18.01 Calculus Jason Starr Fall 2005

Transcription

1 18.01 Clculus Jso Strr Lecture 14. October 14, 005 Homework. Problem Set 4 Prt II: Problem. Prctice Problems. Course Reder: 3B 1, 3B 3, 3B 4, 3B The problem of res. The ciet Greeks computed the res of trigles, qudrilterls, d my other polygos. Their bsic method ws dissectio: dissectig polygol regio exctly ito smller regios, usully trigles, hvig kow res. The re of the lrge regio is the sum of the res of the smll regios. But the ciet Greeks lso kew the re of circle, which cot be dissected exctly ito fiitely my polygol regios. Their method ws exhustio: fidig polygol regios pproximtely equl to the origil regio, d computig the limit of the res of the polygos s the pproximtio improves. Exmple. A regulr N sided polygo iscribed i circle of rdius r hs pothem legth = r cos(π/n ) d chord legth b = r si(π/n ). Thus the re of the polygo is, b A = N = N r si(π/n ) cos(π/n ) = r N si(π/n ) = πr si(π/n ). π/n As N icreses, π/n decreses to 0. Becuse lim t the re of the polygo pproches, 0 si(t)/t equls 1, s N pproches ifiity, lim πr si(π/n ) = π/n N πr. A more sophisticted versio of the method of exhustio gives the Riem itegrl. Here is the bsic problem. Problem (Are). Fid the siged re betwee the grph of y = f (x) d the x xis over the itervl x b.

2 18.01 Clculus Jso Strr For regio bove the x xis, the siged re is simply the re. For regio below the x xis, the siged re is the egtive of the re. For regio prtly bove the x xis d prtly below the x xis, the siged re is the sum of the siged re of the regio bove the x xis d the siged re of the regio below the x xis.. Prtitios. A prtitio of itervl [, b] is fiite decompositio of the itervl s uio of o overlppig subitervls, [, b] = [x 0, x 1 ] [x 1, x ] [x, x ] [x 1, x ]. Sice itervl is determied by its right d left edpoits, to specify prtitio of [, b], it is equivlet to give ordered sequece of icresig umbers, = x 0 < x 1 < x < < x < x 1 < x = b. The k th subitervl of the prtitio is the itervl [x k 1, x k ], hvig legth, Δx k = x k x k 1. A prtitio is fie if the subitervls re smll, d corse if the subitervls re lrge. It my seem the umber of itervls is good mesure of fieess: sice the subitervls of fie prtitio re smll, the umber of subitervls must be lrge. However, prtitio ito my subitervls my iclude few subitervls tht re quite lrge. For istce, the prtitio [0, 1] = [0, 1/] [1/, /] [/, 3/] [( )/, ( 1)/] [ 1/, /] [1/, 1], hs very smll itervls of legth 1/, but hs oe itervl, [1/, 1], of size 1/. The umber 1/ my ot seem lrge, but s icreses, it is quite lrge compred to 1/. Becuse of such exmples, better mesure of fieess is mesh size: The mesh size of prtitio is the mximl legth of y subitervl i the prtitio, mesh = mx Δx k k = 1,...,. 3. Riem sums. Let f(x) be fuctio defied o itervl x b. Give prtitio = x 0 < < x = b of [, b], d give choice, for every k = 1,...,, of elemet x k i the k th subitervl, x k 1 x k x k, the curvilier regio bouded by y = f(x) d the x xis is pproximted by uio of verticl strips. The k th verticl strip lies bove or below the itervl o the x xis, x k 1 x x k, d hs height yk = f(x k ). The width of the verticl strip is Δx k, thus the siged re is, ΔA k = yk Δx k. The totl re of the uio of verticl strips is simply the sum of the res of idividul verticl strips, A = yk Δx k. k=1

3 18.01 Clculus Jso Strr The sum bove is Riem sum. It is pproximtio of the siged re of the curvilier regio. There re my choices of prtitio. Ad for ech prtitio, there re my choices for the umbers xk. However, there re some specil choices. O the kth itervl, the smllest vlue f (x) tkes o is deoted by, y k,mi = mi{f (x) x k 1 x x k+1 }. Similrly, the lrgest vlue f (x) tkes o is deote by, y k,mx = mx{f (x) x k 1 x x k+1 }. For every choice of x k i the k th itervl, yk is trpped betwee these two vlues, Deotig, y k,mi y k y k,mx. ΔA k,mi = y k,mi Δx k, ΔA k,mx = y k,mx Δx k, the re ΔA k is trpped betwee these two vlues, ΔA k,mi ΔA k ΔA k,mx. Deotig the sums of the res by, A mi = k=1 ΔA k,mi = k=1 y k,miδx k, A mx = k=1 ΔA k,mi = k=1 y k,miδx k, the Riem sum A is trpped betwee the two vlues, A mi A A mx. Thus, if A mi d A mx re close to ech other, the vlue of A does ot deped very much o the choices of the umbers x k. 4. The Riem itegrl. The method of the Riem itegrl is to compute both A mi d A mx for sequece of prtitios whose mesh sizes pproch 0. The mesh size mesures the fieess of the prtitio, thus lso the fit of the uio of verticl strips to the curvilier regio. If the two limits, lim A mi, lim A mx, mesh 0 mesh 0 re defied d equl, it is sid the Riem itegrl exists, d the commo limit is clled the Riem itegrl, b f (x)dx = lim A mi = lim A mx. mesh 0 mesh 0 Also, f (x) is sid to be Riem itegrble o the itervl [, b]. Aother me for the Riem itegrl is the defiite itegrl.

4 18.01 Clculus Jso Strr Exmple. Cosider the fuctio f (x) = x o the itervl 0 x L, for some positive umber L. Form the prtitio with subitervls of equl legth, x 0 = 0 = 0L/, x 1 = 1L/, x = L/,..., x k = kl/,... x = L/ = L. Every itervl hs legth Δx k = L/. So the mesh size is L/. The miimum vlue of f (x) o the itervl x k 1 x x k is y k,mi = x k 1 = (k 1)L/. The mximum vlue is y k,mx = x k = kl/. Thus, (k 1)L L L Ami = y k,mi Δx k = = (k 1), k=1 k=1 k=1 d, kl L L A mx = y k,mx Δx k = = k. k=1 k=1 k=1 To evlute these sums, use the well kow formul, ( + 1) k =. k=1 This lso gives, 1 1 ( 1) (k 1) = l = l =, k=1 l=0 l=1 by mkig the substitutio l = k 1. Substitutig the formul gives, L ( 1) = L (1 1 ), A mi = d, L ( + 1) L 1 A mi = = (1 + ). Therefore, L 1 L L lim A mi = lim(1 ) = (1 0) =. 0 Similrly, L 1 L L lim A mx = lim(1 + ) = (1 + 0) =. 0 Sice the two limits re equl, f (x) = x is Riem itegrble o the itervl [0, L], d, L xdx = L /. 0 This grees with the fmilir result from high school geometry: the re of trigle equls oe hlf of the bse times the height, sice both the bse d height of this trigle re L.

5 18.01 Clculus Jso Strr 5. Rules for Riem itegrls. There re severl rules for Riem itegrls, summrized below. b b b (f(x) + g(x))dx = f (x)dx + g(x)dx, b (r b f (x))dx = r f (x)dx, b c c f (x)dx + b f (x)dx = f (x)dx.