Approximating Area under a curve with rectangles. To find the area under a curve we approximate the area using rectangles and then use limits to find

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1 1.8 Approximatig Area uder a curve with rectagles 1.6 To fid the area uder a curve we approximate the area usig rectagles ad the use limits to fid 1.4 the area. Example 1 Suppose we wat to estimate 1. A = the area uder the curve y = 1 x, 0 x 1. f( x = 1 x Left edpoit approximatio curve usig rectagles as follows: 0. To approximate the area uder the curve, we ca circumscribe the We divide the iterval [0, 1] ito 4 subitervals of equal legth, x = 1 0 = 1/4. This divides the iterval [0, 1] ito 4 subitervals [0, 1/4], [1/4, 1/], [1/, 3/4], [3/4, 1] 0.8 each with legth x = 1/4. We label the edpoits of these subitervals as 1 x 0 = 0, x 1 = 1/4, x = /4, x 3 = 3/4, x 4 = Above each subiterval draw a rectagle with height equal to the height of the fuctio at the left ed poit of the subiterval. The values of the fuctio at the edpoits of the subitervals are x i x 0 = 0 x 1 = 1/4 x = 1/ x 3 = 3/4 x 4 = 1 f(x i = 1 x i 1 15/16 3/4 7/ We use the sum of the areas of the approximatig rectagles to approximate the area uder the curve. We get 4 A L 4 = f(x i 1 x = f(x 0 x + f(x 1 x + f(x x + f(x 3 x = 1

2 1 1/4 + 15/16 1/4 + 3/4 1/4 + 7/16 1/4 = 5/3 = L 4 is called the left edpoit approximatio or the approximatio usig left edpoits (of the subitervals ad 4 approximatig rectagles. We see i this case that L 4 = > A (because the fuctio is decreasig o the iterval. There is o reaso why we should use the left ed poits of the subitervals to defie the heights of the approximatig rectagles, it is equally reasoable to use the right ed poits of the subitervals, or the midpoits or i fact a radom poit i each subiterval. Right edpoit approximatio I the picture o the left above, we use the right ed poit to defie the height of the approximatig rectagle above each subiterval, givig the height of the rectagle above [x i 1, x i ] as f(x i. This gives us iscribed rectagles. The sum of their areas gives us The right edpoit approximatio, R 4 or the approximatio usig 4 approximatig rectagles ad right edpoits. Use the table above to complete the calculatio: A R 4 = 4 f(x i x = f(x 1 x + f(x x + f(x 3 x + f(x 4 x = Is R 4 less tha A or greater tha A. Midpoit Approximatio I the picture i the ceter above, we use the midpoit of the itervals to defie the height of the approximatig rectagle. This gives us The Midpoit Approximatio or The Midpoit Rule: x m i = midpoit x m 1 = 0.15 = 1/8 x m = = 3/8 x m 3 = 0.65 = 5/8 x m 4 = 0.65 = 7/8 f(x m i = 1 (x m i 63/64 55/64 39/64 15/64 A M 4 = 4 f(x m i x = f(x m 1 x + f(x m x + f(x m 3 x + f(x m 4 x = Geeral Riema Sum We ca use ay poit i the iterval x i [x i 1, x i ] to defie the height of the correspodig approximatig rectagle as f(x i. I the picture o the right we use radom poits

3 x i i each iterval to produce a rectagle with height f(x i. The sum of the areas of the rectagles gives us a approximatio for A. A 4 f(x i x = f(x 1 x + f(x x + f(x 3 x + f(x 4 x Icreasig The Number of Rectagles ad takig Limits Whe we icrease the umber of rectagles (of equal width used, usig a smaller value for x ( = the width of the rectagles, we get a better approximatio to the area. You ca see this i the pictures below for A = the area uder the curve y = 1 x, 0 x 1. The pictures show the right ed poit approximatios to A with x = 1/8, 1/16 ad 1/18 respectively: R 8 = , R 16 = , R 18 = The pictures below show the left ed poit approximatios to the area, A, with x = 1/8, 1/16 ad 1/18 respectively. L 8 = , L 16 = , L 18 = We see that the room for error decreases as the umber of subitervals icreases or as x 0. Thus we see that A = lim R = lim L = lim f(x i x = lim f(x i x x 0 where x i is ay poit i the iterval [x i 1, x i ]. I fact this is our defiitio of the area uder the curve o the give iterval. (If A deotes the area beeath f(x 0 where f is cotiuous o the iterval [a, b], the each iterval [x i 1, x i ] has legth x = b a for ay give. 3

4 Calculatig Limits of Riema sums The followig formulas are sometimes useful i calculatig Riema sums. I have attached some visual proofs at the ed of the lecture. i = ( + 1, i ( +. 1( + 1 =, [ ] i 3 ( + 1 = Let us ow cosider Example 1. We wat to fid A = the area uder the curve y = 1 x 1.4 o the iterval [a, b] = [0, 1]. 1. f( x = 1 x We kow that A = lim R, where R is the right edpoit approximatio usig approximatig 0.4 rectagles. 0.6 We must calculate R ad tha fid lim 0.8 R We divide the iterval [0, 1] ito strips of equal legth x = of the iterval [0, 1], 1.4 = 1/. This gives us a partitio 1.6 x 0 = 0, x 1 = 0 + x = 1/, x 1.8= 0 + x = /,..., x 1 = ( 1/, x = 1... We will use the right edpoit approximatio R. 3. The heights of the rectagles ca be foud from the table below: x i x 0 = 0 x 1 = 1/ x = / x 3 = 3/... x = / f(x i = 1 (x i 1 1 1/ 1 / 1 3 /... 1 / 4. R = f(x 1 x + f(x x + f(x 3 x + + f(x x = (1 1 1 (1 + 1 ( ( = 1 1 ( ( ( ( 1 = 4

5 5. Fiish the calculatio above ad fid A = lim R usig the formula for the sum of squares ad calculatig the limit as if R were a ratioal fuctio with variable. Also A = lim L From Part 3, we have x = 1/ ad L = 1 + ( ( ( ( ( ( ( groupig the 1 s together, we get = 1 [ ( ] = [ ( 1 ] ( 1 1 ( ( 1 1 = So = i = 1 1 [ (( 1 + 1( 1(( ] = 1 1 [ ( 1( 1( ] 3 6 = 1 ( 1( smaller powers of = 1 6 [ lim L + smaller powers of ] = lim 1 = = /3. 5

6 Riema Sums i Actio: Distace from Velocity/Speed Data To estimate distace travelled or displacemet of a object movig i a straight lie over a period of time, from discrete data o the velocity of the object, we use a Riema Sum. If we have a table of values: time = t i t 0 = 0 t 1 t... t velocity = v(t i v(t 0 v(t 1 v(t... v(t where t = t i t i 1, the we ca approximate the displacemet o the iterval [t i 1, t i ] by v(t i 1 t or v(t i t. Therefore the total displacemet of the object over the time iterval [0, t ] ca be approximated by or Displacemet v(t 0 t + v(t 1 t + + v(t 1 t Displacemet v(t 1 t + v(t t + + v(t t Left edpoit approximatio Right edpoit approximatio These are obviously Riema sums related to the fuctio v(t, hitig that there is a coectio betwee the area uder a curve (such as velocity ad its atiderivative (displacemet. This is ideed the case as we will see later. Whe we use speed = velocity istead of velocity. the above formulas traslate to Distace Travelled v(t 0 t + v(t 1 t + + v(t 1 t ad Distace Travelled v(t 1 t + v(t t + + v(t t Example The followig data shows the speed of a particle every 5 secods over a period of 30 secods. Give the left edpoit estimate for the distace travelled by the particle over the 30 secod period. time i s = t i velocity i m/s = v(t i L = v(t 0 t + v(t 1 t + + v(t 6 t = 50(5 + 60(5 + 65(5 + 6(5 + 60(5 + 55(5 = 5[ ] = 1760m. The above sum is a Riema sum, tellig us that the distace travelled is approximately the area uder the (absolute vale of velocity curve hmmmmm itetrestig... remember speed = v(t = derivative of distace travelled.... 6

7 Appedix Area uder a curve, Summary of method usig Riema sums. To fid the area uder the curve y = f(x o the iterval [a, b], where f(x 0 for all x i [a, b] ad cotiuous o the iterval: 1. Divide the iterval ito strips of equal width x = b a. This divides the iterval [a, b] ito subitervals: [x 0 = a, x 1 ], [x 1, x ], [x, x 3 ],..., [x 1, x = b]. (Note x 1 = a+ x, x = a+ x, x 3 = a+3 x,... x 1 = a+( 1 x, x = a+ x = b.. For each iterval, pick a sample poit, x i i the iterval [x i 1, x i ]. 3. Costruct a approximatig rectagle above the subiterval [x i 1, x i ] with height f(x i. The area of this rectagle is f(x i x. 4. The total area of the approximatig rectagles is (The sum is called a Riema Sum. f(x 1 x + f(x x + + f(x x 5. We defie the area uder the curve y = f(x o the iterval [a, b] as A = lim [f(x 1 x + f(x x + + f(x x] = lim R = lim [f(x 1 x + f(x x + + f(x x] = lim L = lim [f(x 0 x + f(x 1 x + + f(x 1 x]. (I a more advaced course, you would prove that all of these limits give the same umber A which we use as a measure/defiitio of the area uder the curve 7

8 Extra Example Estimate the area uder the graph of f(x = 1/x from x = 1 to x = 4 usig six approximatig rectagles ad x = b a =, where [a, b] = [1, 4] ad = 6. Mark the poits x 0, x 1, x,..., x 6 which divide the iterval [1, 4] ito six subitervals of equal legth o the followig axis: Fill i the followig tables: x i x 0 = x 1 = x = x 3 = x 4 = x 5 = x 6 = f(x i = 1/x i (a Fid the correspodig right edpoit approximatio to the area uder the curve y = 1/x o the iterval [1, 4]. R 6 = (b Fid the correspodig left edpoit approximatio to the area uder the curve y = 1/x o the iterval [1, 4]. L 6 = (c Fill i the values of f(x at the midpoits of the subitervals below: midpoit = x m i x m 1 = x m = x m 3 = x m 4 = x m 5 = x m 6 = f(x m i = 1/x m i Fid the correspodig midpoit approximatio to the area uder the curve y = 1/x o the iterval [1, 4]. M 6 = 8

9 Extra Example Estimate the area uder the graph of f(x = 1/x from x = 1 to x = 4 usig six approximatig rectagles ad x = b a = 4 1 = 1, where [a, b] = [1, 4] ad = 6. 6 Mark the poits x 0, x 1, x,..., x 6 which divide the iterval [1, 4] ito six subitervals of equal legth o the followig axis: Fill i the followig tables: x i x 0 = 1 x 1 = 3/ x = x 3 = 5/ x 4 = 3 x 5 = 7/ x 6 = 4 f(x i = 1/x i 1 /3 1/ /5 1/5 /7 1/4 (a Fid the correspodig right edpoit approximatio to the area uder the curve y = 1/x o the iterval [1, 4]. R 6 = f(x 1 x + f(x x + f(x 3 x + f(x 4 x + f(x 5 x + f(x 6 x = = = (b Fid the correspodig left edpoit approximatio to the area uder the curve y = 1/x o the iterval [1, 4]. L 6 = f(x 0 x + f(x 1 x + f(x x + f(x 3 x + f(x 4 x + f(x 5 x = = = (c Fill i the values of f(x at the midpoits of the subitervals below: midpoit = x m i x m 1 = 5/4 x m = 7/4 x m 3 = 9/4 x m 4 = 11/4 x m 5 = 13/4 x m 6 = 15/4 f(x m i = 1/x m i 4/5 4/7 4/9 4/11 4/13 4/15 Fid the correspodig midpoit approximatio to the area uder the curve y = 1/x o the iterval [1, 4]. 6 M 6 = f(x i x = =

10 Extra Example Fid the area uder the curve y = x 3 o the iterval [0, 1]. We kow that A = lim R, where R is the right edpoit approximatio usig approximatig rectagles. We must calculate R ad tha fid lim R. 1. We divide the iterval [0, 1] ito strips of equal legth x = 1 0 of the iterval [0, 1], = 1/. This gives us a partitio x 0 = 0, x 1 = 0 + x = 1/, x = 0 + x = /,..., x 1 = ( 1/, x = 1.. We will use the right edpoit approximatio R. 3. The heights of the rectagles ca be foud from the table below: x i x 0 = 0 x 1 = 1/ x = / x 3 = 3/... x = / f(x i = (x i 3 0 1/ 3 3 / / / 3 4. R = f(x 1 x + f(x x + f(x 3 x + + f(x x = ( 1 1 ( 3 1 ( ( = = i 3 4 = 1 4 [ i 3 = 1 ( ] 5. [ ] 1 ( + 1 ( + 1 A = lim = lim ( + 1 = lim 4 1 ( + 1 ( + 1 = lim 4 =

11 Extra Example, estimates from data o rate of chage The same priciple applies to estimatig Volume from discrete data o its rate of chage: Oil is leakig from a taker damaged at sea. The damage to the taker is worseig as evideced by the icreased leakage each hour, recorded i the followig table. time i h leakage i gal/h The followig gives the right edpoit estimate of the amout of oil that has escaped from the taker after 8 hours: R 8 = = 363 gallos. The followig gives the right edpoit estimate of the amout of oil that has escaped from the taker after 8 hours: L 8 = = 1693 gallos. Sice the flow of oil seems to be icreasig over time, we would expect that L 8 < true volume leaked < R 8 or the true volume leaked i the first 8 hours is somewhere betwee 1693 ad 363 gallos. Visual proof of formula for the sum of itegers: 11

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