Key questions about the method ors decay with n Are there better methods? As you can see dividing the interval into two reduces the error and there is no reason to stop at just two subinter vals when we can have n subintevals and repeat our midpoint quadrature rule on each subinterval. We obtain the scheme f(x)dx=∑ 2(+) no doubt that we gain, but the key question is by how much? How does this gain scale with the number of subintervals used? And finally, are there clever ways of obtaining better accuracy with less effort? 3.2.3 Numerical Example SLIDE 7 E Note 6 Lets look at an example of integrating f(a)= sin(a) on our domain. We obtain progressively bet ter answers to the integral by increasing the number of subintervals n. The error in evaluating the integral is plotted as a function of the number of subintervals(n). The error appears to be going down as o(-) why From what we have just seen, the error inside the ith subinterval(of length sh=i,is adia for some Si E[=, i]. Hence, for the entire inter val [0, 1] d obtain the error at ion usin￾❀✓❛♦t❋❡✺❀✗❅❇●❖✾❁❉✮❊✺❅❄❍✮♣✭❉❋❡❲●✉●❖P✑❀❈❜❝❀✠●■P✑❉❲❣✁￾ ✂❢✪☎✄ ★✫✍✑✏✗✔❝❱✣✪⑥✔✗✚✣✖✩✖❑✬✛✬✻✪✺✬✻✏❦❱✣✖❑✘❋✍✮✒✆✄❦✢r✔✵✚ ✜✞✝ ✆❯✬✻✖❢✔✵✚✞✖❑✬✻✖✠✟❫✖❑✔✗✔✗✖❑✬❯✯➆✖❑✔✗✚✣✪➣❱✣✏✡✝ ➡➤➢✧➥➧➦✝☛ ♠❅❬❛✮❉❋❡♥❂✗❍❏❊✩❅❖❀✓❀➵❣✑✾❭❨✙✾❆❣❲✾❭❊✺❩❦●❖P✑❀➵✾❭❊❑●❖❀✗◆❖❨✻❍✮❴✴✾❁❊❋●■❉❝●q❃➍❉✆◆❖❀✵❣❲❡✺❂✠❀✵❅✉●❖P✺❀➵❀✓◆■◆■❉✮◆▲❍❏❊✭❣☞●❖P✑❀✗◆❖❀❈✾❆❅▲❊✑❉ ◆■❀✗❍✮❅❖❉✮❊➆●❖❉✲❅q●■❉✮▼➑❍✻● ✏q❡✺❅❇●❦●q❃➍❉⑥❅❖❡✑♣✑✾❁❊❋●■❀✓◆■❨✻❍❏❴❆❅❝❃❄P✑❀✓❊ ❃➍❀❢❂✗❍❏❊ P✭❍✛❨✮❀☞➷ ❅❖❡✑♣✑✾❁❊❑●❖❀✓❨✻❍✮❴❁❅❝❍❏❊✭❣ ◆■❀✓▼✰❀✗❍✻●➃❉❋❡✑◆➃❜❝✾❁❣❲▼✰❉✮✾❁❊❑●➍t❑❡✺❍✮❣✑◆■❍❏●❖❡✑◆■❀➍◆■❡✑❴❁❀❄❉✮❊❝❀✗❍❋❂◗P➫❅❖❡✑♣✑✾❁❊❑●❖❀✗◆❖❨✻❍❏❴➮✈✴➸➤❀▲❉❋♣❲●■❍✮✾❭❊➭●❖P✑❀➉❅❖❂◗P✑❀✗❜❝❀ ✛✢ ✣ ✒ ✕rÏ ✗✬✫❋Ï ✙ ✫✽ ✃✿ ✢ ✴ ➷ ❈❉✝❊❋ ❅❖❡✑♣✑✾❁❊❑●❖❀✓◆■❨✻❍❏❴ ❴❁❀✓❊✑❩✮●❖P ✒ ✕✥Ï✷ ✸✬✗ ❃❄P✑❀✗◆❖❀❈●■P✑❀❝❂✠❀✓❊❑●■◆❖❉❋✾❁❣♥❉✮❵✿●❖P✑❀➭❐✫❒✥❮✆❅❖❡✑♣✑✾❁❊❑●❖❀✓◆■❨✻❍❏❴✣✾❁❅➉Ï✷ ✸✟✙ ✢✯ ✕ ✃☞☛ ✽ ✢ ✱ ✃✽ ✗☞✙ ✃✌☛ ❅✍ ✽ ✈❈➾✉P✑❀✗◆❖❀➵✾❁❅ ❊✑❉♥❣❲❉✮❡✺♣❲●❈●❖P✺❍❏●❈❃➍❀❝❩❋❍✮✾❭❊✴➚✧♣✑❡❲●❈●❖P✺❀➫➴❋❀✓❛➩t❑❡✑❀✵❅q●■✾❭❉❋❊➩✾❆❅❻♣✙❛✩P✑❉✻❃➶❜➭❡✺❂◗P✛ ✦▲❉✻❃ ❣❲❉✙❀✗❅✎●❖P✺✾❁❅ ❩❋❍✮✾❭❊✷❅❖❂✗❍❏❴❁❀➭❃❄✾❭●❖P✷●❖P✑❀➫❊✙❡✑❜➵♣✰❀✓◆❻❉✮❵➍❅❇❡✺♣✑✾❭❊❑●■❀✓◆■❨✛❍✮❴❁❅✎❡✺❅❖❀✗❣✛ ♠❊✭❣✩Ð✺❊✭❍❏❴❁❴❭❛❋➚➣❍❏◆■❀③●❖P✺❀✓◆■❀➫❂✓❴❭❀✗❨✮❀✓◆ ❃✉❍✛❛✙❅⑨❉✮❵✞❉✮♣✑●■❍❏✾❁❊✑✾❁❊✑❩❝♣✭❀✓●❇●■❀✓◆❬❍❋❂✓❂✠❡✺◆■❍❋❂✠❛❦❃❄✾❭●❖P♥❴❭❀✵❅❖❅✉❀✓➯✰❉❋◆❇● ✛ ✍✴→❆➔➣→ ✍ ✎✦✣✯➆✖❑✬✻✢✥✘✮✍✺✼❫✌✌➣✍✑✯✲✱✳✼✥✖ ➙✰➛❑➜➞➝✴➟✑✏ ✛✢ ✣ ✓✽ ❐➮➷✚✕✥Ï✘✗❚✫✮Ï✕✔ ✫✽ ✃✿ ✢ ✴ ➷ ✽ ❐✫➷ ✖ ❐✍❄ ✢✯ ➷ ✘ SMA-HPC ©1999 MIT Normalized 1-D Problem Simple Quadrature Scheme Numerical Example E r r o r n ➡➤➢✧➥➧➦✓✒ ➺➣❀✓●■❅♦❴❭❉✙❉❋➴➑❍❏●☞❍✮❊❺❀✠Ñ✑❍❏❜❝▼✑❴❁❀➩❉✮❵③✾❭❊❑●■❀✓❩✮◆◗❍✻●■✾❭❊✺❩ ✒ ✕rÏ ✗❭✙ ✽ ❐➮➷✚✕✥Ï✘✗✆❉✮❊❺❉❋❡✑◆♥❣❲❉✮❜❦❍❏✾❁❊➣✈ ➸✷❀ ❉✮♣✑●■❍❏✾❁❊☞▼✑◆❖❉❋❩✮◆■❀✗❅■❅❇✾❁❨✮❀✗❴❭❛➫♣✰❀✠●❇●■❀✓◆▲❍✮❊✺❅❇❃⑨❀✓◆◗❅➍●❖❉➫●❖P✺❀❈✾❭❊❑●❖❀✗❩✮◆◗❍❏❴✹♣✙❛❯✾❁❊✺❂✠◆■❀✗❍❋❅❇✾❁❊✑❩➭●❖P✑❀❻❊✙❡✑❜➵♣✰❀✓◆❄❉✮❵ ❅❖❡✑♣✑✾❁❊❋●■❀✓◆■❨✻❍❏❴❆❅❬➷➃✈❈➾✉P✑❀➫❀✓◆■◆❖❉❋◆❬✾❁❊➩❀✗❨✛❍✮❴❭❡✭❍✻●❖✾❁❊✑❩✆●❖P✑❀❝✾❁❊❋●■❀✓❩❋◆■❍✮❴✴✾❆❅➉▼✺❴❭❉✮●❇●❖❀✵❣➩❍❋❅➉❍✆❵r❡✑❊✺❂✠●❖✾❁❉✮❊➩❉✮❵ ●❖P✺❀➉❊✙❡✑❜➭♣✭❀✗◆⑨❉✮❵✴❅❇❡✺♣✑✾❭❊❑●■❀✓◆■❨✛❍✮❴❁❅ ❡r➷❨❢✠✈✿➾✉P✑❀✎❀✓◆■◆❖❉❋◆➍❍❏▼✺▼✭❀✵❍❏◆◗❅✿●■❉➭♣✰❀➉❩❋❉✮✾❁❊✑❩➭❣✑❉✻❃❄❊✆❍✮❅✕✔✗✖ ✽✢ ✍☎✘ ✈ ➺➣❀✓●■❅❄❅❖❀✓❀❻❃❄P✙❛✮✈ ✘✑◆■❉✮❜ ❃❄P✺❍✻●❦❃⑨❀☞P✭❍✛❨✮❀ ✏q❡✺❅q●✆❅❖❀✓❀✗❊➣➚➍●❖P✑❀❢❀✗◆❖◆■❉✮◆➫✾❭❊✭❅❇✾❆❣❲❀♥●❖P✺❀☞❐✫❒✥❮✲❅❖❡✑♣✑✾❁❊❑●❖❀✗◆❖❨✻❍❏❴✂❡✥❉❏❵❻❴❭❀✗❊✑❩❏●■P ✙ ★ ✙ ✽✢ ➻ ❢❄✾❆❅ ❮✛✚ ✯ ✎ ❙ ✍✢✜✤✣✦✥ ✸✌✧ ❙ ❁ ✍ ❵r❉❋◆❬❅❖❉✮❜❝❀✜✷✓✃ ✯ ✱ ✃✌☛ ✽ ✢ ✦ ✃✽ ✶➮✈ ✦❬❀✓❊✺❂✓❀✮➚✭❵r❉✮◆❄●■P✑❀➵❀✗❊❑●❖✾❁◆❖❀❈✾❁❊❑●❖❀✓◆■❨✻❍❏❴ ✱✲ ✦❳✴✝✶ ❃⑨❀➫❂✗❍❏❊⑥❅❇❡✑❜ ●❖P✑❀✵❅❇❀➫❀✓◆■◆❖❉❋◆■❅✎❍❏❊✭❣➩❉❋♣❲●■❍✮✾❭❊➩●❖P✑❀❝❀✓◆■◆■❉✮◆✵➚✺●✽ ❵r❉❋◆❻❍✮❊➤❍✮▼✑▼✑◆■❉✛Ñ❲✾❭❜❦❍✻●■✾❭❉❋❊❢❡✺❅❖✾❭❊✑❩ ★