I nanini ebalulekile. Kwekhompyutha of integrals nanini

Esinye sezigaba eyisisekelo izingabunjalo kuyinto calculus ebalulekile. Ihlanganisa insimu ebanzi kakhulu lezinto, lapho kuqala - kuba nanini ebalulekile. Isikhundla limi njengomthombo ukhiye okungukuthi esafunda esikoleni samabanga aphakeme lembula ezandayo amathemba kanye namathuba, echaza wezibalo ephakeme.

ukubukeka

Uma uthi nhlá, kubonakala sengathi sichitha siqeda ebalulekile yesimanje lefanele, kunika tihlokwana, kodwa umkhuba kuvela ukuthi weza emuva ngo-1800 BC. Ikhaya ukuze kubhekwe ngokusemthethweni Egypt njengoba kwakudliwa kithi ubufakazi ngaphambili khona. It ngenxa yokuntula ulwazi, ngaso sonke isikhathi isebangeni nje into. Wake uyaphinda uqinisekisa ezingeni yentuthuko yesayensi abantu ngalezo zikhathi. Ekugcineni, imisebenzi atholakele le zezibalo lwesiGreki sakuqala, yangekhulu lesi-4 leminyaka BC. Zichaza indlela esetshenziswa lapho nanini ebalulekile, ngokuyisisekelo okwakukhona ukuthola umthamo noma endaweni umumo curvilinear (indiza ngakuthathu mgudumbili, ngokulandelana). ukubala yayisekelwe phezu kwesimiso division of the sibalo ezilimini zokuqala liyiswa olimini izingxenye encanyana, inqobo nje uma ivolumu (endaweni) usevele waziwa kubo. Ngokuhamba kwesikhathi, indlela ukhulile, Archimedes esetshenziswa ukuthola kwendawo parabola. izibalo Okufanayo ngesikhathi esifanayo ukuqhuba ukuqeqeshwa ngendlela okumelwe China lasendulo, lapho kwakuyibona ezimele ngokuphelele kusukela ngesiGreki nabo isayensi.

ukuthuthukiswa

Ukufohla elandelayo XI leminyaka BC isibe umsebenzi isazi Arab "inqola" Abu Ali al-Basri, ngubani kwaphoqa imingcele eyaziwa kakade, zathathwa ifomula ebalulekile ukubala tibalo imali kanye degrees kusuka ngowokuqala yesine, isicelo salokhu sekusaziwa yithi indlela lokungeniswa.
Izingqondo namuhla bayahlonishwa nabaseGibhithe lasendulo wadala ezikhumbuzo okumangalisayo ngaphandle amathuluzi ezikhethekile, ngaphandle ukuthi yezandla zabo, kodwa akuyona amandla ososayensi elihlanyiswa isikhathi engengaphansi kuyisimangaliso? Kuqhathaniswa izikhathi samanje ezimpilweni zabo zibonakala cishe bakudala, kodwa isinqumo of integrals nanini esephetha yonke indawo futhi lisetshenziswe umkhuba ekuthuthukiseni olwengeziwe.

Isinyathelo kwakwenzeka ekhulwini XVI, lapho sezibalo Italian Cavalieri baletha indlela indivisible, okuyinto wacosha Per Ferma. Lezi ubuntu ezimbili wabeka isisekelo sokuba calculus yesimanje ebalulekile, okuyinto okwamanje eyaziwa. Zazibophela imiqondo yokuzihlukanisa kwamangqamuzana ngokwemisebenzi yawo nokuhlanganiswa, okuyinto yayikade itsatfwe njengentfo amayunithi akancike. Ngokwengxenye enkulu, izibalo langaleso sikhathi laliwumfuziselo izinhlayiya wahlukana okutholakele aziphilele ngawodwa abe yizinto, ngokusebezisa esilinganiselwe. Indlela yokuhlanganisa kanye singathola esivumelana ngakho kwaba okuwukuphela kukaNkulunkulu weqiniso. okwamanje, sibonga kuye, esimanjemanje izingabunjalo kwadingeka ithuba ukuze akhule ngendlela.

Nge kwesikhathi sishintsha konke futhi uphawu ebalulekile kanye. Ngokwengxenye enkulu, ke yakhethwa nososayensi abakholelwayo ngendlela yakhe, isibonelo, Newton esetshenziswa isithonjana square, okuyinto wafaka i umsebenzi integrable, noma uma sikubeka kalula ndawonye. Lokhu kungalingani waqhubeka kwaze kwaba ngekhulu XVII, lapho-mlando lonke imfundiso yokuziphendukela zezibalo ukuhlaziywa usosayensi Gotfrid Leybnits wasungula uhlamvu enjalo ajwayelekile kithi. Side "S" empeleni isekelwe le ncwadi ka ezivamile, kusukela lisho isamba primitives. Igama ebalulekile etholwe sibonga Jakob Bernoulli, ngemva kweminyaka engu-15.

Inchazelo esisemthethweni

I ebalulekile nanini incike kwencazelo bakudala, ngakho sicabanga ke kwasekuqaleni.

Antiderivative - umsebenzi inverse we esuselwe, practice ibizwa ngokuthi lasekuqaleni. Kungenjalo: Umsebenzi bakudala of d - kuyinto D umsebenzi, okuyinto esuselwe v <=> V '= v. Ukucinga bakudala kuyinto ukubala ebalulekile esingenasiphelo, inqubo ngokwalo ubizwa ngokuthi ukuhlanganiswa.

Ngokwesibonelo:

Umsebenzi s (y) = y ^3, futhi S yayo bakudala (y) = (y ^4/4).

Isethi wonke primitives umsebenzi - lokhu kuyinto ebalulekile nanini, okhonjiswe it kanje: ∫v (x) dx.

Ngenxa yokuthi V (x) - kuphela umsebenzi ezinye bakudala yokuqala, inkulumo ophethe: ∫v (x) dx = V (x) + C, lapho C - njalo. Ngaphansi njalo ngokungenasizathu isho noma yiluphi ulwazi njalo, kusukela esuselwe salo zero.

izakhiwo

I izakhiwo possessed ebalulekile nanini, ngokuyisisekelo esekelwe definition nezakhiwo amagama ahlobene.
Cabangela amaphuzu enkulumo ayisihluthulelo:

• esuselwe ebalulekile bakudala kuyinto bakudala ngokwayo plus ongumashiqela njalo C <=> ∫V '(x) dx = V (x) + C;
• esuselwe ebalulekile umsebenzi umsebenzi yokuqala <=> (∫v (x) dx) '= v (x);
• njalo ithathwa kusukela ngaphansi uphawu ebalulekile <=> ∫kv (x) dx = k∫v (x) dx, lapho k - kuyinto kokubonisana nomkayo;
• ebalulekile, esithathelwe isamba ncamashí ilingana nenani lezinombolo of integrals <=> ∫ (v (y) + w (y)) dy = ∫v (y) dy + ∫w (y) dy.

Eminyakeni emibili izakhiwo kungenziwa baphetha ngokuthi ebalulekile nanini lwento. Ngenxa lokhu, siba nesimo: ∫ (kV (y) dy + ∫ LW (y)) dy = k∫v (y) dy + l∫w (y) dy.

Ukuze ubone izibonelo ukulungisa izixazululo integrals nanini.

Kufanele uthole ∫ ebalulekile (3sinx + 4cosx) dx:

• ∫ (3sinx + 4cosx) dx = ∫3sinxdx + ∫4cosxdx = 3∫sinxdx + 4∫cosxdx = 3 (-cosx) + 4sinx + C = 4sinx - 3cosx + C.

Kusukela Ngokwesibonelo singaphetha ngokuthi ungazi indlela yokuxazulula integrals kuze kube phakade? Vele ukuthola zonke primitives! Kodwa ucinge izimiso okuxoxwa ngazo ngezansi.

Izindlela kanye Izibonelo

Ukuze ukuxazulula ebalulekile, ungakwazi iphendukele izindlela ezilandelayo:

• ukulungele ukusizakala etafuleni;
• zokuhlanganisa by izingxenye;
• ihlanganiswe ngokufaka esikhundleni variable;
• efingqa ngaphansi isibonakaliso umehluko.

amathebula

I elula kakhulu futhi kujabulise ngendlela. Okwamanje, umqulusizinda kungaba ozibongayo amatafula ngempela ezinde kakhulu, ezingase lelipelwe ifomula eziyisisekelo integrals nanini. Ngamanye amazwi, zikhona izifanekiso etholakala kukuwe futhi ungakwazi kuphela ukusizakala ngazo. Nalu uhlu izikhundla etafuleni main, okuyinto ingaboniswa cishe kuzo zonke izimo, onesixazululo:

• ∫0dy = C, lapho C - njalo;
• ∫dy = y + C, lapho C - njalo;
• ∫y ⁿ dy = (y ^{n +1)} / (n + 1) + C, lapho C - njalo, futhi n - inombolo sihluke ubunye;
• ∫ (1 / y) dy = ln | y | + C, lapho C - njalo;
• ^{∫e y dy = e y + C} , lapho C - njalo;
• ^{∫k y dy = (k y /} ln k) + C, lapho C - njalo;
• ∫cosydy = siny + C, lapho C - njalo;
• ∫sinydy = -cosy + C, lapho C - njalo;
• ∫dy / cos ² y = tgy + C, lapho C - njalo;
• ∫dy / sin ² y = -ctgy + C, lapho C - njalo;
• ∫dy / (1 + y ²⁾ = arctgy + C, lapho C - njalo;
• ∫chydy = + C namahloni, lapho C - njalo;
• ∫shydy = chy + C, lapho C - njalo.

Uma kudingeka, kwenze ambalwa izinyathelo eziholela integrand ukuba umbono tabular futhi ujabulele ukunqoba. ISIBONELO: ∫cos (5x -2) dx = 1 / 5∫cos (5x - 2) d (5x - 2) = 1/5 x isono (5x - 2) + C.

Ngokusho isinqumo kuyacaca ukuthi isibonelo integrand ithebula sintula Okuphindaphinda 5. Thina uyifaka ku parallel nalokhu ukwandiswa by 1/5 nkulumo jikelele awushintshanga.

Integration nge Izingxenye

Cabanga ngemisebenzi emibili - z (y) futhi x (y). Kumele kube njalo differentiable esizindeni salo. Kokunye kwamangqamuzana izakhiwo esinalo: d (xz) = xdz + zdx. Ukuhlanganisa zombili, sithola: ∫d (xz) = ∫ (xdz + zdx) => zx = ∫zdx + ∫xdz.

Ukushintsha Isimo kwesibalo okuholela, sithola ifomula, elichaza indlela ukuhlanganiswa by izingxenye: ∫zdx = zx - ∫xdz.

Kuqakatheke ngani? Iqiniso lokuthi ezinye izibonelo kungenzeka lula, ake sithi, ukunciphisa ∫xdz ∫zdx, uma yokugcina useduze ifomu tabular. Futhi, lokhu ifomula ingasetshenziswa isikhathi esingaphezu kwesisodwa, imiphumela elilungile.

Indlela ukuxazulula integrals nanini ngale ndlela:

• kudingekile ukubala ∫ (s +1) e ^2s ds

∫ (x + ^{1) e 2s ds = {z =} s + 1, Dz = ds, y = 1 ^{/ 2e 2s, dy = e 2x} ds} = ((ama + 1) e ^2s) / 2-1 / 2 ∫e ^2s dx = ((s +1) e ^2s) / 2-e ^2s / 4 + C;

• kufanele abale ∫lnsds

∫lnsds = {z = LNS, Dz = ds / s, y = s, dy = ds} = slns - ∫s x ds / s = slns - ∫ds = slns -S + C = s (LNS-1) + C.

Esikhundleni variable

Lomgomo yokuxazulula integrals nanini akuzona esingaphansi ziphoqa kuka owedlule ezimbili, nakuba kuyinkimbinkimbi. Indlela simi ngale ndlela: Ake V (x) - lo ebalulekile ezinye v umsebenzi (x). Esimweni lapho ngokwako ebalulekile Isibonelo slozhnosochinenny iza, kungenzeka athole bedidekile futhi ukwehla okungalungile endleleni izixazululo. Ukuze ugweme lokhu umkhuba ushintsho kusukela x variable ukuba z, lapho inkulumo jikelele zibukeke lula ebe egcina z kuye x.

Ngamagama zezibalo, lokhu kanje: ∫v (x) dx = ∫v (y (z)) y '(z) Dz = V (z) = V (y -1 (x)), lapho sithi x = y ( z) - esikhundleni. Futhi-ke, le ephambene umsebenzi z = y ^-1 (x) uchaza ngokugcwele ebuhlotsheni nokuhlobana eziguquguqukayo. Isaziso esibalulekile - the dx umehluko ngempela esikhundleni umehluko Dz entsha, njengoba ushintsho okuguquguqukayo ebalulekile nanini kuhilela esikhundleni ke yonke indawo, hhayi nje emazweni integrand.

Ngokwesibonelo:

• kumele uthole ∫ (ama + 1) / (s ² + 2s - 5) ds

Sebenzisa z esikhundleni = (s +1) / (s ² + 2s-5). Khona-ke Dz = 2sds = 2 + 2 (ama +1) ds <=> (s +1) ds = Dz / 2. Ngenxa yalokho, le nkulumo elandelayo, okuyinto lula kakhulu ukubala:

∫ (s +1) / (s ² + 2s-5) ds = ∫ (Dz / 2) / z = 1 / 2ln | z | + C = 1 / 2ln | s ² + 2s-5 | + C;

• kumelwe ukuthola ebalulekile ∫2 ^s e ^s dx

Ukuze kuxazululwe le loba ngesimo esilandelayo:

^{∫2 s e s ds = ∫ (} 2e) s ds.

Thina selisho ngumuntu = 2e (esikhundleni agumenti lesi sinyathelo akuyona, namanje s), sikunikeza zethu ezibonakala kuyinkimbinkimbi ebalulekile tabular eziyisisekelo ifomu:

^{∫ (2e) s ds = ∫a} s ds = a s / lna + C = (2e) s / ln (2e) + C = 2 ^s e ^s / ln (2 + lne) + C = 2 ^s e ^s / (ln2 +1) + C.

Efingqa uphawu umehluko

Ngokwengxenye enkulu, lokhu indlela integrals nanini - umfowabo ayiwele naye isimiso ushintsho variable, kodwa kunomehluko inqubo yokubhalisa. Ake sihlole kabanzi.

Uma ∫v (x) dx = V (x) + C no y = z (x), khona-ke ∫v (y) dy = V (y) + C.

Ngesikhathi esifanayo kumelwe singakhohlwa kugucuka eziwubala ebalulekile, phakathi lapho:

• dx = d (x + a), futhi lapho - ngamunye njalo;
• dx = (1 / a) d (imbazo + isibekiwe b), lapho - njalo futhi, kodwa hhayi zero;
• xdx = 1 / 2d (x ² + b);
• sinxdx = -D (cosx);
• cosxdx = d (sinx).

Uma sicabangela icala jikelele lapho esibala ngayo ebalulekile nanini, izibonelo kungenziwa Akubuyisiwe ngaphansi ifomula jikelele w '(x) dx = DW (x).

izibonelo:

• kumele uthole ∫ (2s + 3) ² ds, ds = 1 / 2d (2s + 3)

∫ (2s + 3) ² ds = 1 / 2∫ (2s + 3) ² d (2s + 3) = (1/2) x ((2s + ^{3) 2)} / 3 + C = (1/6) x (2s + 3) ² + C;

∫tgsds = ∫sins / cossds = ∫d (coss) / coss = -ln | coss | + C.

Online usizo

Kwezinye izimo, iphutha ongaba noma ukuvilapha, noma isidingo esiphuthumayo, ungasebenzisa kukuphi ku-intanethi, noma kunalokho, ukusebenzisa wokubala integrals nanini. Naphezu eyinkimbinkimbi nobumbi imvelo impikiswano we integrals, isinqumo kukhonjelwe algorithm zabo ethize, ogxile isimiso 'uma ungaphenduki ... bese ... ".

Yiqiniso, izibonelo ikakhulukazi oyinkimbinkimbi wokubala ezinjalo ngeke balukhulume, njengoba kukhona izimo lapho isinqumo has ukuthola eyayenziwe "ophoqelelwe" ngokwethula izakhi ezithile kule nqubo, ngoba imiphumela yezindlela ezisobala ukufinyelela. Naphezu isimo impikiswano kwalesi sitatimende, kuyiqiniso, njengoba izibalo, isimiso, isayensi lokuhlaziya, futhi inhloso yayo eyinhloko ibheka isidingo sokubeka imingcele. Ngempela, ukuba bushelelezi run-in the eyeza nezazi zezinkanyezi ezikhuluma kunzima kakhulu ukuya phezulu futhi ziphenduke, ngakho ungacabangi ukuthi izibonelo ukuxazulula integrals nanini, owasinika - lena ukuphakama amathuba. Kodwa emuva ohlangothini lwezobuchwepheshe izinto. Okungenani ukuhlola izibalo, ungasebenzisa isevisi okuyizilimi elalibhalwe ngazo kithi. Uma kukhona isidingo ukucabanga okuzenzakalelayo wokuboniswa eziyinkimbinkimbi ke abanayo iphendukele isofthiwe eyingozi. Kufanele sinake kakhulu imvelo MatLab.

isicelo

Isinqumo integrals nanini unganakile kubonakala akanandaba ngokuphelele ngokoqobo, ngoba kunzima ukubona ukusetshenziswa esobala indiza. Ngempela, ngqo uwasebenzise kuphi awukwazi, bebe bengewona kodwa kudingekile isici Lesisemkhatsini inqubo kokuhoxiswa kokuhlelwa izixazululo elisetshenziswa umkhuba. Ngakho, ukuhlanganiswa yokuzihlukanisa kwamangqamuzana ngokwemisebenzi yawo emuva, ngaleyo ndlela ngenkuthalo iqhaza inqubo ukuxazulula zibalo.
Lezi zibalo nethonya eliqondile ku isinqumo zezinkinga mechanical, trajectory kubala conductivity ezishisayo - ngamafuphi, konke akha khona futhi ekwakheni ikusasa. Nanini ebalulekile, izibonelo esesiye sacabangela ngenhla, ezingatheni Uma unganakile, njengoba base lokufeza ezingasho okutholakale muva.