The indefinite muhiim. Xisaabinta of integrals indefinite

Mid ka mid ah qaybaha asaasiga ah ee falanqaynta xisaabeed ee waa kalkulas u muhiim. Waxaana ay ku aaddan beer aad u balaadhan oo ah shay, halkaas oo marka hore - waa indefinite u muhiim. Sida Wacan ee ay u taagan sidii muhiim ah in uu weli ku jiro dugsiga sare ayaa muujinaya tiro sii kordhaysa oo rajada iyo fursadaha, kaas oo kuu sharaxaya xisaabta sare.

muuqaalka

Jaleecada hore, waxa ay u muuqataa wada muhiim si casri ah, la mariyo, laakiin dhab ahaan ay muuqato in uu ku soo laabtay 1800 yimid BC. Home si rasmi ah loo tixgeliyaa sida Masar na ma gaadhi caddayn hore ee jiritaankeeda. Waxaa sabab u ah xog la'aan, halka oo dhan si fudud u taagan sida ugub ah. mar kale wuxuu Xaqiijiyay heerka horumarka sayniska dadkii waayahaas. Ugu dambeyntii, shuqullada la helay ee xisaabyahannada Giriigga ee hore, shukaansi ka BC qarnigii 4aad. Waxay ku tilmaami habka loo isticmaalo halkaas oo indefinite muhiim, nuxurka kuwaas oo ahaa in la helo mugga ama goobta qaab curvilinear (diyaaradeed saddex-geesood ah oo labada dhinac-cabbir ah, siday u kala horreeyaan). xisaabinta ahayd mid ku salaysan mabda'a qaybinta tirada asalka galay qaybaha infinitesimal, ay shardi tahay in mugga (goobta) waxaa hore u ogaan lahaa iyaga. Waqti ka, habka uu ku koray, Archimedes waxaa loo isticmaalaa si aad u ogaato meesha of parabola ah. xisaabaha la mid ah waqti isku mid ah si ay u qabtaan layliyada ee qadiimiga Shiinaha, halkaas oo ay si buuxda u madaxbanaanayn ka shaqayn jiray sayniska Giriigga.

horumarinta

horumar degdeg ah ee soo socda ee qarnigii line BC waxa uu noqday shuqulka aqoonyahan Carabta "baabuur" Abu Ali al-Basri, kuwaas oo u riixay xuduudaha ay horey u yaqaano, ayaa ka soo jeeda caanaha muhiim waayo xisaabayo wadaraha qadarka iyo darajo ka horeysay si kii afraad, codsanaya this ina ogeysiiyey habka induction.
Minds ee maanta, waa loo riyaaqay by Masriyiintii qadiimiga abuuray taxadiri la yaab leh oo aan wax qalab gaar ah, marka laga reebo in ay gacmahoodu sameeyeen, laakiin aan waa awood saynisyahano waalan waqtiga aan ka yarayn mucjiso ah? Marka la barbar dhigo jeer hadda ka mid ah noloshooda u muuqdaan kuwo ku dhow heer hoose ah, laakiin go'aanka integrals indefinite meel kasta ogaan iyo sida loo isticmaalo si dhab ah u tahay horumarka dheeraad ah.

Tallaabada xigta ka dhacay qarnigii XVI ka, marka xisaab Talyaaniga Cavalieri keenay hab qaybin karo, kaas oo ka soo gaaray Per Ferma. Labadan shakhsiyadda aasaaska u dhigeen, waayo, kalkulas muhiim ah oo casri ah, taas oo la og yahay la joogo. Waxay ku xidhan fikradaha kaladuwan yihiin iyo is-dhexgalka, kaas oo hore loo arkaa unugyada is-jira. By waaweyn iyo, xisaabta ee wakhtigaas ahaa qayb ka burbursan natiijooyinka jira ahaantood, iyadoo la isticmaalayo kooban. Way in ay midoobaan oo ay helaan dhul caadi ahaa runta ah oo keliya la joogo, isaga u mahad, casriga ah falanqaynta xisaabta lahaa fursad ay ku koraan iyo horumarinta.

Iyada oo dhererka waqtiga badala wax kasta oo calaamad u muhiim sidoo kale. By waaweyn iyo, waxaa loo qoondeeyey yahannada jidka isaga u gaar ah, tusaale ahaan, Newton isticmaalaa icon square ah, taas oo ku riday shaqo ah integrable, ama si fudud isugu. kala duwani Tani socday ilaa qarnigii XVII ah, marka durkin ah aragtida oo dhan cilmiga falanqaynta xisaabeed Gotfrid Leybnits soo bandhigay qof sida la yaqaan noo ah. Ifafaalaha "S" dhab ahaantii wuxuu ku salaysan yahay warqaddan ee alifbeetada Roman, tan iyo markii ay tusinaysaa wadarta primitives. Magaca muhiim u mahad helay inuu Jakob Bernoulli, 15 sano ka dib.

Qeexidda rasmi ah

muhiim The indefinite ku xiran tahay qeexidda heer hoose ah, si aan u fiirsan in meesha ugu horeysa.

Antiderivative - waa shaqo-bedelka ee derivative ah, ku dhaqanka waxa la yidhaahdaa waa heer hoose ah. Haddii kale, shaqo heer hoose ah d - waa shaqo D a, taas oo ah V v <=> derivative ah '= v. Search heer hoose waa in la xisaabiyo ah muhiim indefinite, iyo geeddi-socodka laftiisa waxaa loo yaqaan isdhexgalka.

tusaale ahaan:

Shaqada s (y) = y ^3, iyo S heer hoose (y) = (y ^4/4).

set oo dhan primitives shaqada - taasi waa lagama maarmaan aan cayinayn, waxay muujinaysaa sida soo socota: ∫v (x) DX.

By kugula xaqiiqda ah in V (x) - waa uun qaar ka mid ah shaqo asalka heer hoose ah, hadal haya: ∫v (x) DX = V (x) + C, halkaas oo C - joogto ah. Under joogto ah loo aabo yeelin waxaa loola jeedaa in kasta oo joogto ah, tan iyo markii ay derivative waa eber.

guryaha

The guryaha hantiyeen oo ay lagama maarmaan indefinite, muhiimad ahaan ku salaysan qeexidda iyo sifooyinka taagayo.
Tixgeli qodobbada muhiimka ah:

• derivative muhiim ah oo heer hoose ah waxa lagu daray ah loo aabo yeelin joogto C <=> ∫V heer hoose '(x) DX = V (x) + C,
• derivative of muhiim u ah shaqada waa shaqo asalka <=> (∫v (x) DX) = v (x),
• joogto ah waxaa laga soo ka soo hoos calaamada muhiim <=> ∫kv (x) DX = k∫v (x) DX, halkaas oo k - mid ku dhisan;
• lagama maarmaan, taas oo laga soo qaatay soo tirinnay mid ah identically si siman u soo tiriya reer reer integrals <=> ∫ (v (y) + w (y)) haysashada = ∫v (y) haysashada + ∫w (y) haysid.

The laba guri ee la soo dhaafay lagu soo gabagabeeyay karaa in lagama maarmaan ay indefinite waa toosan. Sababo la this, waxaan haysanaa: ∫ (KV (y) haysashada + ∫ lw (y)) haysashada = k∫v (y) haysashada + l∫w (y) haysid.

Si aad u aragto tusaale u ah qamaar xal integrals indefinite.

Waa in aad ka heli ∫ u muhiim (3sinx + 4cosx) DX:

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

From tusaale ahaan waxaannu ku tirinnaa karaa in aanad garanayn sida ay u xaliyaan integrals indefinite? Just ka heli oo dhan primitives ah! Laakiin raadinta mabaadii'da hoos ku wada hadleen.

Hababka iyo Tusaalooyinka

Si ay u xaliyaan muhiim ah, waxaad isticmaali karaan hababka soo socda:

• diyaar u ah inay ka faa'iidaystaan miiska;
• iskudhafka by qaybo;
• midaysan iyadoo la bedelayo variable ah;
• danbaysay ka yar calaamada kala duwan ah.

miisaska

The habka ugu fudud oo farxad leh. Waqtigan xaadirka ah, falanqaynta xisaabta ku faani karo loox arrin ballaaran, taasoo caddeyn formula aasaasiga ah ee integrals indefinite. In si kale loo dhigo, waxa jira arrimo kor u soo jeeda adiga iyo waxaad qaadan kartaa oo kaliya faa'iido iyaga ka mid ah. Halkan waxaa ku qoran liiska jagooyinka miiska ugu weyn, kaas oo la soo bandhigi karo shiidaa tusaale kasta, ayaa xal:

• ∫0dy = C, halkaas oo C - joogto ah;
• ∫dy = y + C, halkaas oo C - joogto ah;
• ∫y ⁿ = haysashada (y ^{n + 1)} / (n + 1) + C, halkaas oo C - joogto ah, iyo n - tiro ka duwan midnimada;
• ∫ (1 / y) haysashada = lihida | y | + C, halkaas oo C - joogto ah;
• ^{∫e y haysashada = e y + C} , halkaas oo C - joogto ah;
• ^{∫k y haysashada = (y k /} lihida k) + C, halkaas oo C - joogto ah;
• ∫cosydy = siny + C, halkaas oo C - joogto ah;
• ∫sinydy = -cosy + C, halkaas oo C - joogto ah;
• ∫dy / sababtoo ah ² y = tgy + C, halkaas oo C - joogto ah;
• ∫dy / dembi ² y = -ctgy + C, halkaas oo C - joogto ah;
• ∫dy / (1 + y ²⁾ = arctgy + C, halkaas oo C - joogto ah;
• ∫chydy = + C xishood, halkaas oo C - joogto ah;
• ∫shydy = chy + C, halkaas oo C - joogto ah.

Haddii loo baahdo, ka dhigi dhowr tallaabo horseedi integrand in view tabular ah oo wuxuu ku raaxaysan guusha. TUSAALE: ∫cos (5x -2) DX = 1 / 5∫cos (5x - 2) d (5x - 2) = 1/5 x dembiga (5x - 2) + C.

Sida laga soo xigtay go'aanka waxaa iska cad in tusaale ahaan integrand miis ka maqan multiplier 5. Waxaan ku dari ay barbar socdaan dhufasho this by 1/5 in ay ra'yi guud oo wax ka beddeli mayso.

Integration by Qaybo

Ka fikir laba hawlood - z (y) iyo x (y). Waa inay noqdaan si joogta ah differentiable ay domain. In mid ka mid ah guryaha ku kaladuwan yihiin waxaan leenahay: d (xz) = xdz + zdx. Is-dhexgalka labada dhinac, waxaan ka heli: ∫d (xz) = ∫ (xdz + zdx) => zx = ∫zdx + ∫xdz.

Muqato isla'egta keentay, waxaan ka heli formula ah, kaas oo kuu sharaxaya habka isdhexgalka by qaybood: ∫zdx = zx - ∫xdz.

Waa maxay sababta ay lagama maarmaan tahay? Xaqiiqada ah in qaar ka mid ah tusaalayaal ah waxaa suurtagal ah in loo fududeeyo, Aynu niraahno, si loo yareeyo ∫xdz ∫zdx, haddii dambe uu ku dhow yahay in qaab tabular ah. Sidoo kale, formula waxaa loo isticmaali karaa hal mar ka badan, natiijooyinka aan fiicnayn.

Sidee si ay u xalliyaan integrals indefinite sidan:

• loo baahan yahay si ay u xisaabiso ∫ (s + 1) ^2s e DS

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

• waa in loo xisaabiyo ∫lnsds

∫lnsds = {z = lns, dz = DS / s, = y s, haysashada = DS} = slns - ∫s x DS / s = slns - ∫ds = slns -S + C = s (lns-1) + C.

Bedelaadda variable ah

Tani waxay mabda'a of xalinta integrals indefinite waa aan ka yarayn in baahida badan labada hore, inkasta oo adag. Habka ugu waa sida soo socota: ha V (x) - maarmaan ah ee qaar ka mid ah v function (x). Haddii ay dhacdo in laftiisa in lagama maarmaan in Tusaale slozhnosochinenny yimaado, waxaa laga yaabaa in la isku buuqo oo waxaad tagtaa xalka jidka qaldan. Si looga fogaado is-beddelkaan dhaqanka ka x variable in z, kuwaas oo ra'yi guud aragga fududeeyay iyadoo la sii z ku xiran x.

In la eego xisaabta, waa sida soo socota: ∫v (x) DX = ∫v (y (z)) y '(z) dz = V (z) = V (y -1 (x)), halkaas oo x = y ( z) - bedelay. Iyo, dabcan, shaqo-bedelka z = y ku ^-1 (x) si buuxda u qeexaya xiriirka iyo xidhiidhka ka mid ah doorsoomayaasha. note Muhiim ah - DX kala duwan daruuri bedelay dz a kala duwan cusub, tan iyo isbedelka of variable in lagama maarmaan ay indefinite lug u bedelay meel kasta, ma ahan oo keliya in integrand ah.

tusaale ahaan:

• waa in la helo ∫ (s + 1) / ⁽² + 2s s - 5) DS

Codso z ka bedelay = (s + 1) / (s ² + 2s-5). Markaas dz = 2sds = 2 + 2 (s + 1) DS <=> (s + 1) DS = dz / 2. Sidaas darteed, hadal soo socda, taas oo ah mid aad u fudud si ay u xisaabiso:

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

• waa in aad hesho muhiim ∫2 ^s e ^s DX

Si loo xaliyo dib u qorid ee foomka soo socda:

^{∫2 s e s DS = ∫ (} DS 2e) s.

Waxaan yeero by a = 2e (bedelka doodda tallaabo tani ma aha, waxaa weli s), waxaan ku siin noo muuqda adag lagama maarmaan in qaab tabular oo aasaasi ah:

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

Gebogebadii calaamad kala duwan ah

By waaweyn iyo, habkan of integrals indefinite - walaalkiis oo mataano ah mabda'a isbedelka of variable, laakiin waxaa jira kala duwanaan ee geedi socodka diiwaangelinta. Ha ka fikiro si aynu si faahfaahsan.

Haddii ∫v (x) DX = V (x) + C iyo y = z (x), ka dibna ∫v (y) haysashada = V (y) + C.

Isla mar ahaantaana waa in aan aanay illoobin transformations muhiim Ciyaalle, kaas oo ka mid ah:

• DX = d (x + a), iyo wixii - kasta oo joogto ah;
• DX = (1 / a) d (faas + b), halkaas oo ah - si joogto ah mar kale, laakiin ma eber,
• xdx = 1 / 2d (x ² + b);
• sinxdx = -d (cosx);
• cosxdx = d (sinx).

Haddii aan ka fiirsan in kiiska guud halkaas oo aanu u xisaabiso ka maarmaan indefinite, tusaalooyin iyohababka karaa caanaha guud ee w '(x) DX = DW (x).

tusaale:

• waa in la helo ∫ (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.

caawimo Online

Xaaladaha qaarkood, gardarrada ah kaas oo noqon kara ama caajisnimo, ama baahi degdeg ah, waxaad isticmaali kartaa Dardar online, ama halkii, in ay isticmalaan xisaabiyaha integrals indefinite. Inkastoo kakanaanta muuqata iyo dabeecadda muranka of integrals ah, go'aanka ku xiran yahay ay isku geynta gaar ah, kaas oo ku salaysan mabda'a "haddii aadan ... ka dibna ...".

Dabcan, tusaale gaar sir ah ee calculator noocan oo kale ah ma yaqaaniin doonaa, maadaama ay jiraan xaalado in go'aanka uu si aad u hesho ah kiimikaysan "qasbay" by bandhigid xubno gaar ah in geeddi-socodka, sababtoo ah natiijada jira siyaabo iska cad in ay gaaraan. Inkastoo nooca muranka bayaankaan, waa run, sida xisaabta, in mabda, sayniska ah aan la taaban karin, oo ay ujeedadu hoose aragto baahida loo qabo in la xoojiyo xuduudaha. Indeed, waayo, a siman orod-in aragtiyaha waa mid aad u adag tahay kor u guurto iyo xuubsiiban, ee ha u qaadan in ay tusaale u ah xalinta integrals indefinite, oo na siiyey - taasi waa height of fursadaha. Laakiin back to dhinaca farsamada ee wax. Ugu yaraan si aad u hubiso xisaabinta, waxaad isticmaali kartaa adeegga taas oo la qoray noo. Haddii ay jirto baahi ah ee xisaabinta si toos ah tibaaxaha adag, ka dibna iyaga oo aan haysan si ay door biday in ay software ah ka sii daran. Waa in fiiro ugu horrayn on deegaanka MatLab.

codsiga

Go'aanka integrals indefinite at jaleecada hore u muuqataa in gebi ahaanba go'ay xaqiiqada, sababtoo ah waa adag tahay in la arko isticmaalka cad ee diyaaradda. Indeed, si toos ah u isticmaali meel aadan awoodin, laakiin ay yihiin arrin muhiim dhexe ee geedi socodka of bixitaanka of xal loo isticmaalo si dhab ah. Sidaas darteed, is-dhexgalka ee kaladuwan dib, sidaas si firfircoon uga qayb-qaataan hawsha ee isla'egta.
Taa baddalkeeda, isla'egyada waxay leeyihiin saameyn toos ah ku go'aanka dhibaatooyin farsamo, xisaabinta dhabbihii iyo conductivity kaamerada - in gaaban, wax kasta oo ka dhigan xaadirka ah iyo qaabeynta mustaqbalka. tusaalooyin oggalaansho muhiim, ee wixii aannu ku kor loo arkaa, kaliya Ciyaalle at jaleecada hore, sida saldhig u fuliyaan ikhtiraac cusub iyo aad u badan.