The picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled

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1 1 Lecture : Area Area ad distace traveled Approximatig area by rectagles Summatio The area uder a parabola 1.1 Area ad distace Suppose we have the followig iformatio about the velocity of a particle, how ca we estimate the distace travelled? Time (secods) Velocity (meters/secod) 5 With this little iformatio, it is impossible to kow the aswer. Oe possibility is to assume that the velocity i each iterval is the same as at the left edpoit. The table below summarizes the calculatio Iterval Legth of iterval Velocity Distace [0, ] s m/s 6 m [, 5] s m/s 8 m [5, 6] 1 s m/s m Total 17 m The picture i figure 1.1 helps us to see that the area represets the distace traveled velocity i meters/secod time i secods Figure 1: Area represets distace travelled Ca you thik of a differet way to estimate the distace? A better way?

2 1. Approximatig area by rectagles I this sectio, we will cosider a way of approximatig the area below the graph of a fuctio. For example, we might be iterested i fidig the area of the regio R = {(x, y) : 1 x, 0 y x } which lies uder the graph of a parabola. 9 y x 1 5 Figure : The left edpoit approximatio to the area uder the parabola y = x for 1 x To do this, we will first divide the regio ito several itervals of equal legth. If we decide to use four itervals, we will use the divisio poits x 0 = 1, x 1 = /, x =, x = 5/, ad x =. For each iterval, we choose a poit i the iterval ad use the value of f at that poit to give the height of a rectagle. Fially, we add the areas of the rectagles to give the value of f. It is ot so importat which value we choose. Evetually, we will take a limit as the itervals become small ad the result of differet choices will be small. For ow, there are three commo choices: for each iterval choose the left edpoit, the right edpoit or the midpoit. These three choices give us three differet approximate values for the area: R = 1 (x 1 + x + x + x ) = 1 ( ) = = 10.75

3 L = 1 ( ) = 7 = 6.75 M = 1 ((5/) + (7/) + (9/) + (11/) ) = 69/8 = We are usig R to deote the sum formed usig the right edpoit ad itervals, L if we use the left edpoit to fix the height of the rectagles ad M if we use the mid-poit. Figure illustrates the computatio of R. 1. Summatio Computig areas requires that we have a efficiet way to evaluate sums. If a k gives a umber for iteger values of k, the we use the otatio For example, Express as a sum a k = a m + a m a. k=m 5 k = = Solutio. Sice the terms differ by, we may use a liear expressio to geerate these umbers. If we wat to start at k = 1 ad have the first term 1 the liear expressio we eed is (k 1) + 1 = k. This is similar to usig the poit-slope form to fid the equatio of a lie. We eed to fid so that = 16 or = 6. Thus, we have = (k 1). Fid the sum 7 (k 1). Solutio. The expressio k 1 will give us the odd umbers. Thus we have 7 (k 1) = = 9.

4 There are several formulae for sums that are quite useful. We give them without proof k = k = k = ( + 1) ( + 1)( + 1) 6 (( + 1)) (1) () () Fid the value of the sum (k + ). Solutio. Usig the distributive law, we have (k + ) = Sice 0 tells us to add together 0 copies of the umber, its value is 0. Accordig to (1), the value of the sum 0 k is 0 1 = 10. Thus we have k + k +. = = A defiitio of area If f is a positive fuctio ad [a, b] is a closed iterval, our goal is to defie the area of R = {(x, y) : a x b, 0 y f(x)}. We take a iterval [a, b] ad divide it ito equal subiterval. The legth of each subiterval is (b a)/. The divisio poits will be x 0 = a, x 1 = a+ b a, x = a+ b a or the kth divisio poit is x k = a + k (b a).,..., x 1 = a+( 1) b a, x = a+(b a) = b

5 Usig these poits, we ca right out the right edpoit, left edpoit, ad midpoit approximatios to area. R = b a L = b a M = b a f(x k ) f(x k 1 ) f( 1 (x k 1 + x k )). Fially, if we let, we expect that the error will be quite small ad disappear i the limit. Thus if R is the regio {(x, y) : a x b, 0 y f(x)}, the ay of the limits lim R, is a good cadidate for the area of R. lim L, lim M Fid the area uder the graph of f(x) = x for 0 x. The area of the regio {(x, y) : 0 y, 0 y x } usig the right edpoit approximatio. Solutio. We use the divisio poits x k = k/. The legth of each iterval is /. Thus the the right edpoit approximatio to the area is R = ( k ). Usig the formula () we have that the sum is R = = 7 9 k k = 7 ( + 1)( + 1). 6 Fially, we may take a limit ad fid a cadidate for the area 7 ( + 1)( + 1) lim 6 = 9.