SYDE 112, LECTURE 2: Riemann Sums

Transcription

1 SYDE, LECTURE : Riema Sums Riema Sums Cosider the problem of determiig the area below the curve f(x) boud betwee two poits a ad b. For simple geometrical fuctios, we ca easily determie this based o ituitio. For a costat fuctio f(x) = c, we kow that the area below the curve betwee ay two poits is give by the legth (b a) times the width (c) so that f(x) = c = area boud betwee a ad b is (b a)c. Now cosider the fuctio f(x) = cx ad, for simplicity, suppose we are iterested i the area boud betwee x = ad x =. This area is triagular, so we ca easily compute the area as oe half base () times height (c) so that f(x) = cx = area boud betwee ad is c. c f(x)=cx c f(x)=c Area= (b-a)c Area= (/)c a b Figure : Areas uder the curve for f(x) = c (boud betwee a ad b) ad f(x) = cx (boud betwee ad ). This process seems simple eough. We kow the area of basic geometric shapes (e.g. squares, rectagles, triagles, parallelograms, circles, etc.) so

2 we might suspect that this kowledge will allow us to compute geeral areas. This ituitio, however, is quickly show to be lackig. If we cosider eve the simple parabola f(x) = x boud betwee x = ad x =, we see that the area below the curve does ot correspod to ay easily computed geometric shape (see Figure ). Figure : The area uder the curve f(x) = x boud betwee ad does ot correspod to ay well-kow, easily-computable geometric shape. It is certaily ot satisfactory to throw our hads i the air ad give up. Suppose we were i the carpettig busiess ad had bee asked how much carpet would be eeded to carpet a room with dimesios give by that i Figure. The customer would ot be very happy if we told him to rebuild his room i the form of a rectagle ad get back to us! I this situatio, we would like to take what we kow about easilycomputable shapes to build a approximatio of the actual area. We might as well start with the easiest shape we kow, the rectagle. Imagie coverig the area give i Figure by two rectagles of equal width (/). If we imagie the height of those rectagles beig give by where the right side hits x, ad where the left side hits x, respectively, the we arrive at Figure 3. We ca easily see that the area i Figure 3(a) overestimates the area, ad the area i (b) uderestimates it, so that uder the curve is x lies betwee the areas of the rectagles give i this picture! That is to say, we have ( ) ( ) ( ) ( ) ( ) () + < Area uder x < < Area uder x < 5 8.

3 (a) (b) Figure 3: The area uder the curve of f(x) = x estimated usig two rectagles. The height of the rectagles is give by the right poit of the fuctio i (a) ad the left poit i (b). Notice that f() = so that the first rectagle i (b) has zero height. This is a pretty big spread. We might woder if we ca do better. Of course we ca! Imagie cuttig estimatig the area with four rectagles istead of two. Now we have the improved estimates for the area give i Figure 4, which computes to < Area uder x < < Area uder x < 5 3. This is a sigificatly better estimate! The differece betwee the upper ad lower boud has bee cut i half, from / to /4. We might woder, based o this, how good of a estimate we ca actually get by takig a greater ad greater umber of rectagles. The aswer is that we ca get as good of a estimate as we like! That is to say, if we wat to estimate the area uder x to the fifth decimal place, there is some umber of rectagles which will give it. If we wat to estimate it to te decimal places, there is some partitio for that as well. Let s state this mathematically. I geeral, we wat to estimate the area betwee x = a ad x = b ad we will take the width of each rectagle to be the same. This meas that, if we have rectagles, each oe has width x = b a. () The height of each rectagle is give by some poit o the curve f(x i ) where the x i is take to lie somewhere o the base of the rectagle. A commo 3

4 (a) (b) Figure 4: The area uder the curve of f(x) = x estimated usig four rectagles. The height of the rectagles is give by the right poit of the fuctio i (a) ad the left poit i (b). Notice that f() = so that the first rectagle i (b) has zero height. choice is to take the x-value to be either the left-had or right-had ed poit, which correspod respectively to or x i = a + (i ) x (Left edpoit of i th rectagle) () x i = a + i x (Right edpoit of i th rectagle). (3) The area of the i th rectagle ca be easily calculated as [base height] to give f(x i ) x (Area of i th rectagle) from which it follows that the area uder the curve f(x) boud betwee x = a ad x = b is give by f(x i ) x. It turs out that, for a appropriate choice of, this is actually a very good approximatio of the area uder a fuctio, but we ca do eve better. We have some ituitio that, if we take a greater ad greater umber of rectagles we are goig to get a better ad better estimate of the area. Well, what happes i the limit as the umber of rectagles approaches ifiity? Is it eve sesible to try to evaluate this limit? After all, i the limit each rectagle has o width (i.e. zero area), but there are a ifiite umber of them. 4

5 It turs out that this limit is (usually) well-defied ad accurately computes the area uder the curve precisely! The Riema sum is defie as lim f(x i ) x (4) where x i is give by either () or (3), ad x is give by (). (We will coect this sum to a cocept called itegratio, which is oe of the foudatioal applicatios of calculus, i a future lecture.) Note: Computig Riema sums ofte requires the use of summatio formulas. The most commo forms ecoutered are = (5) i = i = i 3 = ( + ) (6) ( + )( + ) (7) 6 ( ) ( + ). (8) Example : Evaluate the Riema sum for x boud betwee x = ad x = usig the right-edpoit (3). Solutio: We have so that x = b a = = x i = a + i x = + i ( ) = i. 5

6 This gives us lim f(x i ) x ( i 3 i ) ( ) ( + )( + ) = 3. I other words, the area below x betwee x = ad x = is exactly /3. Example : Evaluate the Riema sum for x 3 boud betwee x = ad x = usig the right-edpoit (3). Solutio: so that This gives us lim We have f(x i ) x [ x = b a x i = a + i x = + i ( + i ) 3 ( ) [( ) [ + + 6( + ) = =. = ( ) i i + 8 i3 3 i 4 3 = ( ) = + i. ] i ] i 3 4( ) ( ) 4 4 ] 6

7 I other words, there is o area uder the curve x 3 boud betwee x = ad x =. Cosiderig the graph of x 3, does this make sese? (Hit: rectagles draw below the x-axis will produce a egative area!) 7