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "We \+ see that the " }{TEXT 284 9 "area is 4" }{TEXT -1 6 ". Our " }{TEXT 282 21 "Accumulation Function" }{TEXT -1 2 ", " }{TEXT 283 4 "A(x)" } {TEXT -1 29 ", needs to return a value of " }{TEXT 286 1 "4" }{TEXT -1 76 " when our inputs for x vary across the same interval [0,2]. Th e graph below" }{TEXT 363 9 " animates" }{TEXT -1 136 " the \"sweeping out\" of the area beneath the triangle as x varies from 0 to 2 along \+ with the plot of A(x). In other words, it shows the " }{TEXT 285 73 " connection between the swept out area and the Accumulation Function, A (x)" }{TEXT -1 28 ". Press \"Enter\" to continue." }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 781 "a:=0:\nb:=2:\nc:=0:\nd:=5:\nn:=20:\nDigits: =4:\nf:x->2*x:\ng:=value(Int(f(t),t=a..x)):\nan1:=plot(f(x),x=a..b,vie w=[a..b,c..d],color=red,thickness=3):\nfor i from 1 to n do \nk:= a+(b-a)*i/n:\nh:=evalf(value(Int(f(x),x=a..k))):\n\nan2[i]:=plot(f(x), x=a..k,filled=true,view=[a..b,c..d],color=AQUAMARINE):\nan3[i]:=plot(g (x),x=a..k,view=[a..b,c..d],color=blue, thickness=3):\nan4[i]:=display (textplot([1.1,3.6,convert(h,string)],font=[TIMES,BOLD,14],color=blue) ):\nend do:\np:=plots[display]([seq(an2[i],i=1..n)],insequence=true):q :=plots[display]([seq(an3[i],i=1..n)],insequence=true):\nr:=plots[disp lay]([seq(an4[i],i=1..n)],insequence=true):\ndisplay(r,an1,p,q,title= \"Triangle and A(x)\",titlefont=[TIMES,BOLD,14],tickmarks=[2,2],TEXT([ 0.7,3.6],'`A(x) = `',FONT(TIMES,BOLD,14),COLOR(RGB,0,0,1)));" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 " " }{TEXT -1 0 "" }{TEXT 289 25 "/*Press Enter to Continue" }{MPLTEXT 1 0 2 " " }{TEXT -1 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "1. \"Click\" on the figure above a nd use the " }{TEXT 290 17 "Animation Toolbar" }{TEXT -1 26 " which wi ll appear at the " }{TEXT 291 15 "top of the page" }{TEXT -1 113 ". Ru n the animation several times.When done, place the cursor back at the \+ prompt > and press \"Enter\" to continue." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 7 "2. The " } {TEXT 294 4 "blue" }{TEXT -1 16 " function is the" }{TEXT 293 1 " " } {TEXT 292 28 "Accumulation Function, A(x)," }{TEXT -1 11 " while the \+ " }{TEXT 295 3 "red" }{TEXT -1 10 " function " }{TEXT 301 20 "defines \+ the triangle" }{TEXT -1 21 " and is the function " }{TEXT 296 9 "f(x) \+ = 2x" }{TEXT -1 89 " over [0,2]. As x increases, the area increases, a nd hence, the height of A(x) increases." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 238 "Q1. As the triangle's area filled, in what manner did th e area function A(x) grow? Explain your answer (You may view the anima tion as often as you like by clicking on it. Place the cursor at the p rompt > and press \"Enter\" when finished )." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "Q2. Fro m the graph determine A(x) for the inputs x = 0, 0.5, 1, 1.5, and 2. \+ Create a t-table for your work." }}{PARA 257 "" 0 "" {TEXT -1 1 " " } {TEXT 297 0 "" }{TEXT 298 20 "x-input | A(x) value" }}{PARA 257 "" 0 " " {TEXT -1 5 "| " }}{PARA 257 "" 0 "" {TEXT -1 5 "| " }}{PARA 257 "" 0 "" {TEXT -1 5 "| " }}{PARA 257 "" 0 "" {TEXT -1 7 "... \+ " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "Q3. If our interval is designated as [a,b], then wha t are the values of A(a) & A(b)? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "Q4. Write out the equati on A(b) - A(a) = . How does this compare to the area?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "Q5. \+ Using the graph and your t-table from Q2, write an equation for A(x). \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "Q6. Compute " }{TEXT 299 5 "A'(x)" }{TEXT -1 53 " and com pare it to the triangle's defining function, " }{TEXT 300 4 "f(x)" } {TEXT -1 22 ". What do you notice? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "The Trapezoid:(Animate d)" }}{PARA 262 "" 0 "" {TEXT 352 128 "We now investigate another fami liar shape: the trapezoid. Execute the following Maple commands by pla cing the cursor in the red " }{TEXT 302 7 "restart" }{TEXT 353 89 " co mmand and pressing \"Enter\". The display is a simple trapezoid over t he interval [1,3]." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 234 "resta rt;\nwith(plots):\na:=1:\nb:=3:\nc:=0:\nd:=20:\nn:=20:\nf:=x->2*x+2:\n plot(f(x),x=a..b,filled=true,title=\"Simple Trapezoid\\nArea = [(base1 +base2) x height]/2\",tickmarks=[3,8], titlefont=[TIMES,BOLD,14],view= [0..3,0..8],color=aquamarine);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "Confirm that the " }{TEXT 305 10 "area is 12" }{TEXT -1 6 ". Our " } {TEXT 303 21 "Accumulation Function" }{TEXT -1 2 ", " }{TEXT 304 4 "A( x)" }{TEXT -1 29 ", needs to return a value of " }{TEXT 308 2 "12" } {TEXT -1 165 " when our inputs for x vary across the same interval [1, 3]. The code below will graph the Accumulation Function, A(x), on the \+ same graph as the trapezoid. The graph " }{TEXT 307 8 "animates" } {TEXT -1 137 " the \"sweeping out\" of the area beneath the trapezoid \+ as x varies from 1 to 3 along with the plot of A(x). In other words, \+ it shows the " }{TEXT 306 73 "connection between the swept out area an d the Accumulation Function, A(x)" }{TEXT -1 28 ". Press \"Enter\" to \+ continue." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 781 "a:=1:\nb:=3: \nc:=0:\nd:=16:\nn:=20:\nDigits:=4:\nf:x->2*x+2:\ng:=value(Int(f(t),t= a..x)):\nan1:=plot(f(x),x=a..b,view=[(a-1)..b,c..d],color=red,thicknes s=3):\nfor i from 1 to n do \nk:=a+(b-a)*i/n:\nh:=evalf(value(Int(f(x) ,x=a..k))):\nan2[i]:=plot(f(x),x=a..k,filled=true,view=[a..b,c..d],col or=AQUAMARINE):\nan3[i]:=plot(g(x),x=a..k,view=[a..b,c..d],color=blue, thickness=3):\nan4[i]:=display(textplot([1.8,11,convert(h,string)],fon t=[TIMES,BOLD,14],color=blue)):\nend do:\np:=plots[display]([seq(an2[i ],i=1..n)],insequence=true):\nq:=plots[display]([seq(an3[i],i=1..n)],i nsequence=true):\nr:=plots[display]([seq(an4[i],i=1..n)],insequence=tr ue):\ndisplay(r,an1,p,q,title=\"Trapezoid and A(x)\",titlefont=[TIMES, BOLD,14],tickmarks=[3,8],TEXT([1.3,11],'`A(x) = `',FONT(TIMES,BOLD,14) ,COLOR(RGB,0,0,1)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 " \+ " }{TEXT -1 0 "" }{TEXT 309 25 "/*Press Enter to Continue" } {MPLTEXT 1 0 2 " " }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "1. \"Click\" on the figure above and use the " }{TEXT 310 17 "Anim ation Toolbar" }{TEXT -1 26 " which will appear at the " }{TEXT 311 15 "top of the page" }{TEXT -1 113 ". Run the animation several times. When done, place the cursor back at the prompt > and press \"Enter\" t o continue." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 7 "2. The " }{TEXT 314 4 "blue" }{TEXT -1 16 " function is the" }{TEXT 313 1 " " }{TEXT 312 28 "Accumulation Functi on, A(x)," }{TEXT -1 11 " while the " }{TEXT 315 3 "red" }{TEXT -1 10 " function " }{TEXT 321 21 "defines the trapezoid" }{TEXT -1 21 " and \+ is the function " }{TEXT 316 13 "f(x) = 2x + 2" }{TEXT -1 89 " over [1 ,3]. As x increases, the area increases, and hence, the height of A(x) increases." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 241 "Q1. As the trapez oid's area filled, in what manner did the area function A(x) grow? Exp lain your answer. (You may view the animation as often as you like by \+ clicking on it. Place the cursor at the prompt > and press \"Enter\" w hen finished ). " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "Q2. From the graph determine A(x) for the inputs x = 1, 1.5, 2, 2.5, and 3. Create a t-table for your w ork." }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{TEXT 317 0 "" }{TEXT 318 20 "x-input | A(x) value" }}{PARA 257 "" 0 "" {TEXT -1 5 "| " }} {PARA 257 "" 0 "" {TEXT -1 5 "| " }}{PARA 257 "" 0 "" {TEXT -1 5 "| " }}{PARA 257 "" 0 "" {TEXT -1 7 "... " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "Q3. If o ur interval is designated as [a,b], then what are the values of A(a) & A(b)? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "Q4. Write out the equation A(b) - A(a) = ...How does this compare to the area?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "Q 5. Using the graph and your t-table from Q2, write an equation for A(x ). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 12 "Q6. Compute " }{TEXT 319 5 "A'(x)" }{TEXT -1 54 " and c ompare it to the trapezoid's defining function, " }{TEXT 320 4 "f(x)" }{TEXT -1 22 ". What do you notice? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 34 "A More Difficult Figure:(Animated)" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "Now we turn our attention to a more challenging area . Consider the function " }{TEXT 322 7 "f(x) = " }{XPPEDIT 323 1 "x^2; " "6#*$%\"xG\"\"#" }{TEXT -1 96 " over the interval [0,4]. Execute the following Maple commands by placing the cursor in the red " }{TEXT 324 7 "restart" }{TEXT -1 31 " command and pressing \"Enter\". " }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 207 "restart;\nwith(plots):\na:=0:\nb:=4:\nc:=0:\nd:=20:\nn:=20:\nf:=x ->x^2:\nplot(f(x),x=a..b,filled=true,title=\"Simple Parabola\\nArea = \+ ?\",tickmarks=[4,8], titlefont=[TIMES,BOLD,14],view=[0..4,0..16],color =aquamarine);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 " " }{TEXT 287 25 "/*Press Enter to Continue" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "We do not have an elementray formula to compute the area \+ of the shaded figure! Our " }{TEXT 325 21 "Accumulation Function" } {TEXT -1 2 ", " }{TEXT 326 4 "A(x)" }{TEXT -1 247 ", needs to return t he correct area value when our inputs for x vary across the same inter val [0,4]. But how can we come up with the correct function? The code \+ below will graph the Accumulation Function, A(x), on the same graph as the parabola and " }{TEXT 328 7 "animate" }{TEXT -1 103 " the \"sweep ing out\" of the area beneath the parabola as x varies from 0 to 4. I t once again shows the " }{TEXT 327 73 "connection between the swept o ut area and the Accumulation Function, A(x)" }{TEXT -1 70 " and should help us find the defining A(x). Press \"Enter\" to continue." }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 756 "a:=0:\nb:=4:\nc:=0:\nd:=25:\nn:=20:\nDigits:=4:\nf:x ->x^2:\ng:=value(Int(f(t),t=a..x)):\nan1:=plot(f(x),x=a..b,view=[a..b, c..d],color=red,thickness=3):\nfor i from 1 to n do k:=a+(b-a)*i/n:\na n2[i]:=plot(f(x),x=a..k,filled=true,view=[a..b,c..d],color=AQUAMARINE) :\nh:=evalf(value(Int(f(x),x=a..k))):\nan4[i]:=display(textplot([1.9,1 5,convert(h,string)],font=[TIMES,BOLD,14],color=blue)):\nan3[i]:=plot( g(x),x=a..k,view=[a..b,c..d],color=blue,thickness=3):\nend do:\np:=plo ts[display]([seq(an2[i],i=1..n)],insequence=true):q:=plots[display]([s eq(an3[i],i=1..n)],insequence=true):\nr:=plots[display]([seq(an4[i],i= 1..n)],insequence=true):\ndisplay(r,an1,p,q,title=\"Parabola and A(x) \",titlefont=[TIMES,BOLD,14],TEXT([1.2,15],'`A(x) = `',FONT(TIMES,BOLD ,14),COLOR(RGB,0,0,1)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "1. \" Click\" on the figure above and use the " }{TEXT 329 17 "Animation Too lbar" }{TEXT -1 26 " which will appear at the " }{TEXT 330 15 "top of \+ the page" }{TEXT -1 113 ". Run the animation several times.When done, \+ place the cursor back at the prompt > and press \"Enter\" to continue. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 7 "2. The " }{TEXT 333 4 "blue" }{TEXT -1 16 " function \+ is the" }{TEXT 332 1 " " }{TEXT 331 28 "Accumulation Function, A(x)," }{TEXT -1 11 " while the " }{TEXT 334 3 "red" }{TEXT -1 10 " function \+ " }{TEXT 337 20 "defines the parabola" }{TEXT -1 20 " and is the funct ion" }{TEXT 338 1 " " }{TEXT 339 7 "f(x) = " }{XPPEDIT 340 0 "x^2;" "6 #*$%\"xG\"\"#" }{TEXT -1 88 "over [0,4]. As x increases, the area incr eases, and hence, the height of A(x) increases." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 232 "Q1. As the parabola fills, in what manner did the a rea function A(x) grow? Explain your answer. (You may view the animati on as often as you like by clicking on it. Place the cursor at the pro mpt > and press \"Enter\" when finished ). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "Q2. From the graph approximate A(x) for the inputs x = 1, 2, 3, and 4? Create a t- table for your work" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{TEXT 335 0 " " }{TEXT 336 20 "x-input | A(x) value" }}{PARA 257 "" 0 "" {TEXT -1 5 "| " }}{PARA 257 "" 0 "" {TEXT -1 5 "| " }}{PARA 257 "" 0 "" {TEXT -1 5 "| " }}{PARA 257 "" 0 "" {TEXT -1 7 "... " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "Q3.Using your data in Q2 and graph, find a suitable A(x). " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "Q4. How does " }{TEXT 341 5 "A'(x)" }{TEXT -1 46 " compar e to the parabola's defining function, " }{TEXT 342 4 "f(x)" }{TEXT -1 81 "? Is this consistent with the results of the rectangle, triangl e, and trapezoid? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "Q5. " }{TEXT 343 98 "Determining a functio n whose derivative is equal to a given function is called antidifferen tiation" }{TEXT -1 159 ". With that in mind, does the Accumulation Fun ction, A(x), which you wrote in Q3 have its derivative equal to the fu nction, f(x)? If it does, then we say that " }{TEXT 344 33 "A(x) is an antiderivative of f(x)" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "Q6. If our i nterval is designated as [a,b], then what are the values of A(a) & A(b )? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "Q7. What is the area beneath the parabola? Explain your answer." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 25 "Generalizing Our Results:" }}{PARA 0 "" 0 "" {TEXT -1 84 "After the previous examples you should \+ be able to complete the following statements." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "1. A(x) is called the _____________________." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "2. A(x) defines a continuous __________________." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "3. A'(x) = __________________." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 114 "4. The process of determining a function whose derivativ e equals a given function is called _____________________." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 43 "5. A(b) - A(a) = _________________________." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 37 "Finding the Antiderivative:(Animated)" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 192 "The Accu mulation Functions (antiderivatives) we previously found were fairly e asy to see from the graphs. In general, the antiderivative is easily d etermined for many elementary functions but " }{TEXT 350 90 "there are examples where finding the antiderivative is either very difficult or impossible" }{TEXT -1 16 ". Furthermore, " }{TEXT 364 53 "what if th e given function, f(x), dips below the axis" }{TEXT -1 217 " somewhere within the given interval? How does that affect the Accumulation Func tion? This next example is one such instance where the function, f(x), takes on both positive and negative values. Consider the function " } {TEXT 345 14 "f(x) = xsin(x)" }{TEXT -1 68 ". Execute the Maple comman ds below by placing the cursor in the red " }{TEXT 346 7 "restart" } {TEXT -1 30 " command and pressing \"Enter\"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 288 "restart;\nw ith(plots):\na:=-1:\nb:=2*Pi:\nc:=-5:\nd:=5:\nn:=20:\nf:=x->x*sin(x): \nan1:=plot(f(x),x=a..b,view=[a..b,c..d],color=aquamarine,filled=true, title=\"The Graph of\\n f(x) = xsin(x)\",titlefont=[TIMES,BOLD,14]):\n an2:=plot(f(x),x=a..b,view=[a..b,c..d],color=red,thickness=3):\ndispla y(an1,an2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 171 "Obviously this fu nction's antiderivative is much more challenging! Not only is the anti derivative going to be difficult to find but how do we interpret A(x)' s values now? " }{TEXT 365 54 "Is it still area if it goes below the h orizontal axis?" }{TEXT -1 120 " Before executing the next set of comm ands conjecture what you think the Accumulation Function's graph might look like. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 789 "a:=-1:\nb:=2*Pi:\nc:=-8:\nDigits:=4:\nd:=5:\nn: =20:\nf:x->x*sin(x):\ng:=value(Int(f(t),t=a..x)):\nan1:=plot(f(x),x=a. .b,view=[a..b,c..d],color=red,thickness=3,axesfont=[TIMES,BOLD,14]):\n for i from 1 to n do \nk:=a+(b-a)*i/n:\nan2[i]:=plot(f(x),x=a..k,fille d=true,view=[a..b,c..d],color=GREEN):\nan3[i]:=plot(g(x),x=a..k,view=[ a..b,c..d],color=blue,thickness=3):\nh:=evalf(value(Int(f(x),x=a..k))) ;\nan4[i]:=display(textplot([3.9,4.2,convert(h,string)],font=[TIMES,BO LD,14],color=blue)):\nend do:\np:=plots[display]([seq(an2[i],i=1..n)], insequence=true):\nq:=plots[display]([seq(an3[i],i=1..n)],insequence=t rue):\nr:=plots[display]([seq(an4[i],i=1..n)],insequence=true):\ndispl ay(r,an1,p,q,title=\"f(x)=xsin(x)\",titlefont=[TIMES,BOLD,16],TEXT([2, 4.2],'`Accumulator = `',FONT(TIMES,BOLD,14),COLOR(RGB,0,0,1)));" }} {PARA 13 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 " 1. \"Click\" on the figure above and use the " }{TEXT 366 17 "Animatio n Toolbar" }{TEXT -1 26 " which will appear at the " }{TEXT 367 15 "to p of the page" }{TEXT -1 112 ". Run the animation several times.When d one, place the cursor back at the prompt > and press \"Enter\" to cont inue" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "Q1. Are the results consistent with what you predicted? E xplain why A(x) looks as it does." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "Q2. Can we still interp ret the result for A(x) as an \"area accumulator\" or is there a bette r way to define it. Please discuss your thoughts." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 120 "The Accumulator Function (antiderivative) is given by \+ executing the code below. Press \"Enter\" to see the antiderivative." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "A(x):=int(f(x),x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "Q3. Now compute A'(x). Does it equal f(x) as it did in the other examples? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "Using the " }{TEXT 347 4 "A(x)" } {TEXT -1 22 " found in Q2, Compute " }{TEXT 348 9 "A(b)-A(a)" }{TEXT -1 95 " on the interval [-1, 2pi]. How does it compare with the accumu latuion value seen on the graph?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 82 "Even though we cannot interpret the endin g value as area, we can refer to it as a " }{TEXT 368 10 "net change" }{TEXT -1 7 " or as " }{TEXT 369 11 "signed area" }{TEXT -1 77 ". A(x) takes on both positive and negative values for differing values of x. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 " " }{TEXT -1 40 " /*Open the next section with the cursor." }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "Summarizing our Results:" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 257 "" 0 "" {TEXT 351 81 "After completing this module it is hoped that you understand the following items." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT -1 64 "1. We can define a cont inuous function, A(x), which accumulates " }{TEXT 370 11 "signed area " }{TEXT -1 24 " over a given interval. " }}{PARA 0 "" 0 "" {TEXT 371 97 "2. The net accumulation of A(x) over a closed interval [a,b] is si mply determined by A(b) - A(a)." }}{PARA 260 "" 0 "" {TEXT -1 3 "3. " }{TEXT 372 3 "If " }{TEXT -1 310 "we can find the Accumulation Functio n, A(x), for a given function, f(x), and f(x) is non-negative over a c losed interval [a,b], then A(b) - A(a) represents the area beneath f(x ) and the horizontal axis over the prescribed interval. This is far ea sier to perform than the Riemann Rectangle Approximation Method." }} {PARA 260 "" 0 "" {TEXT -1 144 "4. There are instances when finding th e Accumulation Function is either very difficult or impossible. This i s the major drawback to this method." }}{PARA 0 "" 0 "" {TEXT 373 80 " 5. The Accumulation Function, A(x), is often called an antiderivativ e of f(x)." }}{PARA 0 "" 0 "" {TEXT 374 94 "6. The Accumulation Functi on, A(x), has the unique property that its derivative A'(x) = f(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{MARK "4" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }