{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 }{PSTYLE "Normal " -1 0 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 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Title" -1 18 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 1 2 2 2 1 1 1 1 }3 1 0 0 12 12 1 0 1 0 2 2 19 1 }{PSTYLE "Heading 2 " -1 256 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 } 3 1 0 0 8 2 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 10 "Calculus I" }}{PARA 256 " " 0 "" {TEXT -1 23 "Lesson 24: Riemann Sums" }}}{PARA 4 "" 0 "" {TEXT -1 37 "Approximating an Area with Rectangles" }}{EXCHG {PARA 0 "" 0 " " {TEXT -1 34 "Consider the area under the curve " }{XPPEDIT 18 0 "f(x ) = x^3-5*x^2+x+10;" "6#/-%\"fG6#%\"xG,**$F'\"\"$\"\"\"*&\"\"&F+*$F'\" \"#F+!\"\"F'F+\"#5F+" }{TEXT -1 25 " on the interval [-1, 4]." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 21 "f:=x->x^3-5*x^2+x+10;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 20 "plot(f(x), x=-1..4);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 678 "How might we find the net area under this graph? If i t were drawn on graph paper, we could get an approximation to the net \+ area by counting the number of rectangles (and parts of rectangles) un der the graph that are above the axis, and subtracting the number that are below the axis . The calculus approach begins with a variant of \+ this simple idea: we approximate the required area by rectangles. Mor e explicitly, we divide the interval [-1, 4] into some number of equal sub-intervals, and then draw a rectangle over each sub-interval whose height is the height of the graph at the right-hand endpoint of the s ub-interval. We can draw the rectangles by taking advantage of " } {TEXT 257 7 "Maple's" }{TEXT -1 1 " " }{TEXT 258 8 "rightbox" }{TEXT -1 26 " command, which is in the " }{TEXT 259 7 "student" }{TEXT -1 9 " package." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(student) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "rightbox(f(x), x=-1..4 , 6);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "In this example, we used 6 sub-intervals. (That's what the last argument in the " }{TEXT 256 8 "rightbox" }{TEXT -1 117 " command is for.) We can add up the areas of the 6 rectangles easily enough (note that each rectangle has base \+ 6/5):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "evalf( 5/6*f(-1+5 /6) + 5/6*f(-1+2*5/6) + 5/6*f(-1+3*5/6) + 5/6*f(-1+4*5/6) + 5/6*f(-1+5 *5/6) + 5/6*f(-1+6*5/6) );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "and we can save typing by using the " }{TEXT 260 3 "sum" }{TEXT -1 9 " co mmand:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "evalf( sum(1*f(-1 + 5/6*k), k=1..6) );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 288 "but it \+ is not likely that this computation by itself will give a very good ap proximation to the area under the graph. To get a better approximati on, we could try using more rectangles. Here are the picture and the \+ sum with 12 rectangles (each of which will then have base length 12/5) :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "rightbox(f, x=-1..4, 1 2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "evalf( sum((5/12)*f( -1 + k*(5/12)), k=1..12) );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 201 "I t seems reasonable that if we take enough rectangles, we can make thei r total area as close as we want to the true area under the graph. (I t will be easier to compare the numerical answers if we use " }{TEXT 261 5 "evalf" }{TEXT -1 14 " on the sums.)" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 28 "rightbox(f(x), x=-1..4, 24);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 45 "evalf(sum((5/24)*f(-1 + k*(5/24)), k=1..24)); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "rightbox(f(x), x=-1..4, 48);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "evalf(sum((5/48)*f (-1 + k*(5/48)), k=1..48));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "rightbox(f(x), x=-1..4, 96);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "evalf(sum((5/96)*f(-1 + k*(5/96)), k=1..96));" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 89 "Here's an animation showing the co nvergence of a Riemann sum to the area under the curve " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 197 "animateRiemann := proc (f, a, b)\n local i, Lb;\n for i from 2 to 33 do\n Lb[i]:=student[rightbox ](f, x=a..b, 3*i):\n end do:\n plots[display]([seq(Lb[i], i=2..33)], insequence=true);\nend proc:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "a nimateRiemann(f, -1, 4);" }}}}{MARK "0 0 0" 1 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }