Clculus Li Vs Defiite Itegrl. The Left d Right Sums The defiite itegrl rises from the questio of fidig the re betwee give curve d x-xis o itervl. The re uder curve c be esily clculted if the curve is give by simple formul. For exmple, if fuctio is positive costt f(x) = c, the re uder this curve o itervl [, b] is the re of the rectgle with sides b d c. Thus the re is A = (b )c. We re iterested i clcultig the re A uder the curve f for x b for y oegtive, cotiuous curve f(x). This, i geerl, rther complex problem c be pproched the followig ide: we divide the itervl [, b] i smll subitervls d pproximte the re uder f(x) o ech subitervl with the re of rectgle with height f(x i ) where x i is poit i i-th subitervl of [, b]. Let us mke this ide more precise usig the sums of rectgles kow s the left d the right sums. Approximtig the re. Step. The prtitio. Divide [, b] ito -pieces. Get the poits = x 0, x,..., x, x = b. This is clled the prtitio of [, b]. The distce betwee ech two poits is h = b. Step. The rectgles. The left d right sums differ i the wy how we choose the poit from the subitervl which will produce the height of the rectgle. To pproximte the re usig the left sum, choose the left edpoit of ech subitervl d evlute the fuctio t it to obti the height of the rectgle. For exmple, x 0 is the left edpoit of the first subitervl [x 0, x ] d so the height of the first rectgle is f(x 0 ).
The re of the first rectgle is A 0 = hf(x 0 ). The left edpoit of the secod itervl [x, x ] is x d the secod height is f(x ). The re of the secod rectgle is A = hf(x ). Cotiuig this process we obti tht the re of the i+-st rectgle is A i = hf(x i ). The process eds with the -th rectgle of the re A = hf(x ). The sum of ll the rectgles is Left sum = hf(x 0 ) + hf(x ) +... + hf(x ) = h (f(x 0 ) + f(x ) +... + f(x )) = b (f(x 0 ) + f(x ) +... + f(x )). Alogously, to pproximte the re usig the right sum, choose the right edpoit of ech subitervl to form rectgle. The poit x is the right edpoit of the first subitervl [x 0, x ] d so the height of the first rectgle is f(x ) d the re is A = hf(x ). The right edpoit of the secod itervl [x, x ] is x, the secod height is f(x ), d the re of the secod rectgle is A = hf(x ). Cotiuig this process we obti tht the re of the i-th rectgle is A i = hf(x i ). is The process eds with the -th rectgle of the re A = hf(x ). The sum of ll the rectgles Right sum = hf(x ) + hf(x ) +... + hf(x ) = h (f(x ) + f(x ) +... + f(x )) = b (f(x ) + f(x ) +... + f(x )). Thus, the re uder curve c be pproximted usig both left d right sums. Are A Left sum = b Are A Right sum = b (f(x 0 ) + f(x ) +... + f(x )) (f(x ) + f(x ) +... + f(x )) Exmple. Approximte the re uder f(x) = x o [0,] usig the left d right sums with subitervls. Solutio. Note tht = 0, b = d =. Thus b = 0 =. The prtitio cosists of 3 poits = x 0 = 0, x = d b = x =. Compute the correspodig y-vlues to be f(0) = 0 = 0, f() = = d f() = = 4. Thus the left sum is L = 0 (0 + ) = d the right sum is L = 0 ( + 4) = 5.
From the grph, we c coclude tht the left sum is uderestimte d the right sum 5 is overestimte. Thus A 5. Note tht the pproximtio is ot very ccurte sice we do ot eve hve the first digit of the exct swer correct. Icresig the umber of subitervls, tht is usig lrger vlues of, icrese the ccurcy of pproximtio s the ext figure illustrtes. This fct suggests tht fidig the exct re uder the curve c be obtied by tkig the limit of y of these sums whe the umber of prtitios icreses to ifiity. If f is icresig d oegtive cotiuous fuctio s o the lst severl figures, the exct re will be sdwiched betwee the left sum from below d the right sum from bove (thus left d right sums re two pieces of bred ). The limit of both left d right sum whe pproches the sme vlue which is the exct vlue of the re uder the curve ( hm ). Are= hm, left d right sums = bred If f is icresig, oegtive, cotiuous fuctio, the left sum lwys produces uderestimte d the right sum overestimte. For decresig, oegtive, cotiuous fuctios this is the opposite. For the fuctios with extreme vlues either hs to be the cse. Whe f is oegtive, cotiuous fuctio, ot ecessry icresig, the re c still be sdwiched betwee the lower sum (obtied by formig rectgles usig poits with the smllest height o ech subitervl) d the upper sum (obtied by formig rectgles usig poits with the highest height o ech subitervl). Usig the Closed Itervl Method, such poits c be foud. 3
Lower Sum Are A Upper Sum closed itervl. This lso illustrtes tht the limit of the sum of rectgles formed by y poit x i from the i- th subitervl [x i, x i ], for i =,,...,, exists. Moreover this limit exists for y cotiuous fuctio (ot ecessrily oegtive) defied o The sum of the rectgles is (f(x ) + f(x ) +... + f(x ))h d c be writte shorter s i= f(x i)h. The legth of the subitervl h = b is lso deoted s x. The limit lim i= f(x i) x is defied to be the defiite itegrl of cotiuous fuctio f(x) o itervl [, b] d it is deoted s b f(x)dx. Thus we hve tht The Defiite Itegrl b f(x) dx = lim i= f(x i ) x. The followig iforml resoig is the bse for the ottio b f(x)dx: the smll vlues of x correspod to dx, whe the umber of subitervls is very lrge d their umber very smll, pickig vlues x i from ech subitervl mouts to pickig ll poits x from the itervl [, b]. I the limitig cse, the sum becomes the itegrl symbol idictig tht we pick vlues x o cotiuous wy from etire itervl [, b]. x dx, f(x i ) f(x), b i= so tht i= f(x i ) x b f(x) dx. Thus, the defiite itegrl provides wy of clcultig the re uder the curve. If f oegtive, cotiuous fuctio o [, b], the re A uder the curve f bove the x-xis, for x b is A = b f(x) dx. Before ddressig the re uder curves which re ot oegtive i the ext sectio, we cosider some pplictios of the left d right sums. Exmple. A skydiver drops from irple. At the ed of the first six secods the diver s speed (i meters per secod) is checked, d it reds s follows: speed 8 4 9 3 5 You c ssume tht the diver s iitil speed is zero. Use the pproprite left d right sums to pproximte the distce the diver flls durig the six-secod period. 4
Solutio. With the restrictios of the mesuremets bove, we pproximte the velocity durig ech secod with costt. Thus the distce trveled i ech secod c be pproximted with the product of velocity d time. If the velocity vlues re represeted s depedet vrible vlues d time vlues s the idepedet vrible vlues, the product of velocity d time correspods exctly to re of rectgles. Sice the velocity vlues re strictly icresig d positive, the left sum provides uderestimte d the right sum the overestimte of the totl distce trveled. Addig the time vlues to the give tble, we obti the followig. time 0 3 4 5 speed 0 8 4 9 3 5 The speed is mesured o the time itervl [0,] so we c deote = 0 d b =. The umber of subitervl is =. Note tht represets the umber of subitervls ot the umber of poits (7 poits determies subitervls). The left sum is L = 0 (0+8+4+9+3+5) = 89 d the right sum is R = 0 (8 + 4 + 9 + 3 + 5 + ) = 5. Thus the diver s speed c be estimted to be betwee 89 d 5 meters (ote tht the uits of the swer re meters sice the term 0 is i secods d the sums i prethesis re i meters per secods). Exmple 3. Approximte the re of the lke usig the show mesuremets of its width which were tke 50 feet prt. 0 88 0 45 80 38 9 93 84 0 Solutio. You c cosider the mesuremet vlues s the depedet vrible vlues t the vlues of the idepedet vrible strtig t 0 d beig 50 feet prt. x 0 50 00 50 00 50 300 350 400 450 y 0 88 0 45 80 38 9 93 84 0 From the tble bove we c see tht = 0 d b = 50 9 = 450. There re 0 poits, thus = 9 subitervls. Altertively, the mesuremets re tke 50 feet prt, thus h = b = 50. The left sum is L = 450 0 (0 + 88 + 0 + 45 + 80 + 38 + 9 + 93 + 84) = 48350 d the right 9 sum is R = 450 0 (88 + 0 + 45 + 80 + 38 + 9 + 93 + 84 + 0) = 48350. So, the re of the lke 9 is estimted to be 48350 squre feet. Try to geerlize the coclusio of this problem: the left d the right sums will be equl wheever f(x 0 ) = f(x ). 5
Left d Right Sum Progrm for TI83 84 The progrm below clcultes left d right sum of give fuctio f o give itervl [, b] with the give umber of subitervls. Whe executig this progrm i order to pproximte b f(x)dx usig left d right sums, the fuctio f should be etered s Y d, b d should be etered whe executig the progrm. Whe eterig the progrm i the clcultor, editig or executig d existig progrm, strt by usig PRGM key. PROGRAM: LFTRGT Disp LOWER BOUND (to disply Disp, choose PRGM the I/0 meu) Iput A (to disply Iput, choose PRGM the I/0 meu) Disp UPPER BOUND Iput B Disp NUMBER OF SUBINTERVALS Iput N (B-A)/N D 0 L For (I, 0, N-) (to disply For, choose PRGM the CTL meu) A+D*I X L+D*Y L (to disply Y, choose VARS, Y-VARS, the : Fuctio) Ed (to disply Ed, choose PRGM the CTL meu) Disp LEFT SUM, L 0 R For (J,, N) A+D*J X R+D* Y R Ed Disp RIGHT SUM, R You c exit the progrm editig by usig QUIT. Exmple 4. Use the clcultor progrm to fid the left d right sums of f(x) = x o itervl [0,] with 50 subitervls. Solutio. Strt by eterig x s your fuctio Y. The execute the progrm by PRGM d choosig LFTRGT. Eter the bouds 0 d d the umber of subitervls 50. You should get.587 for the left d.747 for the right sum. Exmple 5. Use the Left d Right sums to estimte the re uder the curve f(x) = x o itervl [0,] to the first two ozero digits (i.e. mke your estimtes hve the first two ozero digits equl). Solutio. Sice x is positive d icresig o [0,], the left sum is uderestimte d the right sum is overestimte of the exct re 0 x dx. We re give fuctio d the itervl but we re ot give the umber of subitervls. Thus, you c strt by eterig the give fuctio, the edpoits of the itervl d reltively lrge (e.g. 00 or 00) to moitor the left d right sum vlues L d R. Keep icresig if ecessry util L d R both roud to the sme vlue with desired umber of sigifict digits. I tht cse, you obti the vlue of the defiite itegrl you re sked to fid.
Note tht you do ot wt to strt with tht is too smll (e.g. below 0 becuse it will be most likely tht you will hve to icrese it ywy) or tht is too lrge (e.g. 00,000 becuse it my tke log time for your clcultor to produce the swers). Note lso tht you re ot sked to fid the smllest possible vlue of but y vlue of tht will produce the desired ccurcy. I the previous exmple we obtied tht with = 50 the left sum rouds to. d the right sum to.7. Sice we eed to hve the first two digits i the left d right sums equl, we eed to icrese the ccurcy. with = 00 the left sum rouds to.8. d the right sum to.708.7 which is ot ccurte eough. With = 00 we obti.47. for the left d.87.7 for the right which is still ot ccurte eough. With = 300, we obti.533.7 for the left d.800.7 Thus, we coclude tht the re is.7 up to the first two digits. Prctice Problems.. Approximte the re uder the followig curves o the give itervls usig left d right sums usig the prtitio with subitervls. () f(x) = x, [0, ], =. (b) f(x) = x, [0, ], = 4. (c) f(x) = 8 x, [0, 3], = 00. Use the clcultor progrm.. Approximte the itegrl 4 x dx usig the Left-Right Sums clcultor progrm with = 50 subitervls. Note tht the fuctio is decresig o [,4] so tht the left sums gives you the overestimte d the right sum the x uderestimte. 3. Approximte the followig defiite itegrls usig your clcultor progrm to the first two ozero digits. () 0 l(x + ) dx (b) 3 e x x dx 4. The speed of ruer icreses durig the first three secods of rce. His speed (i meters per secod) t hlf-secod itervls is give i the tble bellow. Fid lower d upper estimtes for the distce he trveled durig the first three secods. time 0 0.5.5.5 3 speed 0 3. 5.4 7.4 8.9 9. 0. 5. A chemicl rectio produces compoud X with rte of 3, 9,,, 9, 5, liters per secod t time itervls spced by secod. Approximte the totl volume of the compoud X produced i the secods for which the rte is give usig the left d the right sums. Solutios. 7
. () = d x 0 y 0 (b) = 4, d thus L = 0 x 0 0.5.5 y 0 0.5.5 4 0 (0 + ) = d R = ( + ) = 3. 0 3.5 thus L = (0 + 0.5 + +.5) = 4 R = 0 7.5 (0.5 + +.5 + 4) = = 3.75. 4 (c) f(x) = 8 x, = 0, b = 3 d = 00. Obti tht L = 7.49 d R = 7.9. =.75 d. Use the progrm with f(x) = x, =, b = 4 d = 50. Obti tht L =.759 d R =.74. Thus.74 4 dx.759. x 3. () f(x) = l(x + ), = 0 d b =. With = 00 for exmple, both the left sum d the right sum soud to.4. Thus, 0 l(x + ) dx.4. (b) f(x) = ex x sum roud to 8. Thus, 3, =, d b = 3. With = 300 for exmple, both the left sum d the right dx 8. e x x 4. Similrly s i Exmple, the left d the right sum provide the estimtes for distce trveled. I this problem, = 0, b = 3, = (creful: is ot 7). The left sum is L = 3 0(0+3.+5.4+ 7.4+8.9+9.) = 34 3 0 44. = 7 d the right sum is L = (3.+5.4+7.4+8.9+9.+0.) = =.05. Thus,.05 meters is the upper d 7 meters is the lower estimte for the distce. 5. With the restrictios of the mesuremets bove, we pproximte the rte durig ech secod with costt. Thus the volume dded i ech secod c be pproximted with the product of the rte of icrese d time. If the rte vlues re represeted s depedet vrible vlues d time vlues s the idepedet vrible vlues, the product of the rte d time correspods exctly to re of rectgles. Sice the rte vlues re strictly decresig d positive, the left sum provides overestimte d the right sum the uderestimte of the totl volume produced. Addig the time vlues to the give tble, we obti the followig. time (sec.) 0 3 4 5 rte (l/sec.) 3 9 9 5 Thus, = 0, b = d =. L = 0 (3 + 9 + + + 9 + 5) = 79 liters, R = (9 + + + 9 + 5 + ) = 58 liters. 0 8