Riemann Sums y = f (x)


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1 Riema Sums Recall that we have previously discussed the area problem I its simplest form we ca state it this way: The Area Problem Let f be a cotiuous, oegative fuctio o the closed iterval [a, b] Fid the area bouded above by f (x), below by the xaxis, ad by the vertical lies x = a ad x = b See Figure 11 To solve this problem we will eed to use a A f Figure 11: Fid the area A f uder a oegative cotiuous curve o the iterval [a, b] b Basic Area Properties (Axioms) We assume the followig properties 1 The area of a regio A is a oegative real umber: Area(A) 0 B A Property 2 2 If A is a subset of B, the Area(A) Area(B) 3 If A is subdivided ito two ooverlappig regios A 1 ad A 2, the The area of a rectagle is b h Area(A) = Area(A 1 ) + Area(A 2 ) Property 3 A 1 Figure 12: Properties 2 ad 3 A 2 YOU TRY IT 11 Usig the area properties above, prove that the area of ay triagle is 1 (b h) 2 See Figure 13 Which area properties do you use i your proof? YOU TRY IT 12 How could you use the area formula for a triagle to fid the area of ay polygo? (See Figure 1) What area properties are used to do this? What about curved figures like (semi)circles Why is the area of a circle πr 2 or, equivaletly, the area of a semicircle 1 2 πr2? If we ca solve the geeral area problem, the we will be able to prove that the area of a semicircle is 1 2 πr2 because we kow that the graph of the semicircle of radius r is give by the cotiuous, oegative fuctio f (x) = r 2 x 2 I other words, a semicircular regio satisfies the coditios outlied i the geeral area problem Note: We ll solve the area problem two ways Sice the aswer must be the same, this equality will be the proof for the socalled Fudametal Theorem of Calculus To solve the area problem, we ll eed to use the oly area formula we kow we must use rectagle regios Figure 13: Show A = 2 1 (b h) Figure 1: How you ca fid the area of this polygo? f (x) = r 2 x 2 r r Figure 15: This semicircle satisfies the coditios of the area problem Riema Sums (Theory) The presetatio here is slightly differet tha i your text Make sure that you uderstad what all of the otatio meas Agai, remember what we are tryig to solve:
2 math 131 the area problem ad riema sums, part i 2 The Area Problem Let f be a cotiuous, oegative fuctio o the closed iterval [a, b] Fid the area bouded above by f (x), below by the xaxis, ad by the vertical lies x = a ad x = b As we have just oted, sice the oly area formula we have have to work with is for rectagles, we must use rectagles to approximate the area uder the curve Here s how we go about this approximatio process Step 1 First subdivide or partitio [a, b] by choosig poits {x 0, x 1,, x } where a = x 0 < x 1 < x 2 < < x 1 < x = b Figure 16: A partitio of the iterval [a, b] a = x 0 x 1 x 2 x k 1 x k b = x Step 2 Determie the height of the kth rectagle by choosig a sample poit c k i the kth subiterval so that x k 1 c k x k Use f (c k ) as the height height = f (c k ) Figure 17: f (c k ) is the height of the kth rectagle (see the poit marked with a o the curve) a = x 0 x 1 x 2 x k 1 c k x k b = x Step 3 The width of the base of the kth rectagle is just x k x k 1 We usually call this umber x k (See Figure 18) Step So usig the rectagle area assumptio, the area of the kth rectagle is h b = f (c k ) x k height = f (c k ) Figure 18: x k = x k x k 1 is the width of the kth rectagle So the area of the kth rectagle is f (c k ) x k a = x 0 x 1 x 2 x k 1 c k x k b = x x k Step 5 If we carry out this same process for each subiterval determied by the partitio {x 0, x 1,, x }, we get rectagles The area uder f o [a, b] is approximately the sum of the areas of all rectagles, Area(A) f (c k ) x k
3 math 131 the area problem ad riema sums, part i 3 Figure 19: A rectagular approximatio to the area uder f o the iterval [a, b] a = x 0 x 1 x 2 x k 1 x k b = x DEFINITION 11 (Riema Sum) Suppose f is defied o the iterval [a, b] with partitio a = x 0 < x 1 < x 2 < < x 1 < x = b Let x k = x k x k 1 ad let c k be ay poit chose so that x k 1 c k x k The is called a Riema sum for f o [a, b] f (c k ) x k Notice that i the geeral defiitio of a Riema sum we have ot assumed that f is oegative or that it is cotiuous The defiitio makes sese as log as f is defied at every poit i [a, b] Let s work out a simple example EXAMPLE 101 Estimate the area uder f (x) = (x 1) o the iterval [0, 2] usig the partitio poits x 0 = 0 x 1 = 1 2 ad sample poits x 2 = 3 2 x 3 = 2 c 1 = 1 2 c 2 = 1 c 3 = 7 SOLUTION We use Defiitio 11 ad form the appropriate Riema sum First x 1 = x 1 x 0 = = 1 2 x 2 = x 2 x 1 = = 1 x 3 = x 3 x 2 = = 1 2 c 1 = 1/2 c 2 = 1 c 3 = 7/ So Area(A) 3 f (c k ) x k = f ( 1 2 ) x 1 + f (1) x 2 + f ( 7 ) x 3 = ( 7 8 )( 1 2 ) + (1)(1) + ( 91 6 )( 1 2 ) = x = 0 x 1 = 1/2 x 2 = 3/2 x 3 = 2 Figure 110: A Riema sum for f (x) = (x 1) o the iterval [0, 2] usig three rectagles The height for each rectagle is marked with a Does the approximatio seem to be a uder or overestimate of the true area?
4 math 131 the area problem ad riema sums, part i The Riema sum provides a estimate of as the area uder the curve Yet we do t kow how accurate that estimate is ad we still do t kow the true area uder the curve Further, otice that the use of summatio otatio was ot particularly helpful here If we use Riema sums i a more systematic way, Riema sum otatio ca be very helpful Ad, if we are careful about how we form such sums, we ca eve say whether the sum is a over or uderestimate of the actual area uder the curve Regular Partitios, Upper ad Lower Sums Agai let us assume that is a oegative, cotiuous fuctio o the iterval [a, b] We will ow take a more systematic approach to formig Riema sums for f o [a, b] that will allow us to make more accurate approximatios to the area uder the curve Agai we proceed i a series of steps Step 1 Divide the iterval [a, b] ito equalwidth subitervals The width of each iterval will be x = b a We ca express the partitio poits i terms of a ad x x 0 = a = a + 0 x x 1 = a + x x 2 = a + 2 x x k = a + k x x = b = a + x Equal width partitios are called regular partitios The formula for the kth poit i a regular partitio is x k = a + k x (11) Figure 111: A regular partitio of the iterval [a, b] ito subitervals each of legth x = b a This meas that x k = a + k x x 0 = a x 1 x 2 x k 1 x k x 1 x = b x x x Step 2 Sice f is cotiuous, it achieves a maximum value ad a miimum value
5 math 131 the area problem ad riema sums, part i 5 o each subiterval We use the followig otatio to represet these poits f (M k ) = maximum value of f o the kth subiterval f (m k ) = miimum value of f o the kth subiterval These poits are illustrated i Figure 112 f (M k ) f (m k ) Figure 112: O the kth subiterval the maximum height f (M k ) occurs betwee the two edpoits The miimum height f (m k ) happes to occur at the right edpoit of the iterval, m k = x k x k 1 M k x k x x k 1 x k = m k x Figure 112 shows that we get two differet rectagles for each subiterval depedig o whether we choose the maximum or the miimum value of f as the height These are called the circumscribed ad iscribed rectagles, respectively We see that area of the circumscribed rectagle = f (M k ) x area of the iscribed rectagle = f (m k ) x Step 3 To obtai a approximatio for the area uder the curve, we form a Riema sum usig either the circumscribed (upper) or iscribed (lower) rectagles If we add up all the circumscribed rectagles for a regular partitio with subitervals we get the upper sum for the partitio: Upper Riema Sum = Upper() = f (M k ) x (12) If we add up all the iscribed rectagles for a regular partitio we get the lower sum for the partitio: Lower Riema Sum = Lower() = f (m k ) x (13) Take a momet to review all of the otatio Ok? Let s see how these upper ad lower sums are computed i a simple case EXAMPLE 102 Let = x2 o [0, 2] Determie Upper() ad Lower(), the upper ad lower Riema sums for a regular partitio ito four subitervals SOLUTION We use the steps outlied above Step 1 Determie x Here [a, b] = [0, 2] ad = so x = b a = 2 0 = 1 2 Step 2 Determie the partitio poits, x k Usig (11) ( ) 1 x k = a + k x = 0 + k = k 2 2 (1)
6 math 131 the area problem ad riema sums, part i 6 Step 3 Take a look at the graph of f (x) = x2 o [0, 2] i Figure 113 Sice f is a icreasig fuctio, the maximum value of f o each subiterval occurs at the righthad edpoit of the iterval The righthad edpoit of the i iterval is just x k So M k = x k = k 2 Cosequetly, the maximum value of f o the kth iterval is f (M k ) = f ( ) k = Step Puttig this all together, the upper Riema sum is Upper() = f (M k ) x = f ( ) k 2 = 1 + k2 2 8 ( ) k = ] [1 + k Now use the basic summatio rules ad formulæ to evaluate the sum Figure 113: The upper sum Upper() for the fuctio f (x) = x2 o [0, 2] The maximum value of the fuctio occurs at the righthad edpoit, x k for each subiterval Upper() = [1 + k2 8 ] 1 2 = = 1 2 [(1)] = 31 8 k 2 ( ) (5)(9) 6 The lower sum Lower() ca be calculated i a similar way Agai, because the fuctio is icreasig, the miimum value of f o the kth subiterval occurs at the lefthad edpoit x k 1 Usig the formula i (1) m k = x k 1 = k 1 2 Cosequetly, the miimum value of f o the kth iterval is ( ) k 1 f (m k ) = f = ( ) k 1 2 = 1 + k2 2i = 9 8 k + k2 8 Puttig this all together, the lower Riema sum is Lower() = [ 9 f (m k ) x = 8 k ] + k Agai use the basic summatio rules ad formulæ to evaluate the sum Lower() = [ 9 8 k ] + k = 1 2 = [ ( 9 8 = 23 8 k ( (5) )] k 2 ) ( ) (5)(9) The advatage of upper ad lower sums is that the true area uder the curve is trapped betwee their values Upper() is always a overestimate ad Lower() is a uderestimate More precisely, Figure 11: The lower sum Lower() for the fuctio f (x) = x2 o [0, 2] The miimum value of the fuctio occurs at the lefthad edpoit, x k 1 for each subiterval Lower() area uder f Upper() I this example, Lower() = 23 8 area uder f 31 8 = Upper()
7 math 131 the area problem ad riema sums, part i 7 Here are two questios to thik about: How ca we improve the estimate? Which sum was easier to compute, the lower or the upper? Why? Now let s do the whole process agai This time, though we will use subitervals, without specifyig what the actual value of is This is where the summatio otatio that we have developed really comes to the rescue EXAMPLE 103 Let = x2 o [0, 2] Determie Upper() ad Lower(), the upper ad lower Riema sums for a regular partitio ito subitervals SOLUTION Step 1 Determie x Here [a, b] = [0, 2] so x = b a = 2 0 = 2 Step 2 Determie the partitio poits, x k Usig (11) ( ) 2 x k = a + k x = 0 + k = 2k (15) Step 3 Sice f is a icreasig fuctio, the maximum value of f o each subiterval occurs at the righthad edpoit of the iterval So M k = x k So M k = x k = 2k Cosequetly, the maximum value of f o the kth iterval is f (M k ) = f ( ) 2k = ( ) 2k 2 = 1 + k2 2 2 = 1 + 2k Figure 115: The upper sum Upper() for the fuctio f (x) = x2 o [0, 2] As icreases, Upper() better approximates the area uder the curve (Compare to Figure 112) Puttig this all together, the upper Riema sum is Upper() = f (M k ) x = ] [1 + 2k2 2 2 = k 2 = 2 [(1)] + ( ) ( + 1)(2 + 1) 3 6 = ( 2 2 ) ( = ) 3 2 = The lower sum Lower() ca be calculated i a similar way The miimum value of f o the kth subiterval occurs at the lefthad edpoit: m k = x k 1 = The miimum value of f o the kth iterval is 2(k 1) ( ) 2(k 1) f (m k ) = f = ( ) 2(k 1) 2 = 1 + (k2 2k + 1) = 1 + 2(k2 2k + 1) 2 Puttig this all together, the lower Riema sum is Figure 116: The lower sum Lower() for the fuctio f (x) = x2 o [0, 2] The lower sum is a uderestimate of the area uder f
8 math 131 the area problem ad riema sums, part i 8 Lower() = We kow that f (m k ) x = = 2 [ 1 + 2(k2 2k + 1) 2 ] (k 2 2k + 1) = 2 [(1)] + 3 k k = 2 + [ ] ( + 1)(2 + 1) 3 8 [ ( + 1) [ = ] [ ] = Lower() area uder f Upper() ] + 3 [()1] The formulæ for Upper() ad Lower() are valid for all positive itegers We expect that as icreases the approximatios improve I this case, takig limits lim Lower() area uder f lim Upper(), equivaletly, [ 10 lim ] [ area uder f lim ] 3 2, or 10 3 area uder f 10 3 The oly way this ca happe is if area uder f = 10 3 Takehome Message This is great! We have maaged to determie the area uder a actual curve by usig approximatios by lower ad upper Riema sums The approximatios improve as icreases By takig limits we hoe i o the precise area This is more carefully described i Theorem 11 at the begiig of the ext sectio Fially, agai ask yourself which of the two sums was easier to calculate? Why was it easier? Shortly we will take advatage of this situatio 0 2 Figure 117: The differece betwee the upper sum Upper() for the fuctio f (x) = x2 o [0, 2] ad the lower sum Lower() (shaded) The true area lies betwee the two
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