Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite integrl.
Distnce nd Velocity - 1 Distnce nd Velocity Recll tht if we mesure distnce s function of time t, the velocity is determined by differentiting (t), i.e. finding the slope of the grph. Alterntively, suppose we begin with grph of the velocity with respect to time. How cn we determine wht distnce will be trveled? Does it pper in the grph somehow? Let s begin with the simple cse of constnt velocity... distnce = velocity time
Distnce nd Velocity - 2 For constnt velocity, the distnce trveled in certin length of time ws simply the re of the rectngle underneth the velocity vs. time grph. Wht if the velocity is chnging? We cn t determine the ect distnce trveled, but mybe we cn estimte it. Let s ssume tht the velocity is not chnging too quickly, so over short mount of time it s roughly constnt. We know how to find the distnce trveled in tht short time... Mking mny of these pproimtions, we could come up with rough estimte of the totl distnce. How does this estimte relte to the grph?
Clculting Ares - 1 Clculting Ares It ppers tht the distnce trveled is the re under the grph of velocity, even when the velocity is chnging. We ll see ectly why this is true very soon. If we re simply interested in the re under grph, without ny physicl interprettion, we cn lredy do so if the grph cretes shpe tht we recognize. Emple: Clculte the re between the -is nd the grph of y = + 1 from = 1 nd = 1. y 3 2 1 2 1 1 2 1
Clculting Ares - 2 Emple: Clculte the re between the -is nd the grph of y = 1 2 from = 1 to = 1. 1 y 1 1 1 Wht shpes do you know, right now, for which you cn clculte the ect re?
Estimting Ares - 1 Estimting Ares Unfortuntely, mny or most rbitrry res re essentilly impossible to find the re of when the shpe isn t simple composition of tringles, rectngles, or circles. In these cses, we must use less direct methods. We strt by mking n estimte of the re under the grph using shpes whose re is esier to clculte. Suppose we re trying to find the re underneth the grph of the function f() given below between = 1 nd = 4. Drw this re on the grph below, nd lbel it re A. 5 4 3 2 1 1 1 2 3 4 5
Estimting Ares - 2 We cn mke rough estimtion of the re by drwing rectngle tht completely contins the re, or rectngle tht is completely contined by the re. Clculte this overestimte nd underestimte for the re A. 5 4 3 2 1 1 1 2 3 4 5
Estimting Ares - 3 This one-rectngle estimte is very crude. We cn improve our estimtes by using smller rectngles. E.g. We cn divide the intervl from = 1 to = 4 into 3 intervls, ech of width 1, nd use different rectngle heights on ech intervl. Estimte the re A by using 3 rectngles of width 1. Use the function vlue t the left edge of the intervl s the height of ech rectngle. 5 4 3 2 1 1 1 2 3 4 5
Generlizing Are Estimtes - LEFT nd RIGHT - 1 Generlizing Are Estimtes - LEF T (n) nd RIGHT (n) We cn repet the rectngle-building process for ny number of rectngles, nd we epect tht our estimtion of the ect (curved) re will get better the more rectngles we use. The method we used erlier, choosing the height of the rectngles bsed on the function vlue t the left edge, is clled the left hnd sum, nd is denoted LEFT(n) if we use n rectngles. Suppose we re trying to estimte the re under the function f() from = to = b vi the left hnd sum with n rectngles. Then the width of ech rectngle will be = b n. b
Generlizing Are Estimtes - LEFT nd RIGHT - 2 If we lbel the endpoints of the intervls to be = 0 < 1 < < n 1 < n = b, then the formul for the left hnd sum will be LEF T (n) =f( 0 ) + f( 1 ) +... + f( n 1 ) n = f( i 1 ). i=1 b 1 2 3... n 1 0 n
Generlizing Are Estimtes - LEFT nd RIGHT - 3 We hve similr definition for the right hnd sum, or RIGHT(n), clculted by tking the height of ech rectngle to be the height of the function t the right hnd endpoint of the intervl. RIGHT (n) =f( 1 ) + f( 2 ) +... + f( n ) = n f( i ). i=1 b 1 2 3... n 1 0 n
Generlizing Are Estimtes - LEFT nd RIGHT - 4 Clculte LEFT(6) nd RIGHT(6) for the function shown, between = 1 nd = 4. You will need to estimte some rectngle heights from the grph. 5 4 3 2 1 5 4 3 2 1 1 2 3 4 5 1 2 3 4 5 1 1
Generlizing Are Estimtes - LEFT nd RIGHT - 5 In generl, when will LEFT(n) be greter thn RIGHT(n)? When will LEFT(n) be n overestimte for the re? When will LEFT(n) be n underestimte?
Riemnn Sums - 1 Riemnn Sums Are estimtions like LEF T (n) nd RIGHT (n) re often clled Riemnn sums, fter the mthemticin Bernhrd Riemnn (1826-1866) who formlized mny of the techniques of clculus. The generl form for Riemnn Sum is f( 1) + f( 2) +... + f( n) n = f( i ) where ech i is some point in the intervl [ i 1, i ]. For LEF T (n), we choose the left hnd endpoint of the intervl, so i = i 1 ; for RIGHT (n), we choose the right hnd endpoint, so i = i. i=1 b 1 2 3... n 1 0 n
Riemnn Sums - 2 The common property of ll these pproimtions is tht they involve sum of rectngulr res, with widths ( ), nd heights (f( i )) There re other Riemnn Sums tht give slightly better estimtes of the re underneth grph, but they often require etr computtion. We will emine some of these other clcultions little lter.
The Definite Integrl - 1 The Definite Integrl We observed tht s we increse the number of rectngles used to pproimte the re under curve, our estimte of the re under the grph becomes more ccurte. This implies tht if we wnt to clculte the ect re, we would wnt to use limit. The re underneth the grph of f() between = nd = b is equl to n lim LEF T (n) = lim f( i 1 ), where = b n n n. i=1 b b b b
The Definite Integrl - 2 This limit is clled the definite integrl of f() from to b, nd is equl to the re under curve whenever f() is non-negtive continuous function. The definite integrl is written with some specil nottion. Nottion for the Definite Integrl The definite integrl of f() between = nd = b is denoted by the symbol b f() d We cll nd b the limits of integrtion nd f() the integrnd. The d denotes which vrible we re using; this will become importnt for using some techniques for clculting definite integrls. Note tht this nottion shres the sme common structure with Riemnn sums: sum ( sign) widths (d), nd heights (f())
The Definite Integrl - 3 Emple: Write the definite integrl representing the re underneth the grph of f() = + cos between = 2 nd = 4.
The Definite Integrl nd LEFT vs RIGHT - 1 The Definite Integrl - LEF T (n) vs RIGHT (n) s n We might be concerned tht we defined the re nd the definite integrl using the left hnd sum. Would we get the sme nswer for the definite integrl if we used the right hnd sum, or ny other Riemnn sum? In fct, the limit using ny Riemnn sum would give us the sme nswer. Let us look t the left nd right hnd sums for the function 2 on the intervl from = 1 to = 3. Clculte LEF T (2) RIGHT (2) for 3 1 2 d. Tht is, how big is the difference between these two estimtes of the re under y = 2 over = 1... 3?
The Definite Integrl nd LEFT vs RIGHT - 2 Clculte LEF T (4) RIGHT (4) for 3 1 2 d.
The Definite Integrl nd LEFT vs RIGHT - 3 Clculte LEF T (n) RIGHT (n) for 3 1 2 d.
The Definite Integrl nd LEFT vs RIGHT - 4 Wht will the limit of this LEF T (n) RIGHT (n) difference be s n? Wht does this tell us bout wht would hppen if we defined the definite integrl in terms of the right hnd sum? b f() d = lim n LEF T (n) vs. lim n RIGHT (n)?
Estimting Are Between Curves - 1 More on the Definite Integrl nd Are We cn use the definite integrl to clculte other res, s well. Suppose we wnt to find the re between the curves y = 2 nd the line y = 2. It is esy to see tht the two intersect s shown in the following grph. y 4 3 2 y = 2 1 y = 2 3 2 1 1 1 We cn gin clculte this re by estimting vi rectngles nd the tking the limit to get the definite integrl.
Estimting Are Between Curves - 2 y 4 3 2 y = 2 1 y = 2 3 2 1 1 1 If we estimte this re using the left hnd sum, wht will be the height of the rectngle on the intervl [ i, i+1 ]? 1. height = 2 i (2 i) 2. height = (2 i ) 2 i 3. height = (2 i ) + 2 i
Estimting Are Between Curves - 3 Write the formul for LEF T (n), using this height s the function vlue. Wht will be? Write the definite integrl representing the re of this region. 4 3 y 2 y = 2 1 y = 2 3 2 1 1 1
Negtive Integrl Vlues - 1 Negtive Integrl Vlues So fr we hve only delt with positive functions. Will the definite integrl still be equl to the re underneth the grph if f() is lwys negtive? Wht hppens if f() crosses the -is severl times? Emple: Suppose tht f(t) hs the grph shown below, nd tht A, B, C, D, nd E re the res of the regions shown. If we were to prtition [, b] into smll subintervls nd construct corresponding Riemnn sum, then the first few terms in the Riemnn sum would correspond to the region with re A, the net few to B, etc.
Negtive Integrl Vlues - 2 Which of these sets of terms hve positive vlues? Which of these sets hve negtive vlues?
Negtive Integrl Vlues - 3 Epress the integrl nd E. () b b f(t) dt = A + B + C + D + E f(t) dt in terms of the (positive) res A, B, C, D, (b) (c) (d) b b b f(t) dt = A - B + C - D + E f(t) dt = -A + B - C + D - E f(t) dt = -A - B - C - D - E
Negtive Integrl Vlues - 4 If f(t) represents velocity, wht do the negtive res in B nd D represent? () The res B nd D represent negtive positions. (b) The res B nd D represent bckwrds motion. (c) The res B nd D represent distnce trvelled bckwrds.
Estimting Integrls - Midpoint Rule - 1 Better Approimtions to Definite Integrls We sw how to pproimte definite integrls using LEF T (n) nd RIGHT (n) Riemnn sums. Unfortuntely, these estimtes re very crude nd inefficient. Even when we hve sophisticted techniques for evluting integrls, these methods will not pply to ll functions, for emple: e 2 d - used in probbility sin() d - used in optics
Estimting Integrls - Midpoint Rule - 2 To evlute definite integrls of such functions, we could use left or right hnd Riemnn sums. However, it would be preferble to develop more ccurte estimtes. But more ccurte estimtes cn lwys be mde by using more rectngles. More precisely, we wnt to develop more efficient estimtes: estimtes tht re more ccurte for similr mounts of work.
Estimting Integrls - Midpoint Rule - 3 Midpoint Rule The ccurcy of Riemnn sum clcultion will usully improve if we choose the midpoint of ech subdivision rther thn the right or left endpoints. y y y b b b Compre the ccurcy of the left-hnd, right-hnd nd midpoint rules for estimting the re on the intervl.
Estimting Integrls - Midpoint Rule - 4 For wht kinds of functions f will the midpoint rule lwys give vlue tht is ectly equl to the integrl?
Estimting Integrls - Trpezoidl Rule - 1 The Trpezoidl Rule The Midpoint Rule is only one possible vrition on Riemnn sums. Another pproch is to use shpe other thn rectngle to estimte the re on n intervl. For the ppropritely nmed Trpezoidl Rule, we use trpezoid on ech intervl. Sketch trpezoidl pproimtion to the re under the grph, nd then write formul for the re of the single trpezoid. y i 1 i
Estimting Integrls - Trpezoidl Rule - 2 Write formul for the full Trpezoid Rule (written T RAP (n)), estimting the entire re under grph with multiple intervls. b 1 2 3... n 1 0 n
Estimting Integrls - Trpezoidl Rule - 3 Alterntive: Epress the trpezoidl rule T RAP (n) in terms of the left nd right hnd Riemnn sums (LEF T (n) nd RIGHT (n)). b 1 2 3... n 1 0 n
Estimting Integrls - Trpezoidl Rule - 4 The trpezoidl rule is especilly ccurte for functions f for which f () is smll for ll. Eplin from n intuitive point of view why you would epect this sttement to be correct. b 0 1 2 3... n 1 n
Estimting Integrls - Trpezoidl Rule Emples - 1 Use the trpezoidl rule to estimte vlues in the following tble for f(). 10 0 2 4 6 8 10 f() 1 3 4 5 4 2 0 f() d, if we hve mesured the f() 5 4 3 2 1 0 0 2 4 6 8 10