I yesiko elingenammiselo. Kubalwa integrals elingenammiselo

Enye amacandelo angundoqo uhlalutyo zezibalo ke calculus ezifunekayo. Igubungela intsimi olubanzi kakhulu izinto, apho kuqala - oko ke elingenammiselo ibalulekileyo. Indawo limi njengoko isitshixo nangoku kwisikolo semfundo ephakamileyo kubonisa inani ngamathuba kunye namathuba, nto leyo ichaza nemathematika eziphezulu.

ukubonakala

Ekuqaleni, kubonakala phantsi njengendibaniselwano mihla, mlonyeni, kodwa ngokwesiqhelo kungcono ninikane ukuba ke ngowe-1800 BC. Ekhaya ukuze ingqalelo ngokusesikweni eYiputa njengoko asifikeleli ukuba ubungqina ngaphambili bobukho bayo. It ngenxa yokungabikho kwengcaciso, lonke elo xesha ibekwe nje nje senzeko. Ephinde engqina inqanaba yophuhliso abantu baloo maxesha zenzululwazi. Ekugqibeleni, imisebenzi afunyanwa le zezibalo yamandulo yamaGrike, niyathandana ukususela BC ngenkulungwane yesi-4. Bachaza indlela esetyenziswayo apho elingenammiselo elipheleleyo, eli leyo ukufumana umthamo okanye indawo yemilo wegophe (moya onamacala-abathathu-mbini, ngokulandelelanayo). ukubala yayisekelwe kumgaqo kwisahlulo mzobo original zibe kumacandelo omncane, ukuba umthamo (indawo) sele eyaziwa kubo. Ekuhambeni kwexesha, indlela liye lakhula, Archimedes walisebenzisa ukufumana kummandla parabola. izibalo ezifanayo ngaxeshanye ukuba baqhube imisebenzi e China yamandulo, apho yedwa ngokupheleleyo leyona amanye Greek.

uphuhliso

Sesona sisombululo elandelayo kwinkulungwane XI BC uye waba umsebenzi umphengululi Arabhu "wagon" Abu Ali al-Basri, ngubani nizihlabe imida sele eyaziwa, zathathwa ifomula apheleleyo yokubala izibalo ze-mali kunye izidanga ukusuka ekuqaleni ukuya yesine, isicelo oku kuthi indlela kuFacebook.
Ezingqondweni namhlanje bayinqule ngamaYiputa amandulo wadala ezikhumbuzo emangalisayo ngaphandle naziphi na izixhobo ezizodwa, ngaphandle koko wezandla zabo, kodwa akukho amandla izazinzulu mad ixesha akukho ngaphantsi ummangaliso? Xa kuthelekiswa la maxesha angoku ebomini babo babonakala phantse likumgangatho ophantsi, kodwa isigqibo integrals elingenammiselo kwakuthe kwafunyanwa kuyo yonke indawo, kwaye esetyenziswa practice malunga nophuculo.

Inyathelo elilandelayo senzeka kwinkulungwane XVI, xa sezibalo Italian Cavalieri wazisa indlela ayahlukahlukenanga, leyo iphucuka Per Funa. Ezi ubuntu ezimbini wabeka isiseko calculus mihla ibalulekileyo, nto leyo eyaziwa ngalo mzuzu. Bayifaka ingqiqo iyantlukwano ukuhlanganiswa, leyo ebisakuthatyathwa neeyunithi liqulathwe-ngokwalo. Ubukhulu becala, imathematika yelo xesha amasuntswana ngokwahluka iziphumo ezikhoyo ngokwazo, kunye ukusebenzisa. Indlela liwenze ukufumana into enivumelana yaba kuphela okwangoku, ibulela kuye, loo mihla uhlalutyo zemathematika waba ithuba ukuze akhule kwaye aphuhle.

Ngokuhamba kwexesha itshintsha yonke kunye uphawu ezifunekayo ngokunjalo. Ubukhulu becala, ukuba yakhethwa iinzululwazi ngendlela yakhe, umzekelo, Newton wasebenzisa icon isikwere, awawabeka khona umsebenzi integrable, okanye nje ubeke kunye. Le kungalingani waze ngenkulungwane XVII, xa yimbalasane yonke imfundiso ukuxhomekeka kolungiso Isazinzulu Gotfrid Leybnits laqalisa umlinganiswa enjalo eliqhelekileyo kuthi. Side "S" Kwandile isekelwe kule ncwadi oonobumba yamaRoma, kuba ibonisa ukuba sum of primitives. Igama elipheleleyo zifumaneka Jakob Bernoulli, emva kweminyaka eli-15.

Inkcazelo esesikweni

I yesiko elingenammiselo ixhomekeke kwinkcazelo kudidi, ngoko siqwalasela kwindawo yokuqala.

Antiderivative - lo ngumsebenzi uguqulo lonikezelo lwe ithe, xa kusenziwa oko kuthiwa kwakudala. Kungenjalo: umsebenzi ukuzazi of d - ngumsebenzi D, nto leyo V v <=> esukela '= v. Ukufuna ukuzazi ukuba abale efunekayo elingenammiselo, yaye inkqubo ngokwalo ibizwa ngokuba nomanyano.

umzekelo:

Umsebenzi s (y) = y ^3, kwaye S yayo abangekaphuhli (y) = (y ^4/4).

Isethi zonke primitives lo msebenzi - oku kuyinto ebaluleke elingenammiselo, zibonakaliswe oko ngolu hlobo lulandelayo: ∫v (x) DX.

ngenxa yokuba V (x) Ngu - kuphela ezinye nobu- umsebenzi yokuqala, ibinzana ibamba: ∫v (x) DX = V (x) + C, apho C - rhoqo. Phantsi rhoqo engenasizathu ubhekiselela kuye nabani rhoqo, ekubeni ithe yayo ngu zero.

izakhiwo

Mihlaba ubephethwe elipheleleyo elingenammiselo, Luyimfuneko olusekelwe kwinkcazelo neempawu nezakhi.
Qwalasela iingongoma eziphambili:

• livela efunekayo kudidi yi abangekaphuhli ngokwayo kunye engenasizathu rhoqo C <=> ∫V '(x) DX = V (x) + C;
• yemvelaphi ye nengxam komsebenzi ngumsebenzi original <=> (∫v (x) DX) '= v (x);
• rhoqo kuthathwa phantsi kophawu efunekayo <=> ∫kv (x) DX = k∫v (x) DX, apho k - is aliqonde;
• yesiko, nto leyo ithathwa kwisixa se alinganayo ukukhetha nokubeka isixa integrals <=> ∫ (v (y) + w (y)) dy = ∫v (y) dy + ∫w (y) dy.

Iimpawu ezimbini zokugqibela kuqukunjelwe ukuba elipheleleyo elingenammiselo lwento. Ngenxa oku, kufuneka: ∫ (kV (y) dy + ∫ Lw (y)) dy = k∫v (y) dy + l∫w (y) dy.

Ukuze ubone imizekelo ukulungisa izisombululo integrals elingenammiselo.

Kufuneka ufumane ∫ efunekayo (3sinx + 4cosx) DX:

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

Ukususela mzekelo sinokugqiba ukuba abazi indlela yokucombulula integrals elingenammiselo? kufumana nje zonke primitives! Kodwa ke ukufuna imigaqo kuxoxwa ngezantsi.

Iindlela kunye Imizekelo

Ukuze zokusombulula elipheleleyo, ungenza kubhenela ndlela zilandelayo:

• ukulungele ukuthatha ithuba etafileni;
• ukudibanisa ziindawo;
• ihlanganiswe ngokususa variable;
• UPawulos phantsi umqondiso umahluko.

iitafile

Eyona ndlela ilula kwaye kube mnandi. Okwangoku, uhlalutyo zezibalo liyi iitafile kakhulu ezinde, wayenza yacaca formula ezingundoqo integrals elingenammiselo. Ngamanye amazwi, kukho itemplates eyenziwe phezulu kuwe kwaye ungakwazi basebenzisa kuphela kubo. Nalu uluhlu lwe zithuba itafile engundoqo, leyo ukuboniswa phantse zonke iimeko, has isisombululo:

• ∫0dy = C, apho C - njalo;
• ∫dy = y + C, apho C - njalo;
• ∫y ⁿ dy = (y ^{n + 1)} / (n + 1) + C, apho C - a rhoqo, kwaye n - inani yahlukile umanyano;
• ∫ (1 / y) dy = ln | y | + C, apho C - njalo;
• ^{∫e y dy = e y + C} , apho C - njalo;
• ^{∫k y dy = (k y /} ln k) + C, apho C - rhoqo;
• ∫cosydy = siny + C, apho C - njalo;
• ∫sinydy = -cosy + C, apho C - njalo;
• ∫dy / cos ² y = tgy + C, apho C - rhoqo;
• ∫dy / isono ² y = -ctgy + C, apho C - rhoqo;
• ∫dy / (1 + y ²⁾ = arctgy + C, apho C - njalo;
• ∫chydy = + C oneentloni, apho C - rhoqo;
• ∫shydy = chy + C, apho C - rhoqo.

Ukuba kuyimfuneko, enze phambi kwam amanyathelo kukhokelela integrand kwimbono tabular aze onwabele uloyiso. UMZEKELO: ∫cos (5x -2) DX = 1 / 5∫cos (5x - 2) d (5x - 2) = 1/5 x isono (5x - 2) + C.

Ngokutsho isigqibo kucacile ukuba umzekelo integrand itafile uswele lingabali ezinye 5. Siya yidibanise ngaxeshanye kunye nale phinda ngo 1/5 ukuya ibinzana ngokubanzi ayizange iguquke.

Integration ngu Iinxalenye

Cinga imisebenzi emibini - z (y) kunye x (y). Bamele differentiable ngokuqhubayo yaso. Xa iimpawu omnye umahluko esinayo: d (xz) = xdz + zdx. Ukuhlanganiswa omabini, sifumana: ∫d (xz) = ∫ (xdz + zdx) => zx = ∫zdx + ∫xdz.

Yeremiya inxaki onesiphumo, sifumana ifomula, echaza indlela ukuhlanganiswa ziindawo: ∫zdx = zx - ∫xdz.

Kutheni kuyimfuneko? Into yokuba eminye yemizekelo kusenokwenzeka ukuba lula, masithi, ukunciphisa ∫xdz ∫zdx, ukuba umphathi lo kufutshane kwifom tabular kulo. Kwakhona, le fomyula ingasetyenziswa ngaphezu kanye, kuba iziphumo ezigqwesileyo.

Indlela yokusombulula integrals elingenammiselo ngale ndlela:

• kuyimfuneko ukubala ∫ (s + 1) e ^{zi 2} DS

∫ (x + ^{1) e 2 DS = {z =} s + 1, DZ = DS, y = 1 ^{/ 2e 2, Dy = e 2x} DS} = ((s + 1) e ²⁾ / 2-1 / 2 ∫e ^2s DX = ((s + 1) e ²⁾ / 2-e ^2/4 + C;

• mazibale ∫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.

Lithatha variable

Lo mgaqo yokusombulula integrals elingenammiselo azikho ngaphantsi zifunwa ngaphezu ezimbini ezidlulileyo, nangona ezintsonkothileyo. Le ndlela ilandelayo: Makhe V (x) - indlela edibeneyo ezinye umsebenzi v (x). Kwimeko ukuba ngokwayo ebalulekileyo Umzekelo slozhnosochinenny uyeza, kungenzeka ukuba bedidekile behle izisombululo endleleni engalunganga. Ukuze uphephe oku utshintsho nendlela x kwi z, apho ibinzana jikelele abangaboniyo lula lo gama kugcinwe lo z ngokuxhomekeke x.

Ngokwalo zemathematika, oku ngolu hlobo lulandelayo: ∫v (x) DX = ∫v (y (z)) y '(z) DZ = V (z) = V (y -1 (x)), apho ux = y ( z) - endaweni. Kwaye, kakade, isichasi umsebenzi z = y ^-1 (x) uchaza ngokupheleleyo ulwalamano kunye nobudlelwane nesebenzayo. Ingqalelo ebalulekileyo --DX umahluko endaweni mfuneko a DZ umahluko entsha, ekubeni utshintsho variable kwi elipheleleyo elingenammiselo kubandakanya ulibuyisele yonke indawo, hayi nje kuphela kweli integrand.

umzekelo:

• kufuneka afumane ∫ (s + 1) / (s + 2 ² - 5) DS

Faka endaweni Z = (s + 1) / (s ² + 2-5). Emva koko DZ = 2sds = 2 + 2 (s + 1) DS <=> (s + 1) DS = DZ / 2. Ngenxa yoko, ibinzana elithi ilandelayo, kulula kakhulu ukuba ukubala:

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

• kufuneka ufumane i yesiko ∫2 ^s e ^s DX

Ukusombulula i phinda ubhale ngale ndlela ilandelayo:

^{∫2 s e s DS = ∫ (} 2e) s DS.

Thina abonisa yi = 2e (replacement Kwempikiswano eli nyathelo akunjalo, nangoku s), sinikela zethu ezibonakala nzima njengendibaniselwano uhlobo esisiseko tabular:

^{∫ (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.

UPawulos uphawu umahluko

Ubukhulu becala, le ndlela integrals elingenammiselo - umzalwana amawele umgaqo wokutshintsha variable, kodwa kukho umahluko kwinkqubo yobhaliso. Makhe sihlolisise ngokweenkcukacha.

Ukuba ∫v (x) DX = V (x) + C no y = z (x), ngoko ∫v (y) dy = V (y) + C.

Ngelo xesha ke kufuneka ungalibali ukuba utshintshwano ezingenamsebenzi efunekayo, phakathi kwawo:

• DX = d (x + a), emasikwini - ngalinye rhoqo;
• DX = (1 / a) d (izembe + b), apho a - rhoqo kwakhona, kodwa hayi zero;
• xdx = 1 / 2D (x ² + b);
• sinxdx = -D (cosx);
• cosxdx = d (sinx).

Xa siqwalasela imeko jikelele apho abale elipheleleyo elingenammiselo, imizekelo nga subsumed phantsi indlela jikelele w '(x) DX = DW (x).

imizekelo:

• kufuneka afumane ∫ (2 + 3) ² DS, DS = 1 / 2D (2 + 3)

∫ (2 + 3) ² DS = 1 / 2∫ (2 + 3) ² d (2 + 3) = (1/2) x ((2 + 3) ²⁾ / 3 + C = (1/6) x (2 + 3) ² + C;

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

sinoncedo oluselayinini

Kwezinye iimeko, kona ongaba okanye ukonqena, nokuba kukho imfuno engxamisekileyo, ungasebenzisa kukuphi-intanethi, okanye mandithi, ukusebenzisa isixhobo sokubala integrals elingenammiselo. Nangona nobuntsompothi okubonakala kunye nohlobo impikiswano ke integrals, isigqibo kuxhomekeke algorithm zazo, nto leyo esekelwe kumgaqo "xa musa ... ngoko ...".

Kakade ke, imizekelo ingakumbi zintsonkothe ye ikhaltyhuleyitha akayi kakuhle, njengoko iimeko apho isigqibo kufuneka ukufumana ngezandla "ngenkani" ngokwazisa iinkalo ezithile kwinkqubo apho, ngenxa yokuba iziphumo iindlela ezicacileyo ukufikelela. Nangona impikiswano yale ngxelo, enyanisweni, njengoko imathematika, umgaqo, i inzululwazi abstract, kwaye injongo yayo ephambili eqwalasela imfuneko yokuxhobisa imida. Eneneni, ukuba egudileyo run-kwi ngcamango kunzima kakhulu ukuya phezulu yaye bathungelane, ngoko musa ukucinga ukuba imizekelo yokusombulula integrals elingenammiselo, owasinika - oku ukuphakama amathuba. Kodwa emva kwicala yobugcisa yezinto. Ubuncinane ukukhangela izibalo, ungasebenzisa i inkonzo apho kuyo kwathiwa kuthi. Ukuba kukho imfuneko ukubala oluzenzekelayo amabinzana ezintsonkothileyo, ngoko ke musa ukuba ukubhenela software ezimandundu. Kufuneka ahlawule ingqalelo ngokuyintloko bume MatLab.

isicelo

Isigqibo integrals elingenammiselo xa uqala kubonakala elufeleni ngokupheleleyo abakuso, ngokuba kunzima ukubona ukusetyenziswa ke lowo moya. Eneneni, sebenzisa ngqo yiku wena awukwazi, kodwa into efunekayo intermediate kwinkqubo zokurhoxa izisombululo asetyenziswe practice. Ngenxa yoko, ukuhlanganiswa umahluko umva, ngaloo ndlela ngokuthatha inxaxheba kule nkqubo zokusombulula zibalo.
Kwelinye icala, ezi zibalo kuba nefuthe ngqo kwi nesigqibo iingxaki mechanical, indlela yokubala kunye conductivity thermal - ngokufutshane, yonke into eyenza langoku ekubumbeni elizayo. imizekelo Nanini engundoqo, apho siye sithethe ngayo ngasentla, engenamsebenzi kuphela xa uqala, njengokuba isiseko ukwenza izinto ezintsha ngakumbi nangakumbi.