#VLJUBCOIFSIBLLTBLMESWF\":%*/:\":*/-\"3*OBBJUUJSTBZMZBTBOOIÐLÐNMFSJOF HËSFLJUBCOEÐ[FOJ NFUOJ TPSVWFõFLJMMFSJLTNFOEFPMTBIJ¿CJSõFLJMEFBMOQZBZNMBOB- NB[ GPUPLPQJZBEBCBõLBCJSUFLOJLMF¿PóBMUMBNB[ :BZO4PSVNMVTV $BO5&,÷/&- %J[HJ–(SBGJL5BTBSN *4#//P \"ZEO:BZOMBS%J[HJ#JSJNJ :BZOD4FSUJGJLB/P #BTN:FSJ ÷MFUJöJN &SUFN#BTN:BZO-UEõUJr \":%*/:\":*/-\"3* JOGP!BZEJOZBZJOMBSJDPNUS 5FMr 'BLT 0533 051 86 17 aydinyayinlari aydinyayinlari * ÜNwİwVwE.ayRdinSyaİyTinlaEri.YcoEm.trHAZIRLIK ·/÷7&34÷5&:&)\";*3-*, %¸O¾P.DSDáñ MATEMATİK - 2 7. MODÜL Alt bölümlerin Karma Testler İNTEGRAL EDĜOñNODUñQñL©HULU KARMA TEST - 2 ÷OUFHSBM Modülün sonunda tüm alt bölümleri ³ Belirsiz İntegral t 2 1. y r y= 3x 3. 6 sin x x2 + y2 = 16 # dx x r x4 + x2 + 1 ³ İntegral Alma Kuralları t 8 – O 6 :VLBSEBLJõFLJMEFY2 + y2 =¿FNCFSJWF y = 3 x EPóSVTVWFSJMNJõUJS JOUFHSBMJOJO TPOVDV BöBôEBLJMFSEFO IBOHJTJOF ³ Değişken Değiştirme Yöntemi t 17 5BSBMCÌMHFOJOBMBOOWFSFOJOUFHSBMBöBôEBLJ- FöJUUJS MFSEFOIBOHJTJEJS ³ Belirli İntegral - I t 24 1 4r2 3r - 1 5r A) # a 3 x - 16 - x2 k dx A) B) C) 0 2 5 2 6 B) # a 3 x - 16 - x2 k dx 0 D) 1 E) 0 1 ³ Belirli İntegral - II t 32 C) # a 16 - x2 - 3 x k dx 0 ³ Riemann Toplamı t 40 6ñQñIð©LðĜOH\\LĜ 2 4. #JSIBSFLFUMJJMLI[OEBOCBõMBZQEPóSVTBMCJSõF- L©HUHQNDUPDWHVWOHU ³ Belirli İntegral ile Alan Hesabı t 44 D) # a 16 - x2 - 3 x k dx LJMEFI[MBOBSBLTOEFNTI[BVMBõNõEBIB \\HUDOñU www.aydinyayinlari.com.tr 0 TPOSBI[OEPóSVTBMCJSõFLJMEFB[BMUBSBLTOEF ·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- π NTI[BEÐõÐSNÐõEBIBTPOSBTBCJUI[MBTO 3 HJUNJõWFEBIBTPOSBEPóSVTBMCJSõFLJMEFZBWBõMB- ³ İki Fonksiyonun Grafiği Arasında Kalan #&-÷34÷;÷/5&(3\"- E) # a 16 - x2 - 3 x k dx ZBSBLTOEFEVSNVõUVS 0 TANIM ÖRNEK 2 %XE¸O¾PGHNL¸UQHN #VIBSFLFUMJOJO[0, 15]TBOJZFMFSBSBTOEBBME- Bölgenin Alanı t 48F ( x ) fonksiyonunun türevi f ( x ) olsun. f ( x ) # (f (x) + x2 - 4) dx = x.f (x) VRUXODUñQ©¸]¾POHULQH ÷OUFHSBM ôUPQMBNZPMLBÀNFUSFEJS DNñOOñWDKWDX\\JXODPDVñQGDQ GPOLTJZPOVOVO UÐSFWJ BMONBEBO ËODFLJ Iº- PMEVôVOBHÌSF f ( x ) fonksiyonunun x =OPLUBTO- XODĜDELOLUVLQL] ³ Doğrusal Hareket Pli roolanbFle(xm) folenkrsiiytonun5a 4f(x) fonksiyonunun EBLJUFôFUJOJOFôJNJLBÀUS \" # $ % & ters türevi EFOJS#JSGPOLTJZPOVOUFSTUÐSFWJOJ <HQL1HVLO6RUXODU<(1m1(6m/6258/$5 CVMNBJõMFNJOFJOUFHSBMBMNBJöMFNJEFOJS ³ Karma Testler t 55 ' Y GPOLTJZPOVOVO UÐSFWJ G Y PMNBL Ð[FSF F( x ) fonksiyonuna f ( x ) fonksiyonunun integra- 2. y 1. ôFLJMEFLJUBSBMCËM- ôFLJM5. E:FB UFBõZLFEOÑB[SMFNEFI3B.SFL#FJUSFNEÐFIOFCOEJSJTDJCTJNSZJOÐ[JNWNFFI-BWV[VUBTBSMBZQCVOMBS ³ Yeni Nesil SoMJErFuOJlSa r t 63 ±SOFóJO UÐSFWJGh Y =PMBOG Y GPOLTJZPOV HFOJOBMBO пHFOCJ¿J[NBJNOEBFOLJECFJSOLMFNJB U F=(xU)-=-YWF+U=YWFBHO OY- = 4x -YGPOLTJZPO- 5x + Y- Y õFLMJOEFCJSGPOLTJZPO- k ln f b p JMF CVMB- IBWV[VO ECBJSIL[FOBSNT BMEôMZBPSMOOBNSBTPOMEEVBôLJVTOBOSHMÌCSËFM H FEFPMBDBLõFLJMEFLP- a EVS#VEVSVNEBG Y = 5x +D DTBCJU PMBSBL y= k NFUSFECJSV D#JVTNIJBO- ZPM – [BNBPOSEEJOFBOULTMFJTNUFJN BJöOBEôFEBCLJJSMJNFSJ-NFUSFPMBDBLõFLJMEF x CJMEJóJOFHËSF, JGBEFFEJMFCJMJS WV[VO UBCEBFOOOIOBOUHBJ-TJEJS NPEFMMJZPS 0RG¾O¾QJHQHOLQGH\\RUXP NFUSF \\DSPDDQDOL]HWPHYE O ab x \"ôCFÌLJMMH FTJOJOBMB- NBNOO GBZBOTMBSMB EHFHULOHUL¸O©HQNXUJXOX OLBÀCJSJNLBSF- VRUXODUD\\HUYHULOPLĜWLU #JS G Y GPOLTJZPOV1OVO CFMJSTJ[ JOUFHSBMJ ÖRNEK 3 EJS LBQMBNBTA )S(t) =MJSBt3 + t2 – 8t F(x) $\\UñFDPRG¾OVRQXQGD # f (x) dx CJ¿JNJOEF JGBEF FEJMJS #V JOUFHSBMJO # x.f (x) dx = x4 - x3 - 2x2 CVMVONBT J¿JO CJS 'h Y = G Y PMBDBL õFLJMEF y $ MO F UVUNBLUBES 2 CJS' Y GPOLTJZPOVBSBõUSMSWFJOUFHSBMTBCJ- PMEVôVOBHÌSF f ( x ) fonksiyonunun yerel minimum O & MO F UJPMBODCV' Y GPOLTJZPOVOBFLMFOJS#VEV- OPLUBTOOBQTJTJLBÀUS y = 2x #OV 3JIFBNWBVO[BOV)OSUP(QtU)BMBC=NB-t23 – t2 + 8t + 20 :Ð[NF SVNEB ôFLJM LNJBHOJCUJóHFZMOBJCõõM)JFóLSJJM(t)NE=FFUt--3 - t2 + 8t + 3 IBWV[V # f (x) = F (x) + c olur. y= 2 x g(x) A SF PMBO BMU BSBMLMBSB x BZSMBSBL PDM)VõSUV(tS)V=MBOt3 + t2 - 8t + 3 2 #VSBEB G Y FJOUFHSBOU JOUFHSBMJBMOBOGPOL- $OW%¸O¾P7HVWOHUL \" MO # MO EJLEËSUHFOMFS ZBSE- TJZPO EYFJOUFHSBMEFóJõLFOJ DZFJTFinteg- ral sabitiEFOJS % MO F NZMBLBQMEBO) BSDB(tL)U=St4 – 3t 2 4t :Ñ[NFIBWV[VOVOEFSJOMJôJ NFUSFPMBDBôOB #JS'POLTJZPOVO(SBGJôJJMFY<FOJ\"SBTOEB,BMBO#ÌMHFOJO\"MBO - +H2Ì0SF UBTBSMBEôZÑ[NFIBWV[VLBÀN3TVBMS TEST - 21 \" # $ % & 3. E 4. C 5. B 1. D 2. D#VOB HÌSF öFLJM EFLJ HJC5J6 CJS LBQMBNB JÀJO 1. ôFLJMEF41 42 43CVMVOEVLMBSCËMHFMFSJOBMBOMBS- 4. ôFLJMEF G Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõUJS ÌEFOFDFLUVUBSLBÀMJSBPMNBMES ES y \" # $ % & y ÖRNEK 4 ÖRNEK 1 S3 # f (x) dx = - x2 - 4x - 2 S1 y = f(x) Her alt bölümün # f (x) dx = x4 - 3x2 + 5x + 6 –1 S1 xx 2 x VRQXQGDRE¸O¾POHLOJLOL 4 WHVWOHU\\HUDOñU S2 7 10PMEVôVOB HÌSF f ( x ) fonksiyonunu–3n yerel maks2i- S2 4 PMEVôVOBHÌSF f ( x ) fonksiyonunu bulunuz. Nf(Vx)NEFôFSJLBÀUS WDPDPñ\\HQLQHVLOVRUXODUGDQ 41 =CS2 42 =CS2WF 41 WF 42 J¿JOEF CVMVOEVóV CËMHFMFSJO BMBOMBSO 2. 4 m ôFLJMEFLJ QBSL ¿JNMFOEJSJ- 4. ôFLJMEF LPPSEJOBU TJTUFNJOEF NPEFMMFONJõ BMU WF ROXĜDQWHVWOHUEXOXQXU HËTUFSN FLÐ[FSF ÐTU TOSMBS QBSBCPM õFLMJOEF PMBO CJS UÐOFM HËSÐM- 10 MFDFLUJS)FSCJSLTNBSB- NFLUFEJS,PPSEJOBUTJTUFNJOEFCJSJN NFUSF # f(x) dx = 16 4 8 m TOEBLJV[BLMLMBSCJSCJSJOF LBCVMFEJMJQËM¿FLMFOEJSJMNJõUJS –1 # f (x) dx = 6 ve S1 = 9 br2 -3 y PMEVô VOBHÌSF S3BMBOLBÀbr2EJS PMEVôVOBHÌSF 42 kBÀCS2EJS 36 32 A ) 11 B ) 15 C ) 16 D ) 17 E ) 18 12 m FõJU WF 2 3 NFUSFEJS ¥JNMFOEJSNF JõMFNJ 3JF- A ) 21 B ) 18 C ) 12 D ) 6 E ) 3 16 m NBOO ÐTU UPQMBN NBOU- 1. 4x23.– 6xô+FL5JMEFZ=G Y JOHSBGJóJWFS2JMNJõUJS 3 12 m óZMBZBQMSTB9NFUSFLB- y 3. SF 3JFNBOO BMU UPQMBN 2. –3 4. 4 x 8 46 8 m NBOUóZMB ZBQMSTB : –6 –4 O NFUSFLBSFLTN¿JNMFOEJ- y = f(x) A 12 C x SJMNJõPMVZPS 5ÑOFMJOZBOZÑ[FZMFSJOJONFUSFLBSFGJZBU5- –3 B 57 6 4m PMBO LBQMBNB NBM[FNFTJ JMF LBQMBONBT EVSV- y NVOEBLBQMBNBNBMJZFUJLBÀ5-PMVS 5. S2 5 x #VOBHÌSF 9-:BöBôEBLJMFSEFOIBOHJTJEJS YFLT FOJOJO \"#ZBZJMFTOSMBEóCËMH FOJOBMBO –6 S1 1 y = f(x) A) 8 3 # 12 3 $ 16 3 \" # $ CS2 #$ZBZJMFTOSMBE óCËMHFOJOBMBOCS2 % 18 3 & 24 3 % & 6 ôFLJMEF 41 =CS2 42 =CS2JTF 1. D 2. E 63 3. A 4. D PMEVôVOBHÌSF # f (x) dx LBÀUS 5 –3 # af^xh + f^xh kdx A ) 60 B ) 30 C ) 17 D ) 10 E ) 7 -6 JOUFHSBMJOJOEFôFSJLBÀUS A ) -12 B ) -6 C ) 6 D ) 12 E ) 18 3. y ôFLJMEF \"1, A2 CVMVOEVLMBS CËM- –3 –2 A1 O 4x HFOJO BMBOO HËTUFSNFLUFEJS A2 A1 =CS2, A2 =CS2WF 6. Z= x2 –2 4 # f(x) dx = - 5 ise # f_ x i dx QBSBCPMÑJMFZ= Y=WFY=EPôSVMBSBSB- TOEBLBMBOBMBOLBÀCS2EJS –3 –3 A ) 18 B ) 20 C ) 21 D ) 24 E ) 27 JOUFHSBMJOJOEFôFSJLBÀUS E ) 16 A ) -10 B ) -8 C ) -5 D ) 8 1. \" 2. & 3. \" 47 4. & 5. D 6. C
ÜNwİwVwE.ayRdinSyaİyTinlaEri.YcoEm.trHAZIRLIK ·/÷7&34÷5&:&)\";*3-*, MATEMATİK - 2 7. MODÜL İNTEGRAL ³ Belirsiz İntegral t 2 ³ İntegral Alma Kuralları t 8 ³ Değişken Değiştirme Yöntemi t 17 ³ Belirli İntegral - I t 24 ³ Belirli İntegral - II t 32 ³ Riemann Toplamı t 40 ³ Belirli İntegral ile Alan Hesabı t 44 ³ İki Fonksiyonun Grafiği Arasında Kalan Bölgenin Alanı t 48 ³ Doğrusal Hareket Problemleri t 54 ³ Karma Testler t 55 ³ Yeni Nesil Sorular t 63 1
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr #&-÷34÷;÷/5&(3\"- TANIM ÖRNEK 2 F ( x ) fonksiyonunun türevi f ( x ) olsun. f ( x ) # (f (x) + x2 - 4) dx = x.f (x) GPOLTJZPOVOVO UÐSFWJ BMONBEBO ËODFLJ Iº- li olan F ( x ) fonksiyonuna f ( x ) fonksiyonunun PMEVôVOBHÌSF f ( x ) fonksiyonunun x =OPLUBTO- ters türevi EFOJS#JSGPOLTJZPOVOUFSTUÐSFWJOJ EBLJUFôFUJOJOFôJNJLBÀUS CVMNBJõMFNJOFJOUFHSBMBMNBJöMFNJEFOJS fh( 1 ) bize soruluyor. ' Y GPOLTJZPOVOVO UÐSFWJ G Y PMNBL Ð[FSF ( x.f ( x ) )h = f ( x ) + x2 - 4 F( x ) fonksiyonuna f ( x ) fonksiyonunun integra- 1.f ( x ) + x.fh( x ) = f ( x ) + x2- 4 MJEFOJS x.fh( x ) = x2 - 4 j x = 1 koyarsak 1.fh( 1 ) = 1 - 4 = -3 j fh( 1 ) = -3 ±SOFóJO UÐSFWJGh Y =PMBOG Y GPOLTJZPOV 5x + Y- Y õFLMJOEFCJSGPOLTJZPO- ÖRNEK 3 EVS#VEVSVNEBG Y = 5x +D DTBCJU PMBSBL JGBEFFEJMFCJMJS # x.f (x) dx = x4 - x3 - 2x2 #JS G Y GPOLTJZPOVOVO CFMJSTJ[ JOUFHSBMJ PMEVôVOBHÌSF f ( x ) fonksiyonunun yerel minimum OPLUBTOOBQTJTJLBÀUS # f (x) dx CJ¿JNJOEF JGBEF FEJMJS #V JOUFHSBMJO ( x4 - x3 - 2x2 ) h = x.f ( x ) CVMVONBT J¿JO CJS 'h Y = G Y PMBDBL õFLJMEF 4x3 - 3x2 - 4x = x.f ( x ) j f ( x ) = 4x2 - 3x - 4 CJS' Y GPOLTJZPOVBSBõUSMSWFJOUFHSBMTBCJ- UJPMBODCV' Y GPOLTJZPOVOBFLMFOJS#VEV- 3 SVNEB fh( x ) = 8x - 3 = 0 j x = # f (x) = F (x) + c olur. 8 #VSBEB G Y FJOUFHSBOU JOUFHSBMJBMOBOGPOL- TJZPO EYFJOUFHSBMEFóJõLFOJ DZFJTFinteg- ral sabitiEFOJS ÖRNEK 1 ÖRNEK 4 # f (x) dx = x4 - 3x2 + 5x + 6 # f (x) dx = - x2 - 4x - 2 PMEVôVOBHÌSF f ( x ) fonksiyonunu bulunuz. x2 PMEVôVOB HÌSF f ( x ) fonksiyonunun yerel maksi- f ( x ) fonksiyonunun integrali x4 - 3x2 + 5x + 6 oldu- ôVOEBOY4- 3x2 + 5x + 6 ifadesinin türevi f ( x ) olur. NVNEFôFSJLBÀUS ( x4 - 3x2 + 5x + 6 )h =f ( x ) = 4x3 - 6x + 5 f- 2 l f(x) x - 4x - 2 p = x 2 -x - 4 = f(x) j -x2 - 4x = f ( x ) x fh( x ) = -2x - 4 = Y= -2 j f ( -2 ) = -4 + 8 = 4 1. 4x3 – 6x + 5 2 2. –3 3. 3 4. 4 8
www.aydinyayinlari.com.tr ÷/5&(3\"- 7. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 5 ÖRNEK 8 # ^ x + f (2x) h dx = 4x3 + x2 + 6 N= 4x + 5 PMEVôVOBHÌSF EYJGBEFTJOJOFöJUJOJCVMVOV[ 2 PMEVôVOBHÌSF G LBÀUS dm dm = 4dx j dx = 2l f 4x3 + x + 6 p = x + f (2x) 4 2 ÖRNEK 9 12x2 + x = x + f ( 2x ) f ( 4 ) = 12.22 = 48 f(x) = x2 + 2 x - 1 x %JGFSBOTJZFM,BWSBN PMEVôVOBHÌSF EG Y JGBEFTJOJOFöJUJOJCVMVOV[ TANIM df ( x) = d x2 + 2 x- 1 l Türevlenebilir bir f ( x ) fonksiyonu için x n dx d ^ f (x) h =Gh Y JGBEFTJG Y GPOLTJZPOVOVO dx df ( x) = f 2x + 2· 11 2 + p dx UÐSFWJPMNBLÐ[FSF d^ f (x) h = f' (x) dx JGBEFTJ- 2 ne f ( x ) fonksiyonunun diferansiyeli EFOJS x x 4POV¿PMBSBLCJSG Y GPOLTJZPOVOVOEJGFSBOTJ- ZFMJPMBOE G Y JGBEFTJ GPOLTJZPOVOUÐSFWJJMF = f 2x + 11 EYJO¿BSQNOBFõJUUJS + p dx 2 x x ÖRNEK 6 #FMJSTJ[÷OUFHSBMJO²[FMMJLMFSJ f ( x ) = x3 - x2 - 4x + 6 %m/*m fonksiyonunun diferansiyelini bulunuz. # f' (x) dx = f (x) + c df ( ( x ) ) = f' ( x ) dx = ( 3x2 - 2x - 4 ) dx # d^ f(x) h = f(x) + c ÖRNEK 7 # c.f(x) dx = c # f(x) dx , ^ c ! R h y = u4 - 2 d # f (x) dx = f (x) u dx PMEVôVOBHÌSF EZOJOFöJUJOJCVMVOV[ d # f (x) dx = f (x) dx # f d f (x) p dx = f (x) + c dx # ^ f(x) ± g(x) h dx = # f(x) dx ± # g(x) dx dy = d 4 - 2 l & dy = f 3 + 2 u u n du 4u 2 p du u 3 dm 9. f 2x + 11 8. + p dx 6. (3x2 – 2x – 4)dx 32 2 5. 48 4 x x 7. f 4u + 2 pdu u
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr ÖRNEK 10 ÖRNEK 14 d # (x2 + 5x - 1) dx # > d ^ x3 - 4x hdxH dx dx JGBEFTJOJOFöJUJOJCVMVOV[ JGBEFTJOJOFöJUJOJCVMVOV[ #d a x 2 + 5x - 1 k dx = 2 + 5x - 1 x dx #= d ^ 3 h Gdx 3 - 4x = - 4x + c dx x x ÖRNEK 11 ÖRNEK 15 # d^ x h # d f x2 + 4x p JGBEFTJOJOFöJUJOJCVMVOV[ x+2 JGBEFTJOJOFöJUJOJCVMVOV[ # d^ x h = x + c ÖRNEK 12 #df x2 + 4x p= 2 + 4x +c d # d ^x3 + x2h x dx x+2 x+2 JGBEFTJOJOFöJUJOJCVMVOV[ #d d^ 3 + 2 h = ( 3x2 + 2x ) x x dx ÖRNEK 13 ÖRNEK 16 d # (3x2 - 2x + 1) dx d # a d # ^ x2 - 4x h dx k JGBEFTJOJOFöJUJOJCVMVOV[ JGBEFTJOJOFöJUJOJCVMVOV[ # # #ddd^ 2 - 4x hdx n = d ^ x2 - 4x hdx x #d 2 - 2x + 1) dx = 2 - 2x + 1) dx = ^ 2 - 4x hdx (3x (3x x 10. x2 + 5x – 1 11. x + c 12. 3x2 + 2x 13. (3x2 – 2x + 1)dx 4 14. x3–4x+c 2 16. (x2 – 4x) dx x + 4x 15. + c x+2
#FMJSTJ[÷OUFHSBM TEST - 1 1. # f (x) dx = x2 - 3x + 1 5. # f (x) dx = 2x2 + 8x - 3 PMEVôVOBHÌSF G Y GPOLTJZPOVOVOFöJUJBöBô- x EBLJMFSEFOIBOHJTJEJS PMEVôVOBHÌSF G Y GPOLTJZPOVOVOZFSFMNJOJ- A) x3 - 3x2 + x + c # Y2 - 3x + 1 NVNEFôFSJLBÀUS 32 A) - # - $ % & C) ( 2x - EY % Y- 3 E) x3 - 3x2 6. # x.f(x) dx = 2x + 1 + c x+1 4 PMEVôVOBHÌSF f-1 Y BöBôEBLJMFSEFOIBOHJTJ- EJS 2. # x.f (x) dx = x3 - 3x2 + x + 1 A) 1 # 2 C) - 2 1 - 2x 2x + 1 2x + 2 PMEVôVOBHÌSF G LBÀUS A) - # -2 C) - % & % 1 E) 1 2x - 1 2x + 2 3. f (x) = # ^ x3 + x2 + 4 h dx 7. # x3.f' (x) dx = x5 - 4x4 + c PMEVôVOBHÌSF G Y GPOLTJZPOVOBY=BQTJTMJ 5 OPLUBTOEBOÀJ[JMFOUFôFUJOFôJNJLBÀUS PMEVôVOBHÌSF Gh - LBÀUS \" # $ % & A) - # -21 C) - % -18 E) -17 4. # x.f (x) dx = x3 - 6x2 8. G Y EPóSVTBMGPOLTJZPOPMNBLÐ[FSF PMEVôVOBHÌSF G +Gh LBÀUS g(x) = d # f(x) dx + # d f (x) \" # $ % & dx f ( 1 ) = f ( 2 ) = 5 ve g ( 3 ) = -4 PMEVôVOBHÌSF H-1 LBÀUS \" # $ % & 1. D 2. B 3. C 4. \" 5 5. \" 6. D 7. & 8. D
TEST - 2 #FMJSTJ[÷OUFHSBM 1. # f(x) - 1 x-3 + x +c 5. # x2.f (x) dx = x4 - 15x3 + c dx = x+2 ^ x - 1 h2 2 PMEVôVOBHÌSF f ( - LBÀUS A) - # -11 C) - % -45 E) -89 FöJUMJôJOJTBôMBZBOG Y GPOLTJZPOVJÀJO f ( 5 ) de- ôFSJLBÀUS A) - 25 # - 7 C) 45 % 38 9 E) 18 12 16 11 2 2. f(x) = # mx - 1 dx PMNBLÐ[FSF 6. f(x) = d # d 9 # (x3 - 4x) dxC x3 + 4x2 - x + 5 dx PMEVôVOBHÌSF Gh LBÀUS f ( x ) fonksiyonunuO HSBGJôJOJO Y = - BQTJTMJ OPLUBTOEBLJOPSNBMJZ= 3x -EPôSVTVPMEV- \" # $ % & ôVOBHÌSF NLBÀUS A) - 13 # - 7 C) 14 % 23 48 E) 2 10 9 5 7 3. :FSFMFLTUSFNVNOPLUBMBSOEBOCJSJOJOBQTJTJ 7. # x.f(x) dx = d # (2x3 + x)dx PMBOG Y GPOLTJZPOVJÀJO dx PMEVôVOBHÌSF G LBÀUS f (x) = # (4x2 - m) dx \" # $ % & PMEVôVOBHÌSF NLBÀUS \" # $ % & 4. f (x) = # d (3x2 + 6x) PMNBLÐ[FSF 8. d # d (x2 + 6x) f ( 1 ) = PMEVôVOB HÌSF G Y GPOLTJZPOVOVO JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS BMBCJMFDFôJFOLÑÀÑLEFôFSLBÀUS A) x2 +Y # Y+ 6 A) - # -6 C) - % -8 E) -9 C) ^ x2 + 6x hdx % Y+ EY E) x2 + 6x +D 1. & 2. C 3. D 4. C 6 5. & 6. B 7. C 8. D
%JGFSBOTJZFM,BWSBN TEST - 3 1. # ^ x + 2 h f (x) dx = x4 + 2x3 - 8x2 - 24x + 5 5. # d^ x3 + x h + d # ^ x3 + x hdx PMEVôVOBHÌSF G Y GPOLTJZPOVOBY= -BQTJT- dx MJOPLUBTOEBOÀJ[JMFOUFôFUJOFôJNJLBÀUS JöMFNJOJOTPOVDVBöBôEBLJMFSEFOIBOHJTJEJS A) - # -10 C) - % -6 E) -4 A) x4 + x3 + x2 + x + c 42 2. # ^ x + 1 h.f' (x) dx = x3 - 4x2 - 11x + 10 # 2x3 + 2x PMEVôVOBHÌSF G Y GPOLTJZPOVOVOFLTUSFNVN C) 2x3 + 2x +D OPLUBTOOBQTJTJLBÀUS % 6x + 2 E) 6x2 + 2 A) 8 # $ 10 % 11 E) 4 3 33 6. d # ^ x2 + 5x hdx = # f(x) dx dx PMEVôVOBHÌSF G LBÀUS \" # $ % & 3. # 2.f' (x) dx = 4f (x) - 6x2 + 10 PMEVóVOBHËSF G Y GPOLTJZPOVBöBôEBLJMFSEFO 7. # f d (x3 + x) pdx = # f'(x) dx IBOHJTJPMBCJMJS dx A) x2 -Y # Y2 + 4x C) 3x2 - 2x PMEVôVOBHÌSF f ( x ) fonksiyonunun artan oldu- ôVFOHFOJöBSBMLBöBôEBLJMFSEFOIBOHJTJEJS % Y2 + 4 E) 2x2 + 4 \" mÞ > # < Þ $ -Þ Þ % <- Þ & -Þ -4 ) 4. f ( x ) = x2 -YPMEVôVOBHÌSF 8. d # 9d # x2 dxC f' (x) JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS d ^ f(x) h JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS A) x3 + c C) x2 3 E) x2EY #) 2x \" # Y- $ EY % 1 E) x2 - 6x % Y2 +D dx 1. B 2. D 3. D 4. D 7 5. C 6. D 7. C 8. &
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr ÷/5&(3\"-\"-.\",63\"--\"3* ÷OUFHSBM\"MNB,VSBMMBS ÖRNEK 4 %m/*m D`3PMNBLÐ[FSF # x2. 3 x dx # dx = x + c JOUFHSBMJOJOFöJUJOJCVMVOV[ # a.dx = ax + c 2 2 1/3 7/3 Oá-1 ve n `2PMNBLÐ[FSF . x .x x dx # xn dx = xn + 1 + c # # #x 3 x dx = dx = n+1 a `3PMNBLÐ[FSF x 10/3 + c = 3 ·x10/3 + c = # a.xn dx = a # xn dx 10 10 # ^ f(x) ± g(x) h dx = # f(x) dx ± # g(x) dx 3 ÖRNEK 1 ÖRNEK 5 # 3 dx JOUFHSBMJOJOFöJUJOJCVMVOV[ # 4 dx # 3 dx = 3x + c 5 x2 JOUFHSBMJOJOFöJUJOJCVMVOV[ # #4 -2/5 4.x 3/5 dx = 4.x dx = 3 +c 52 x 5 = 20 3/5 + c 3 x ÖRNEK 2 ÖRNEK 6 # x2 dx # 3 x2 dx JOUFHSBMJOJOFöJUJOJCVMVOV[ 4 x3 3 JOUFHSBMJOJOFöJUJOJCVMVOV[ #2 x x dx = + c 3 ÖRNEK 3 32 # #x # ^ 4x3 + 2x - 6 h dx dx = 2/3 –3/4 JOUFHSBMJOJOFöJUJOJCVMVOV[ x .x dx 43 x 2/3–3/4 –1/12 x dx x dx # #= = 42 11/12 3 4x 2x x + c = 12 ·x11/12 + c #a + 2x - 6 kdx = + - 6x + c = 4x 42 11 11 = x4 + x2 - 6x + c 12 3 8 4. 3 x10/3 + c 5. 20 x3/5 + c 6. 12 11/12 + c 1. 3x + c x 3. x4 + x2 – 6x + c 10 3 11 x 2. + c 3
www.aydinyayinlari.com.tr ÷/5&(3\"- 7. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 7 ÖRNEK 10 # ^ 2x - 1 h ^ x + 3 h dx # x2 y dx + # x2 y dy JOUFHSBMJOJOFöJUJOJCVMVOV[ JOUFHSBMJOJOFöJUJOJCVMVOV[ # #^ 2x - 1 h ^ x + 3 h dx = a 2x2 + 5x - 3 kdx # #2 2 3 22 x xy +c x y dx + x y dy = y + 32 3 2 2x 5x = + - 3x + c 32 ÖRNEK 8 ÖRNEK 11 # 2x3 - 4x2 + 3x dx # m3 - m2 + 2 dm x m2 JOUFHSBMJOJOFöJUJOJCVMVOV[ integralinJOFöJUJOJCVMVOV[ # #2x3 - 4x2 + 3x a 2 - 4x + 3 kdx x dx = 2x = 2 x3 - 2x2 + 3x + c 32 3 # #m - m + 2 dm = ^ m - 1 + –2 h dm 2 2m m 2 –1 2 m 2m m 2 = -m+ +c= -m- +c 2 -1 2 m UYARI öOUFHSBMBMNBJõMFNJGBSLMEFóJõLFOMFSFHËSFEF uygulanabilir. ÖRNEK 9 ÖRNEK 12 # u2 du + # u2 dm # x - 4 dx JOUFHSBMJOJOFöJUJOJCVMVOV[ x+2 integralinJOFöJUJOJCVMVOV[ 3 # #x - 4 dx = ^ x - 2 h^ x + 2 h # #2 2 u 2 dx u du + u dm = + u .m + c 3 x+2 ^ x+2h 3/2 #= a x1/2 - 2 k dx = x - 2x + c 3 2 2 3/2 = x - 2x + c 3 2x3 5x2 2 u3 +u2.m+ c 9 3 22 2 + - 3x + c m2 2 3/2 7. 8. x3–2x2+3x + c 9. xy xy +c 11. - m - + c 12. x - 2x + c 10. + 32 3 3 32 2m 3
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr ÖRNEK 13 ÖRNEK 16 # ^ x6 - 1 h dx Gh Y = 3x2 + 6x + 5 ve f ( 1 ) = 3 PMEVôVOBHÌSF G LBÀUS ^x4 - x3 + x - 1h integralinJOFöJUJOJCVMVOV[ f (x ) = x3 + 3x2 + 5x + c f ( 1 ) = 1 + 3 + 5 + c = 3 j c = -6 # #^ x6 - 1 h ^ 3 - 1 h^ 3 + 1 h f ( 3) = 27 + 27 + 15 - 6 = 63 dx = x x dx ^ x 4 - 3 + x - 1 h 3 ^ x - 1 h + x - 1 x x # #= f ^ x3 - 1 h^ x3 + 1 h p dx = ^ x2 + x + 1 hdx ^ 3 h^ + 1 x - 1 h x 32 xx = + +x+c 32 ÖRNEK 17 Ghh Y = 6x2 +Y Gh = 0 ve f ( -1 ) = 1 PMEVôVOBHÌSF f ( x ) fonksiyonunu bulunuz. ÖRNEK 14 f' ( x ) = 2x3+ 3x2 + c f' ( 1 ) = 2 + 3 + c = D= -5 # 3 x - 6 dx + # 6+2 x f' ( x ) = 2x3 + 3x2 - 5 dx 43 xx 2x 3x f(x) = + - 5x + c 43 integralinJOFöJUJOJCVMVOV[ 1 7 f^ -1 h = - 1 + 5 + c = 1, c =- 22 # #3 x - 6 + 6 + 2 x 5x dx = 5x + c 4 dx = f^ x h = x + x3 - 5x - 7 22 xx ÖRNEK 15 ÖRNEK 18 Gh Y = 3x2 - 4x + 1 ve f ( 1 ) = 2 y =G Y FóSJTJOJOÐ[FSJOEFLJ5 OPLUBTOEBO¿J[JMFO PMEVôVOBHÌSF G Y GPOLTJZPOVOVCVMVOV[ UFóFUEPóSVTVOVOFóJNJES f (x ) = x3 - 2x2 + x + c olur. f'' ( x ) =PMEVôVOBHÌSF G LBÀUS f(1) = 1 - 2 + 1 + c = 2 c=2 f ( 3 ) = Gh = Ghh Y = 4 f ( x ) = x3 - 2x2 + x + 2 f' ( x ) = 4x +D Gh = 12 + c = D= -6 f'( x ) = 4x - 6 j f ( x ) = 2x2 - 6x + c f ( 3 ) = 18 - 18 + c = D= 7 f ( x ) = 2x2 - 6x + 7 f ( 5 ) = 50 - 30 + 7 = 27 32 4 x3 7 xx 14. 5x+c 15. x3 – 2x2 + x + 2 10 16. 63 17. + x - 5x - 18. 27 13. + + x + c 22 32
www.aydinyayinlari.com.tr ÷/5&(3\"- 7. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 19 ÖRNEK 22 y = f ( x ) fonksiyonunun r Y Z OPLUBTOEBLJUFóFUJOJO f ( x ) = 4x3 + 6x2 + 2x - 1 FóJNJ POPLUBOOBQTJTJOJOLBUOBFõJUUJS fonksiyonunun ilkeli g ( x ) ve g ( -1 ) = -PMEVôVOB f ( 3 ) =PMEVôVOBHÌSF G LBÀUS HÌSF H LBÀUS f'( x ) = 2x j f ( x ) = x2 + c g ( x ) = x4 + 2x3 + x2 - x + c g ( -1 ) = 1 - 2 + 1 + 1 + c = - D= -2 f ( 3 ) = 9 + c = D= 1 g ( x ) = x4 + 2x3 + x2 - x - 2 f ( x ) = x2 + G = 5 g ( 1 ) = 1 + 2 +1 - 1 - 2 = 1 ÖRNEK 20 ÖRNEK 23 # x2.d (x2) EG Y = ( 3x2 + EYWFG = 0 JOUFHSBMJOJOFöJUJOJCVMVOV[ PMEVôVOBHÌSF G LBÀUS 2x 4 f' ( x ) dx = ( 3x2+ 4 ) dx f' ( x ) = 3x2 + 4 j f ( x ) = x3 + 4x + c # #2dx = 3 dx = 4 +c f ( 1 ) = 1 + 4 + c = 0 j c = -5 x .2x f ( x ) = x3 + 4x - 5 2x f ( 0 ) = -5 4 x = +c 2 ÖRNEK 21 ÖRNEK 24 # x.f (x) dx = ax2 + 4x + b # 3 3 x2 FõJUMJóJWFSJMJZPS x+ dx f ( 1 ) =PMEVôVOBHÌSF BLBÀUS x x f ( x ) = 2ax + 4 1.f ( 1 ) = 2a + 4 = 0 a = -2 JOUFHSBMJOJOFöJUJOJCVMVOV[ 1/3 2/3 # # #x + x dx = x 1/3 – 1/2 dx + x 2/3–1/2 dx x 1/2 # #–1/6 1/6 = x dx + x dx 5/6 7/6 6x 6x = + +c 57 4 11 22. 1 23. –5 24. 6 5/6 + 6 7/6 + c 19. 5 x 21. –2 x x 20. + c 57 2
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr ÖRNEK 25 ÖRNEK 28 f (x) = # (3x2 - 2x + 1) dx ve f ( 2 ) = 8 Gh Y = 3x2 - 6x + 4 ve f ( 1 ) = 7 PMEVôVOBHÌSF G LBÀUS PMEVôVOBHÌSF G LBÀUS f ( x ) = x3 - 3x2 + 4x + c f ( x ) = x3 - x2 + x + c f(1) = 1 - 3 + 4 + c = 7 j c = 5 f(0) = 5 f(2) = 8 - 4 + 2 + c = 8 jc=2 f ( x ) = x3 - x2 + x + 2 j f ( 1 ) = 1 - 1 + 1 + 2 = 3 ÖRNEK 26 ÖRNEK 29 Ghh Y =YPMNBLÐ[FSF f (x) = # x4 - 1 dx ve f ( 1 ) = 3 f ( x ) fonksiyonuna A ( - OPLUBTOEBO¿J[JMFOUFóFUJO FóJNJPMEVóVOBHËSF G LBÀUS x-1 PMEVôVOBHÌSF G LBÀUS f''( x ) = 6x j f' ( x ) = 3x2 + c f ( -1 ) = Gh -1) = 3 j f'(-1) = 3 + c = 3 j c = 0 432 f' ( x ) = 3x2 j f ( x ) = x3+ c x x x f(-1) = -1 + c = D= 3 j f ( x ) = x3 + 3 #f(x) = ^ 3 + x 2 + x + 1 h dx = + + +x+c f(1) = 1 + 3 = 4 x 432 f^ 1 h = 1 + 1 + 1 + 1 + c = 3 432 26 48 22 11 j f(0) = 11 +c= j c= = 24 24 24 12 12 ÖRNEK 27 ÖRNEK 30 # ^ a b + b a h da f (x) = # (3x2 + 4x + 3) dx ve f ( 1 ) = 10 JOUFHSBMJOJOFöJUJOJCVMVOV[ PMEVôVOBHÌSF f ( x ) fonksiyonunun x =OPLUBTO- EBLJUFôFUJOJOEFOLMFNJOJCVMVOV[ 2 3/2 # ^ a b + b a h da = a b.a .2 f(x) = x3+ 2x2 + 3x + c b+ +c f(1) = 10 j c = G = 4 23 f'(x) = 3x2 + 4x + 3 j f'(0) = 3 y - 4 = 3x j y = 3x + 4 a2 b b.a3/2.2 12 28. 5 29. 11 30. y = 3x + 4 25. 3 26. 4 27. + +c 12 23
÷OUFHSBM\"MNB,VSBMMBS TEST - 4 1. G Y GPOLTJZPOVJ¿JOGh Y = 6x2 - 2x + 5 ve 5. P (x) = # a 6x 5 + 4 - 4x 3 k dx veriliyor. f ( 1 ) = 8 PMEVôVOBHÌSF G LBÀUS 5x \" # $ % & #VOB HÌSF 1 Y QPMJOPNV JÀJO 1 - 1 LBÀUS A ) - # - $ % & 2. f ( x ) fonksiyonunun diferansiyeli d ( f ( x ) ) ol- 6. # x2.fl (x) dx = x4 - 2x3 + c ve f^ 1 h = - 5 NBLÑ[FSF 42 9 - x3E G Y = 6x ve f ( 3 ) = 2 PMEVôVOB HÌSF G Y BöBôEBLJMFSEFO IBOHJTJ- EJS PMEVôVOBHÌSF G LBÀUS \" # $ % & A ) x2 - 12x + 6 # x2 - 4x + 12 2 2 3. # ^ x - 2 h2.x dx C ) x3 - 12x2 + 6 % 2x3 - 18x2 + 1 2 6 JOUFHSBMJOJO TPOVDV BöBôEBLJMFSEFO IBOHJTJ- EJS E ) x2 - 3x + 1 A) x3 - 4x2 + 4x +D # Y3 + 4x2 - 4x +D 7. y = G Y FóSJTJ Ð[FSJOEFLJ IFS Y Z OPLUBTOEBO C) x4 - 4x3 + 4x +D % x4 - 4x3 + 2x2 + x + c FóSJZF¿J[JMFOUFóFUMFSJOFóJNJ POPLUBJMFZFLTFOJ 43 BSBTOEBLJV[BLMóOLBSFTJOFFõJUUJS E) x4 - 4x3 + 2x2 + c 43 f ( 1 ) =PMEVôVOBHÌSF G LBÀUS A ) 10 # 11 $ % 14 E ) 16 33 33 4. # 3 x x3 x dx JOUFHSBMJOJO TPOVDV BöBôEBLJMFSEFO IBOHJTJ- 8. :FSFM FLTUSFNVN OPLUBMBSOEBO CJSJ \" -2 ) EJS # 3 x9/4 + c PMBOG Y GPOLTJZPOVJÀJO A) 4 x11/3 + c 7 f (x) = # (3x2 - a) dx 7 C) 9 x9/7 + c % 14 x9/14 + c PMEVôVOBHÌSF f ( 3 ) LBÀUS 14 9 E) 9 x14/9 + c \" # $ % & 14 1. \" 2. B 3. & 4. & 13 5. D 6. \" 7. & 8. &
TEST - 5 ÷OUFHSBM\"MNB,VSBMMBS 1. Gh Y = 6x2 + 4x ve f ( 1 ) = 5 5. f(x) = # d 9 # ^ 3x2 - 2 h dxC ve f ( 1 ) = 0 PMEVôVOBHÌSF G - LBÀBFöJUUJS PMEVôVOBHÌSF G LBÀUS \" # $ % & A) - # $ % & 6. # f(x) dx + # x.f'(x) dx 2. f : R Z3PMNBLÐ[FSFG Y GPOLTJZPOVOVO JOUFHSBMJOJO TPOVDV BöBôEBLJMFSEFO IBOHJTJ- EJS \" - OPLUBTOEBLJOPSNBMJOJOFóJNJ- 1 ve 3 f2^ x h # Y+ f ( x ) +D A) + f (x) + c f( x ) = 12x2 -PMEVóVOBHËSF f ( - LBÀUS 2 C) x.f ( x ) +D % Y2+ f ( x ) +D A ) -7 # - 9 C ) - 7 % -3 E ) - 5 E) f2 (x) + x2 f (x) + c 22 2 2 7. # x.f' (x) - 2f (x) dx x3 3. Ghh Y = 12x +PMNBLÐ[FSF integraliOJO TPOVDV BöBôEBLJMFSEFO IBOHJTJ- G Y GPOLTJZPOVOVO \" OPLUBTOEB ZFSFM EJS ekstremumVPMEVôVOBHÌSF G LBÀUS f(x) f(x) f(x) A) + c # + c C) + c xx 2x x2 \" # $ % & f(x) E) + c % YG Y +D x3 4. # ^hlogh ^xh .gl^xh dx 8. # x .^ x + 1 h.^ x - x h^ x2 + x h dx JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS \" IhPH Y +D # HhPI Y +D A) x5 + x3 + c # x5 - x3 + c 53 53 $ HhPIh Y +D % IPH Y +D C) x6 + x4 + c % x6 - x4 + c 64 64 & HPI Y +D E) x9 + x5 + c 95 1. B 2. \" 3. D 4. D 14 5. & 6. C 7. C 8. B
÷OUFHSBM\"MNB,VSBMMBS TEST - 6 1. P (x) = # r' (x) .k (x) dx + # r (x) .k' (x) dx 5. f(x) = # ^x2 - 2hf 1 + 2 p dx x4 P ( 4 ) = S L = 6 ve r ( 3 ) . k ( 3 ) = 10 PMEVôVOBHÌSF 1 LBÀUS fonksiyonu veriliyor. \" # $ % & f ( 1 ) = -PMEVóVOBHËSF f ( - LBÀUS \" # 4 C) 5 % & 7 3 3 3 2. f(x) = # f x.g'(x) - g(x) p dx 6. Gh Y = 4x -NPMNBLÐ[FSF x2 G Y GPOLTJZPOVOVO \" OPLUBTOEB ZFSFM fonksiyonu veriliyor. NJOJNVNVWBSTBG LBÀUS 3.f ( 3 ) = g ( 3 ) - 6 ve g ( 2 ) = 8 \" # $ % & PMEVôVOBHÌSF G LBÀUS \" # $ % & 3. f (x) = # ^ x4 - x3 + 4x hdx 7. f(x) = # x9 - 8 dx ve f ( 2 ) = 5 x6 + 2x3 + 4 PMEVôVOBHÌSF G Y GPOLTJZPOVOVOY= -1 nok- UBTOEBLJ UFôFUJOJO FôJNJ BöBôEBLJMFSEFO IBO- PMEVôVOBHÌSF G - LBÀUS HJTJEJS \" # $ % & A) - # -2 C) - % & 4. # f(x) dx = 4x2 + 3x 8. # ^ x - 2 h P (x) dx = x3 + mx2 - 4x x2 FöJUMJôJOJTBôMBZBO1 Y QPMJOPNVJÀJO1 EF- ôFSJLBÀUS PMEVôVOBHÌSF f ( x ) fonksiyonunun yerel eks- USFNVNOPLUBMBSOOBQTJTMFSJUPQMBNLBÀUS A) - 1 # - 1 $ % 1 E) 1 \" # $ % & 42 2 4 1. C 2. \" 3. B 4. \" 15 5. & 6. C 7. D 8. C
TEST - 7 ÷OUFHSBM\"MNB,VSBMMBS 1. Q Y L Y S Y UÐSFWMFOFCJMFOGPOLTJZPOMBSPMNBL 5. %JGFSBOTJZFMJ Y+ EYPMBOG Y GPOLTJZPOVJMFEJ- Ð[FSF feransiyeli ( 3x2 - EYPMBOH Y GPOLTJZPOVOVO LFTJNOPLUBMBSOEBOCJSJZFLTFOJÐ[FSJOEFEJS p (x) = # k' (x) dx k (x) = # r (x) dx G Y JMFH Y GPOLTJZPOMBSOOLFTJöUJLMFSJEJôFS OPLUBMBS\" Q S WF# L N PMEVôVOBHÌSF PMEVôVOBHÌSF BöBôEBLJMFSEFOIBOHJTJZBOMö- Q+LLBÀUS US \" Qh Y =Lh Y A) - # - $ % & # S Y =Qh Y $ Lh Y = r( x ) 6. # x2 dx - # 4x - 4 dx % Q Y -L Y TBCJUGPOLTJZPOEVS x+2 x2 + x - 2 & Q( x ) =Sh( x ) JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS 2. Diferansiyeli ( 2x - EYPMBOGPOLTJZPOBöBô- A) x2 - 2x + c # x2 + 2x + c 2 2 EBLJMFSEFOIBOHJTJEJS C) x3 - 4x + c % x4 - 2x + c A) 2x - # $ Y2 +D 3 4 % Y2- 4x +D & Y2- 4x +D E) x4 - x2 + c 42 3. \"öBôEBLJFöJUMJLMFSEFOIBOHJTJZBOMöUS A) # eda = e.a + c # # ndn = n2 + c 7. G Y FóJNJQP[JUJGPMBOEPóSVTBMCJSGPOLTJZPOPMNBL C) # p2r dr = p3 .r + c 2 Ð[FSF 3 % # u.t2 du = u2 t2 + c # x2.f (x) .f' (x) dx = 4x4 + 24x3 2 FõJUMJóJWFSJMJZPS E) # k3 dk = k4 + c #VOBHÌSF G Y GPOLTJZPOVOVOZFLTFOJOJLFT- UJôJOPLUBOOPSEJOBULBÀUS m 4m \" # $ % & 4. f (x) = 3x + 1 g (x) = 2x + 1 x-2 x-3 GPOLTJZPOMBSWFSJMJZPS 8. d^ h (x) h = 2x + 4 ve p (x) = # h (x) dx # ^ fog h (x) dx dx FõJUMJLMFSJWFSJMJZPS JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS I =WFQ = 5 PMEVôVOBHÌSF Q LBÀ- A) x +D # 3x + 1 + c C) x2 + c US x-2 2 \" # $ % & % x - 2 + c E) x2+D 3x + 1 1. & 2. D 3. C 4. C 16 5. D 6. \" 7. C 8. &
www.aydinyayinlari.com.tr ÷/5&(3\"- 7. MODÜL ·/÷7&34÷5&:&)\";*3-*, %&ó÷õ,&/%&ó÷õ5÷3.&:²/5&.÷ TANIM ÖRNEK 3 öOUFHSBMBMNBLVSBMMBSJMFBMONBT[PSPMBOCB- # ^ 1 - 4x2 h25.4x dx [JOUFHSBMMFSEFóJõLFOEFóJõUJSNFZËOUFNJLVM- MBOMBSBLEBIBCBTJUJOUFHSBMMFSIBMJOFHFUJSJMEJL- JOUFHSBMJOJOFöJUJOJCVMVOV[ UFOTPOSBLPMBZDBJOUFHSBMJBMOBCJMJS #^ 1 - 2 h25 .4x dx V= 1 - 4x2 EV= -8x dx Oá Oá-1 ve n `2PMNBLÐ[FSF 4x # ^ F^ x hhn.F'^ x h dx CJ¿JNJOEFLJ JOUFHSBMMFSEF j du = -2.4x dx TSBTZMBBõBóEBLJBENMBSVZHVMBOS u = F^ x h PMTVOEFOJS #j u25.d - du n 2 du = F'^ x h dx olur. #VEVSVNEBJOUFHSBMJNJ[ #1 25 1 26 - 1 ^ 1 - 2 h26 # un.du CJ¿JNJOFEËOFS =- u du = - u +c= 4x +c # un.du = un + 1 + c olur. 2 2 26 52 n+1 VZFSJOF' Y ZB[MSTB ÖRNEK 4 F^ x hn + 1 # ^ x6 + 3x2 h3^ x5 + x hdx = +c JOUFHSBMJOJOFöJUJOJCVMVOV[ n+1 õFLMJOEF¿Ë[ÐNÐCVMNVõPMVSV[ #a 6 + 2 k3.^ 5 + x hdx V= x6 + 3x2 ÖRNEK 1 x 3x x # (x + 3) 7dx JOUFHSBMJOJOFöJUJOJCVMVOV[ du = (6x5 + 6x) dx j du = 6(x5+ x) dx 4 du 1 u3 du = 1 · u + c # #j ^ 3 h· = u 66 64 4 a x6 + 3x2 k4 +c u = +c= 24 24 # ^ x + 3 h7 dx V= x + EV= dx #j 7 du = 8 +c= ^ x + 3 h8 +c u u 88 ÖRNEK 5 ÖRNEK 2 # x2 dx # 2 2x - 1 dx JOUFHSBMJOJOFöJUJOJCVMVOV[ x3 + 4 inteHSBMJOJOFöJUJOJCVMVOuz. #x 2 dx V= x3 + EV= 3x2 dx 3 x +4 # 2 2x - 1 dx V= 2x - EV= 2dx & # du 3u u 3/2 + c = 2 ^ 2x - 1 h3/2 + c 1/2 # #j 1/2 #1 –1/2 1 u2 3 u du = du = = u du = +c= u 33 x +4+c 13 33 2 2 ^ x + 3 h8 2. 2 ^ 2x - 1 h3/2 + c 17 - ^ 1 - 2 h26 a x6 + 3x2 k4 23 1. + c 3 3. 4x + c 4. + c 5. x +4+c 8 52 24 3
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr ÖRNEK 6 ÖRNEK 9 # 3x dx # x d^ x3 h 5 - x2 4 + x4 inteHSBMJOJOFöJUJOJCVMVOuz. inteHSBMJOJOFöJUJOJCVMVOuz. # 3x dx V= 5 - x2 EV= -2x dx # #x.d^ 3 h 3 5 - x2 x = 3x dx V= 4 + x4 EV= 4x3 dx # #3 du - 3 –1/2 4 4 4+x 4+x j- = u du 2 u2 du 3 3 4 + x4 = +c - 3 1/2 2 # #3 –1/2 du = j u 4 u4 2 = 2u + c = - 3 5 - x + c 2 ÖRNEK 7 ÖRNEK 10 # F ›^ 3x + 1 h .F^ 3x + 1 h dx rY WFG Y PMNBLÐ[FSF G Y G h Y = 2x3 + 12x2 + 16x tir. inteHSBMJOJOFöJUJOJCVMVOuz. f ( 2 ) = 12 PMEVôVOBHÌSF G LBÀUS f ( x ).f' ( x ) = 2x3 + 12x2 +Y G = G = # #f^ x h.f'^ x h dx = 32 2x + 12x + 16x dx #u = f^ x h x 4 u du = du = f'^ x h dx & 2 + 4x3 + 8x2 + c # F'^ 3x + 1 h.F^ 3x + 1 h dx V= F(3x + 24 du = F'(3x + 1).3 dx u x + 4x3 + 8x2 + c = 22 2 u.du 1 u + c = 1 ^ F^ 3x + 1 h h2 + c #j =· 2 ^ x h 4 3 32 6 j f = x + 4x3 + 8x2 + c 22 x=2j 2 ^ x h = 8 + 32 + 32 + c j 72 = 72 + c j c = 0 f 2 f2 ^ 3 h = 81 + 108 + 72 = 441 & f^ 3 h = 21 22 2 ÖRNEK 8 ÖRNEK 11 # f3 (x) .f› (x) dx # dx 3 f (x) inteHSBMJOJOFöJUJOJCVMVOuz. ^ x + 2 h4 inteHSBMJOJOFöJUJOJCVMVOuz. #3 ^ x h.f'^ x h dx f u =G Y EV= f'( x ) dx 3 f^ x h 3 11/3 # dx u .du u8/3du = 3u V= x + EV= dx = ^ x + 2 h4 # #j 11 +c 1/3 u = 3 ^ f^ x h h11/3 + c # #du –4 u –3 -1 11 j 4 = u du = +c= +c –3 3^ x + 2 h3 u 6. - 3 5 - x2 + c 7. 1 ^ F^ 3x + 1 h h2 + c 8. 3 ^ f^ x h h11/3 + c 18 3 4 + x4 10. 21 -1 6 11 9. + c 11. + c 2 3^ x + 2 h3
www.aydinyayinlari.com.tr ÷/5&(3\"- 7. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 12 ÖRNEK 15 # ^ 3x2 - 2x + 5 h3.^ 3x - 1 h.dx g^ x h = # fl^ x h.f^ x hdx ve f^ 1 h = g^ 1 h = - 3 tür. inteHSBMJOJOFöJUJOJCVMVOuz. f ( 2 ) = 6 ise H LBÀUS # a 3x2 - 2x + 5 k3 ^ 3x - 1 hdx #g ( x ) = Gh Y G Y EY G = g ( 1 ) = - G = g ( 2 ) = u = 3x2 - 2x + EV= 6x -EY EV= 2(3x - 1) dx #u =G Y EV=Gh Y EY H Y = u du #= 3 du 4 1 a 3x2 - 2x + 5 k4 + c g^ x h = f2^ x h +c u· = 1u +c= 8 2 · 2 24 9 - 15 , g^ 2 h = 36 - 15 -3 = +c & c = 2 2 22 g^ 2 h = 21 2 ÖRNEK 13 ÖRNEK 16 # fl (x) .fm (x) dx # ^ 1 + x2 h11.x dx inteHSBMJOJOFöJUJOJCVMVOuz. JOUFHSBMJOJOFöJUJOJCVMVOV[ # f' (x) .f' ' (x) dx V=Gh Y EV= f'' ( x ) dx #^ 1 + 2 h11 .x dx V= 1 + x2 EV= 2x dx x #2 ^ f' (x) h2 #1 ^ u h11.du = 1 · 12 +c= ^ 1 + x2 h12 +c u j u.du = + c = +c & u 22 2 2 12 24 ÖRNEK 14 ÖRNEK 17 # 20.^ x2 - 4 h19.x dx # 1+3 k dk inteHSBMJOJOFöJUJOJCVMVOuz. 3 k2 JOUFHSBMJOJOFöJUJOJCVMVOV[ #20. ^ 2 - 4 h19 .x dx V= x2 - EV= 2x dx # 1 + 3 k dk u = 1 + 3 k du = 1 dk x #1 ^ u h19 du = 10.u 20 32 32 20· k 3k 2 20 +c u 3/2 .2 = 2^ 1 + 3 k h3/2 + c u du = 3 #3 3 = 1 ^ x2 - 4 h20 + c 2 a 3x2 - 2x + 5 k4 + c 13. ^ f'^ x h h2 +c 14. 1 ^ x2 - 4 h20 + c 19 21 ^ 1 + x2 h12 17. 2^ 1 + 3 k h3/2 + c 2 15. 16. + c 12. 24 82 2
TEST - 8 %FôJöLFO%FôJöUJSNF:ÌOUFNJ 1. # ^ 2x - 3 h6 dx 5. # ^ x4 - 2 h3 .x3 dx JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS A) 1 ^ x4 - 2 h3/2 + c # 1 ^ x4 - 2 h5/2 + c ^ 2x - 3 h7 ^ 2x - 3 h6 4 10 A) + c # + c C) 1 ^ x4 - 2 h5/2 + c % 1 ^ x4 - 2 h3/2 + c 12 16 7 6 E) 1 ^ x4 - 2 h3/2 + c ^ 2x - 3 h7 ^ 2x - 3 h6 12 C) + c % + c 6. # 3x - 3 dx 14 12 3 x2 - 2x + 6 2.^ 2x - 3 h7 integralJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS E) + c A) ( x2 - 2x + 6 )2/3 +D # Y2 - 2x + 6 )3/2 +D 7 C) 4 ^ x2 - 2x + 6 h3/2 + c 2. # ^ x2 - 4x h5.^ x - 2 h dx 9 % 4 ^ x2 - 2x + 6 h2/3 + c JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS 9 A) ^ x2 - 4x h6 + c # ^ x2 - 4x h6 + c E) 9 ^ x2 - 2x + 6 h2/3 + c 6 12 4 ^ x - 2 h6 % 2^ x2 - 4x h6 + c C) + c 3 6 E) ^ x2 - 4x h5 + c 10 3. # ^ x3 + 1 h3.x2 dx JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS 7. # x2.f' ^ x3 h dx A) ^ x3 + 1 h4 + c # ^ x3 + 1 h4 + c JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS 12 6 A) f ( x3 ) +D # G Y3 ) +D C) ^ x3 + 1 h4 + c % Y3 + 1)4 +D 3 C) f^ x3 h + c % Ghh Y3) +D 3 E) 3 ^ x3 + 1 h4 + c E) f'^ x3 h + c 4 2 4. # x2 + 4 xdx 8. # 4x.f' ' ^ x2 h dx JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS A) x2 + 4 + c # 2 x2 + 4 + c A) f'^ x2 h + c # Ghh Y2 ) +D 2 2 C) 2 ^ x2 + 4 h3/2 + c % 1 ^ x2 + 4 h3/2 + c C) f' '^ x2 h + c % Gh Y2) +D 33 2 E) 1 ^ x2 + 4 h2/3 + c & Ghhh(x2) +D 3 1. C 2. B 3. \" 4. D 20 5. B 6. & 7. C 8. D
%FôJöLFO%FôJöUJSNF:ÌOUFNJ TEST - 9 1. # ^ 2x3 + 3x h.^ 2x2 + 1 h dx 4. # x dx JOUFHSBMJOJOEFôFSJOFEJS x-3 A ) 1 ^ 2x3 + 3x h2 + c # 1 ^ 2x3 + 3x h2 + c 24 JOUFHSBMJOJO TPOVDV BöBôEBLJMFSd en hangisi- C ) 1 ^ 2x3 + 3x h2 + c % 1 ^ 2x2 + 1 h2 + c 66 EJS E ) 1 ^ 2x2 + 1 h2 + c 2 ^ x - 3 h3 A) + x - 3 + c 3 ^ x - 3 h3 # + 6 x - 3 + c 3 C) 2 f ^ x - 3 h3 x - 3 p+ c +3 3 2. # ^x2 + 3hdx 2^ x - 3 h3 % + 3 x - 3 + c x3 + 9x + 8 3 JOUFHSBMJOJO TPOVDV BöBôEBLJMFSEFO IBOHJTJ- E) 2 3 x - 3 + 6 x - 3 + c EJS A) 1 3 # 2 3 5. # ^ 3x - 1 h4 ^ 2x + 1 h dx x + 9x + 8 + c x + 9x + 8 + c integralinin sPOVDV BöBôEBLJMFSEFO IBOHJTJ- 33 EJS C) 2 +c % 3 x3 + 9x + 8 + c A) 1 f ^ x - 1 h6 + ^ 3x - 1 h5 p + c 3 x3 + 9x + 8 2 93 E ) 3 x3 + 5x + 8 + c # 1 f ^ 3x - 1 h6 + ^ 3x - 1 h5 p + c 3. # x11dx 93 1 + x6 C) 1 f ^ 3x - 1 h6 + ^ 3x - 1 h5 p + c integralinin sonucu aöBôEBLJMFSEFO IBOHJTJ- EJS 64 % 1 f ^ 3x - 1 h3 + ^ 3x - 1 h6 p + c 93 E) 1 f ^ 3x - 1 h6 + ^ 3x - 1 h5 p + c 94 A) 2a 1 + x 6 3 + 2 1 + x6 + c 1- x k dx # 3 6. # 2 a 1 + x6 k - 2 1 + x6 + c 1+ x 3 integralinde x = t2EÌOÑöÑNÑZaQMSTB BöBô- 1a 3 EBLJJOUFHSBMMFSEFOIBOHJTJFMEFFEJMJS k C) 1 + x 6 + 2 1 + x6 + c 3 A ) # 1 - t dt # # t.^ t - 1 h dt 2a 3 1+t 1+t % 1 + x 6 k - 2 1 + x6 + c 3 # t^1 -th % 2 # t^ 1 - t h dt E) 1 a 3 1 C ) .dt 1+t 1 + x 6 k - 1 + x6 + c 1 +t 93 E ) 2 # 1 - t dt 1+t 1. C 2. B 3. & 21 4. C 5. B 6. D
TEST - 10 %FôJöLFO%FôJöUJSNF:ÌOUFNJ 1. # x2.f' (x3) dx 5. f^ x h = # x4 - 8x3 .^ x3 - 6x2 hdx veriliyor. JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS f ( -1) =PMEVóVOBHËSF G LBÀUS A) f ^ x3 h + c # 3 f ^ x3 h + c C) f ^ x3 h + c % x3.f^ x3 h + c A) 1 # 1 C) 1 % & 3 3 8 4 2 E) 2 f ^ x3 h + c 3 6. # dx 9x2 - 12x + 4 JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS 2. g^ x h = # x.f' ^ x2 + 1 h dx H = G = A) - 1 + c # - 1 + c 9x - 6 6x - 4 f (10) =PMEVôVOBHÌSF H LBÀUS C) 1 + c % 1 + c 9x - 6 6x - 4 \" # $ % & E) 3 + c 6x - 4 7. # 3 x2 - 6x ^ x - 3 h dx 3. # f' ' '^ x h.f' ^ f' '^ x hh dx integralinde t3 = x2 -YEFôJöLFOEFôJöUJSNFTJ ZBQMODBBöBôEBLJJOUFHSBMMFSEFOIBOHJTJFMEF JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS FEJMJS \" Gh Y +D # Ghh Y +D $ Ghhh Y +D A) 2 # t3 dt # 3 # t3 dt % G Gh Y +D & G Ghh Y +D 3 2 C) 2 # t2 dt % 9 # t2 dt 9 2 E) 9 # t3 dt 4 4. # 3 2x + 5 - 2x + 5 dx 6 2x + 5 8. # dx integralinde 2x + 5 = u6 deôJöLFOEFôJöUJSNFTJ 6 2x + 3 ZBQMODBBöBôEBLJJOUFHSBMMFSEFOIBOHJTJFMEF integralinde m6 = 2x +EFôJöLFOEFôJöUJSNFTJ FEJMJS ZBQMODBBöBôEBLJJOUFHSBMMFSEFOIBOHJTJFMEF A) 3 # ^ u7 - u6 h du # 3 # ^ u6 - u7 h du FEJMJS A) 3 # m6 dm # 3 # m5 dm C) 1 # ^ u7 - u6 h du %) 1 # ^ u6 - u7 h du C) 3 # m4 dm % 5 # m5 dm 3 3 E) 3 # ^ u5 - u6 h du E) 5 # m4 dm 1. C 2. & 3. & 4. B 22 5. C 6. \" 7. B 8. C
%FôJöLFO%FôJöUJSNF:ÌOUFNJ TEST - 11 1. # x2.^ x - 2 h ^ x2 + 2x + 4 h dx 5. # 3x + 3 dx x3.^ x + 2 h3 JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS A) ^ x3 - 8 h + c # ^ x3 - 8 h2 + c integrBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS 6 6 A) - 3 + c # - 4 + c C) x3 - 8 + c % ^ x3 - 8 h + c 4^ x2 + 2x h2 3^ x2 + 2x h2 3 2 C) - 3 + c % - 2 + c E) ^ x3 - 8 h3 + c 2^ x2 + 2x h2 3^ x2 + 2x h2 6 E) - 3 + c 4^ x + 2 h2 f'f 2 p x 2. # dx 4x2 6. # f' '^ x h dx integraliOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS ^ f'^ x hh3 A) - 1 ff 2 p + c # - 1 ff 2 p + c JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS 2x 8x C) 1 ff 2 p + c % 1 ff 2 p + c A) - 2^ f'^ x hh2 + c # -^ f'^ x hh2 + c 2x 8x - f'^ x h -^ f'^ x hh2 E) 2ff 2 p + c C) + c % + c x 2 2 E) - 1 + c 2^ f'^ x hh2 3. # dx x .^ x + 4 h2 7. # x5 ^ x2 + 1 h dx JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS integralinde u = x2 + EFôJöLFO EFôJöUJSNFTJ A) - 4 + c # - 3 + c ZBQMODBBöBôEBLJJOUFHSBMMFSEFOIBOHJTJFMEF x+4 x+4 FEJMJS C) - 2 + c % 2 + c A) 1 # u3 du # 1 # ^ u - 1 h2 du x+4 x+4 2 2 E) 3 + c C) 1 # ^ u - 1 h2 u du % 2 # ^ u - 1 h3 du x+4 2 E) 2 # ^ u - 1 h2 u du 4. # ^ 3 x + 2 h3 8. # ^ f^ 3x + 1 hh2.f'^ 3x + 1 h dx dx 3 x2 JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS JOUFHSBMJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS A) 2 ^ 3 x + 2 h4 + c # 2 ^ 3 x + 2 h3 + c A) 1 f^ 3x + 1 h + c # 1 ^ f^ 3x + 1 hh3 + c 3 3 27 27 C) 3 ^ 3 x + 2 h4 + c % 3 ^ 3 x + 2 h3 + c C) 1 ^ f^ 3x + 1 hh3 + c % 1 ^ f^ 3x + 1 hh3 + c 4 4 93 ^ 3 x + 2 h4 E) 3 ( f ( 3x + 1 ) )3 +D E) + c 4 1. B 2. B 3. C 4. C 23 5. \" 6. & 7. C 8. C
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr #&-÷3-÷÷/5&(3\"-* TANIM ÖRNEK 1 # f^ x h dx = F^ x h + c PMNBLÐ[FSF 2 b # ^ 2x - 3 h dx inteHSBMJOJOTPOVDVLBÀUS # f^ x h dx JGBEFTJOe f ( x ) inBEBOCZFbelirli 1 a 22 integrali EFOJS #^ 2x - 3 h dx = 2 - 3x x b 11 # f^ x h dx = F^ x h b = F^ b h - F^ a h = (4 - 6) - (1 - 3) = 0 a a ile bulunur. ÖRNEK 2 UYARI 2 öOUFHSBMTBCJUJPMBODTBEFMFõUJóJJ¿JOCFMJSMJJOUFHSBM- # ^ x2 - x + 2 h dx inteHSBMJOJOTPOVDVLBÀUS EFZB[MNB[ –1 2 32 2 x x #^ 2 - x + 2 h dx = - + 2x x 32 –1 –1 = d 8 - 2 + 4 n - d - 1 - 1 - 2 n = 15 3 32 2 #FMJSMJ÷OUFHSBMJO²[FMMJLMFSJ b bb ÖRNEK 3 1. # 6f^ x h ± g^ x h@ dx = # f^ x h dx ± # g^ x h dx 4 x+1 dx inteHSBMJOJOTPOVDVLBÀUS a aa # 0x bb 44 2. # c.f^ x h dx = c. # f^ x h dx # #x + 1 dx = x + –1/2 dx aa x a 0x 0 3. # f^ x h dx = F^ a h - F^ a h = 0 = 2 3/2 + 1/2 4 16 +4 n-0 = 28 a x 2x =d ba 3 03 3 4. # f^ x h dx = - # f^ x h dx ab 5. a <D<CPMNBLÐ[FSF ÖRNEK 4 b cb 3 # f^ x h dx = # f^ x h dx + # f^ x h dx # 3x2 dy inteHSBMJOJOTPOVDVLBÀUS a ac 2 6. G Y TÐSFLMJWF¿JGUGPOLTJZPOJTF 3 aa 3 # f^ x h dx = 2. # f^ x h dx #2 dy = 2 = 2 ^ 3- 2h 3x –a 0 3x y 3x a 2 7. G Y TÐSFLMJWFUFLfonksiyon ise # f^ x h dx = 0 2 –a 24 1. 0 15 28 4. 3x2^ 3 - 2 h 2. 3. 23
www.aydinyayinlari.com.tr ÷/5&(3\"- 7. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 5 ÖRNEK 8 10 0 dx inteHSBMJOJOTPOVDVLBÀUS # ^ 5y - 2 h dt inteHSBMJOJOTPOVDVLBÀUS # -4 1 - 2x 1 01 10 10 # #dx 1 du V= 1 -Y EV= -2dx j - # ^ 5y - 2 h dt = ^ 5y - 2 ht 2 –4 1 - 2x u 11 9 = ( 5y - 2 ) . ( 10 - 1 ) = 9. ( 5y - 2 ) #1 11 1 =- –1/2 du = - 1/2 = -(1 - 3) = 2 u ·2u 22 9 9 ÖRNEK 6 52 # f (x) dx = 9 PMEVôVOBHÌSF # f (3x - 1) dx ifade- 21 TJOJOTPOVDVLBÀUS 5 2 # f^ x h dx = 9 # f ^ 3x - 1 h dx 2 1 V Ym EV EY ÖRNEK 9 5 ôFLJMEFZ G Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõUJS y #1 f^ u h du = 1 ·9 = 3 4 y = f(x) 33 2 ÖRNEK 7 2 x 5 –3 –2 O1 4 # f (3x + 2) dx = 45 PMEVóVOBHËSF –2 0 3 # f^ 5x + 2 hdx JOUFHSBMJOJOTPOVDVLBÀUS 0 53 1 # #f^ 3x + 2 h dx = 45 f^ 5x + 2 h dx LBÀUS #VOBHÌSF # f›^ 4x hdx JOUFHSBMJOJOTPOVDVLBÀUS 00 –1 2 hh u = 3x + 2 u = 5x + 2 17 #1 du = 3 dx f^ u h du 14 du =EY # #1 17 5 f' (4x) dx V=Y EV= 4dx j f'^ u h du 2 #1 f^ u h du = 45 4 1 –2 3 – 2 2 17 1 4 1 ^0-4h= 1 ·-4 =-1 #j f^ u h du = 135 1 = f^ u h = & 135· = 27 4 –2 4 4 2 5 5. 9(5y – 2) 6. 3 7. 27 25 8. 2 9. –1
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr ÖRNEK 10 ÖRNEK 14 4 5 # 2x2dx JOUFHSBMJOJOTPOVDVLBÀUS # f (4x - 1) dx = A PMEVóVOBHËSF –4 -2 4 43 4 16 # #2 2 x 256 # f (x + 3) dx integSBMJOJOTPOVDVLBÀUS 2x dx = 4 x dx = 4 = 3 –2 3 –4 0 0 5 16 # #f^ 4x - 1 hdx = A f^ x + 3 hdx = ? –2 –12 u = 4x - 1 u=x+3 ÖRNEK 11 du = 4dx du = dx 100 19 19 # 4x3dx JOUFHSBMJOJOTPOVDVLBÀUS #1 f^ u h du = A # f^ u h du = 4A –100 4 –9 100 –9 #3 ÖRNEK 15 4x dx = 0 ÷ÀUFLJGPOLTJZPO UFLGPOLTJZPO –100 GWFHTÐSFLMJGPOLTJZPOMBSPMNBLÐ[FSF 22 5 # f(x) dx = 5 # f(x) dx = 4 # g(x) dx = 3 ÖRNEK 12 –1 5 –1 2 FõJUMJLMFSJWFSJMJZPS # x 2x2 + 1 dx JOUFHSBMJOJOTPOVDVLBÀUS 5 0 #VOBHÌSF # ^ 2 f (x) - 3 g^ x hh dx integralinin sonu- DVLBÀUS –1 2 5 55 # x. 2 + 1 dx V= 2x2 + EV= 4x dx # # #^ 2f^ x h - 3g^ x h h dx = 2 f^ x h dx - 3 g^ x h dx 2x –1 –1 –1 0 9 = 2 ( 5 - 4 ) - 3.3 = 2 - 9 = -7 9 1 #1 1 3/2 2 = u du = u · 4 43 1 = 1 ^ 27 - 1 h = 26 = 13 6 63 ÖRNEK 13 ÖRNEK 16 :BOEBLJöFLJMEFWFSJMFO- MFSFHÌSF 5 x dx JOUFHSBMJOJOTPOVDVLBÀUS y y = f(x) x 3 # 2 2 x-1 –2 3 # fl (x) .f (x) dx integrali- -2 OJOTPOVDVLBÀUS 5 3 # x dx V2 = x - VEV= dx # f'^ x h.f^ x h dx V=G Y EV= f'(x) dx 2 x-1 2 ^ 2 h 3 2 #= 2 u + 1 u du = 2f u +up –2 3 u f^ 3 h 2 2 2 11 # #u =2-0=2 u du = u du = = 2 d d 8 + 2 n - d 1 + 1 n n = 2d 7 + 1 n = 20 2 0 f^ –2 h 0 33 33 256 11. 0 13 20 26 14. \" 15. –7 16. 2 10. 12. 13. 3 3 3
www.aydinyayinlari.com.tr ÷/5&(3\"- 7. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 17 ÖRNEK 20 3EFUBONMGGPOLTJZPOV- OVO HSBGJóJ õFLJMEFLJ HJCJ- f ( 1 ) = 1 ve f ( 3 ) =PMEVóVOBHËSF y EJS 4 3 x a Ob # f^ x h. fl^ x h dx integralinin sonucu kaÀUS 1 3 # f^ x h.f'^ x h dx V=G Y EV= f'(x) dx 1 b 42 4 # f^xh.fl^xh dx JOUFHSBMJOJOTPOVDVLBÀUS #u 1 15 a u du = =8- = 2 22 1 1 b # f^ x h.f'^ x h dx V=G Y EV= f' ( x ) dx a ÖRNEK 18 42 4 G Y GPOLTJZPOVOVOBOBMJUJLEÐ[MFNEFY1 =OPLUBTO- #u =8-0=8 EBLJUFóFUJOFóJNJ Y2 =OPLUBTOEBLJUFóFUJOJOFóJ- NJ-ES = u du = 0 2 7 0 #BunaHÌSF fll ^xh fl ^xh dx JOUFHSBMJOJOTPOVDVLBÀ- ÖRNEK 21 US 4 7 7 # ^x2 + x5 + x7 + x11 + x15hdx # f' '^ x h.f'^ x h dx V=Gh Y EV= f'' ( x ) dx -7 4 integralinin sonucu LBÀUS –6 2 –6 #u 9 27 = 18 - = & u du = 2 22 3 3 77 2 5 7 11 15 2 x x x x x x dx # #^ + + + + h dx = –7 –7 3 7 343 - ^ - 343 h 686 x == ÖRNEK 19 = –7 3 3 y 3 F(x) :BOEBLJõFLJMEF' Y GPOL- TJZPOVOVO HSBGJóJ WFSJMNJõ- tir. –1 3 x O2 4 –2 ÖRNEK 22 3 7 #VOBHÌSF # ^x + F' (x)hdx integralinin sonucuLBÀ- # ^x7 - x5 - 5x3h21dx US -1 -7 integralinin sonucu LBÀUS 32 3 3 #7 7 - 5 - 3 k21 dx = 0 –1 a # ^ x + F'^ x h h dx = x +F^ x h x x 5x 2 –1 –1 =d 9 - 1 n+^-2-0h= 4-2 = 2 –7 22 15 27 19. 2 27 20. 8 686 22. 0 17. 18. 21. 2 2 3
TEST - 12 #FMJSMJ÷OUFHSBM a 3 #1. ^3x2 - 2x + ch dx = a3 - a2 + 5a 5. # 9x2.f' (x) + 2x.f (x)C dx = 18 0 0 PMEVôVOBHÌSF DEFôFSJLBÀUS PMEVôVOBHÌSF G LBÀUS A ) -B # mB $ % B & A ) 22 # $ 16 % & 3 3 k 6. ôFLJMEF G Y GPOLTJZPOVOVO WF OPLUB- 2. # ^4x - 2hdx = 54 ve k -N= 3 MBSOEBLJFLTUSFNVNOPLUBMBSWFSJMNJõUJS y m (4, 7) (1, 2) PMEVôVOBHÌSF L+NLBÀUS O1 f(x) \" # $ % & x 4 4 integralinin so- #VOB HÌSF # f' ^ x h f' ' ^ x h dx 1 nucu kaÀUS A ) - # - $ % & k 3. # ^3x2 + 6x - 2hdx = k3 + 1 0 FöJUMJôJOFHÌSFLOJOBMBCJMFDFôJEFôFSMFSUPQMB- NLBÀUS 59 7. # f(x) dx = 7 ve # f(x) dx = 16 \" # 2 C ) 1 % - 1 E ) –1 22 3 33 5 PMEVôVOB HÌSF # f^ x h dx integralinin sonucu 9 LBÀUS A) - # - $ % & 5 11 8. 3 3 dx 4. # f^ x h dx = 12 # f^ x h dx = - 3 # x +x 15 2 0 x +1 11 JOUFHSBMJOJOTPOVDVLBÀUS PMEVôVOBHÌSF # 2f^xhdx integralinin sonucu A) - # 7 C) 8 % 10 1 3 3 E) LBÀUS 3 A ) - # - $ % & 1. & 2. \" 3. B 4. & 28 5. & 6. C 7. \" 8. B
#FMJSMJ÷OUFHSBM TEST - 13 1 5. y ôFLJMEFZ= f( x ) fonk- #1. ^2x + 1ht dx = 40 6A TJZPOVOB Ð[FSJOEFLJ \" t +1 OPLUBTOEBO¿J[JMFOUF- 0 PMEVôVOBHÌSF ULBÀUS óFUEEPóSVTVEVS \" # $ % & f(x) x 4 10 d 4 #f( 1 ) =PMEVôVOBHÌSF f› (x) .f›› (x) dx integ- 1 SBMJOJOTPOVDVLBÀUS 2 \" m # - 3 C ) –1 % - 1 E ) 0 22 2. # x2 f^ x3 h dx = 20 0 8 PMEVôVOB HÌSF # f (x) dx integralinin sonucu 0 LBÀUS \" # $ % & 6. 2 x3 dx # –2 x4 + x2 + 6 integralinin soOVDV BöBôEBLJMFSEFO IBOHJTJ- EJS A) - # - $ % & 9 #3. 1 + x dx JOUFHSBMJOJOTPOVDVLBÀUS 0 A) 232 # 242 C) 202 7. ôFLJMEF G Y GPOLTJZ POVOVO\" Z0 OPLUBTOE BLJ 15 15 15 teóFUJZ= x +EPóSVTVEVS % 191 E) 12 15 y y=x+2 A y = f(x) O2 x –3 11 G Y GPOLTJZPO VOVO HSBGJôJ Z FLTFOJOJ -3 ) 4. # f^xh dx = K PMEVóVOBHËSF 2 3 OPLUBTOEB LFTUJôJOF HÌSF # x . f' '^ x h.dx in- 5 0 # f^ 2x + 1 h dx tegralinin sonucu LBÀUS 1 JOUFHSBMJOJOTPOVDVLBÀUS A) K # , $ , % , & , \" m # m $ m % m & 2 1. B 2. & 3. \" 4. \" 29 5. B 6. C 7. \"
TEST - 14 #FMJSMJ÷OUFHSBM 1 3x2 2x3 i4._ x2 i m+1 3m 1. 4. f # 2x dx p = # 6x5 dx #_ - x - dx 0 00 JOUFHSBMJOJOTPOVDVLBÀUS olEVôVOBHÌSF NLBÀUS A) 1 # 1 C) 1 % 1 1 A) - 1 # - 1 C) 1 % 1 E) 1 60 50 40 30 E) 4 24 2 20 10 3 2. # x2 + 6 .x dx 5. # F_ 4x - 1 i dx = 48 PMEVóVOBHËSF 3 1 integralinde u2 = x2 + EFôJöLFOEFôJöUJSNFTJ 2 ZBQMSTB BöBôEBLJ JOUFHSBMMFSEFO IBOHJTJ FMEF FEJMJS # F_ 3x + 5 i dx 10 10 4 2 – A) # u.du # # u2.du C) # u2.du 3 3 3 3 JOUFHSBMJOJOTPOVDVLBÀUS 4 E) 1 4 \" # $ % & % 2 # u2 .du # u2.du 23 3 6. ôFLJMEF Z = ' Y GPOLTJZPOVOVO HSBGJóJ WFSJMNJõ- tir. 3. ôFLJMEF Z = ' Y GPOLTJZPOVOVO HSBGJóJ WFSJMNJõ- y = F(x) y tir. Ap Bm y 6 135° 4 O –2 –1 O x x F ( x ) in A ( - Q OPLUBTOEB ZFSFM FLTUSFNVNV 3 PMVQ ' Y F# - N OPLUBTOEBO¿J[JMFOUFóFUY FLTFOJJMFQP[JUJGZËOEFMJLB¿ZBQNBLUBES y = F(x) –1 3 #VOBHÌSF # F'_ x i.F' '_ x i dx integralinin so- #VOBHÌSF # F'_ x i dx JOUFHSBMJOJOTPOVDVBöB- –2 0 OVDVLBÀUS ôEBLJMFSEFOIBOHJTJEJS A) 5 $ 3 % & 1 A) - # -3 C) - % & 2 2 2 # 1. D 2. C 3. C 30 4. B 5. & 6. &
#FMJSMJ÷OUFHSBM TEST - 15 57 4. 16 x - 16 dx 1. # 3 x5 + 1 .x4 dx # 4 x+4 0 integralinde x5 + 1 = u3EFôJöLFOEFôJöUJSNFTJ integraliniOTPOVDVLBÀUS ZBQMSTB BöBôEBLJ JOUFHSBMMFSEFO IBOHJTJ FMEF A) - 64 # - 32 - 16 FEJMJS 3 3 C) 7 # 3 7 % - 8 3 50 3 A) # u3 du # u3 du E) - 4 3 0 3 2 % 3 7 C) # u3 du # u3 du 51 51 E) 5 2 31# u3 du 2. ôFLJMEFLJHSBGJLUFZ=' Y GPOLTJZPOVOVO\"WF# 5. y =' Y GPOLTJZPOVOVOHSBGJóJ\" WF# OPLUBMBSOEBLJUFóFUMFSJ¿J[JMNJõUJS OPLUBMBSOEBOHF¿NFLUFEJS y y = F(x) #VOBHÌSF 10 A 4 x.F'_ x i - F_ x i B6 dx # 2 F2 _ x i JOUFHSBMJOJOTPOVDVLBÀUS x A) - 1 # - 1 C) - 1 % 1 E) 1 6 9 8 66 9 –4 –2 O2 6 #VOB HÌSF # x.F' '_ x i integralinin sonucu –2 LBÀUS \" # $ % & 7 44 3. # F_ 3x + 1 i dx = 10 PMEVóVOBHËSF 6. # x.F'_ x i dx = 2m, # F_ x i dx = n 3 0 03 5 PMEVôVOBHÌSF ' Ñn m ve OUÑSÑOEFOFöJUJ BöBôEBLJMFSEFOIBOHJTJEJS # F_ 2x i dx A) 3m + n # 2m + n C) 6m + n 11 6 12 12 integralinin sonucu LBÀUS % 6m + n E) 3m + 2n 18 12 \" m # m $ % & 1. C 2. & 3. B 31 4. B 5. C 6. C
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr #&-÷3-÷÷/5&(3\"-** 1BSÀBM'POLTJZPOMBSO#FMJSMJ÷OUFHSBMJ ÖRNEK 2 TANIM æ F^ x h = * 2 - x , x < 3 ise 2x - 7 , x ≥ 3 ise H Y WFI Y JOUFHSBMMFOFCJMJSJLJGPOLTJZPOWF 3 a #D#CPMNBLÐ[FSF PMEVôVOB HÌSF # F^ x + 1 h dx integralinin deôFSJ LBÀUS 1 F_ x i = * g_ x i , x < c ise h_ x i , x $ c ise CJ¿JNJOEF UBONM ' Y GPOLTJZonunun [B C] 3 u = x + EV= dx BSBMóOEBLJ JOUFHSBMJOJ CVMNBL J¿JO JOUFHSBM # F^ x + 1 h dx = GPOLTJZPOVOVO LVSBMOO EFóJõUJóJ D OPLUBTOB 1 HËSF 43 4 b cb # # #F^ u h du = ^ 2 - x h dx + ^ 2x - 7 h dx # F_ x i dx = # g_ x i dx + # h_ x i dx 22 3 a ac 2 3 4 2 x + x - 7x 2x - 23 2 biçiNJOEFJLJJOUFHSBMJOUPQMBNPMBSBLZB[MNB- = =d 6 - 9 n - ^ 4 - 2 hG + 7^ 16 - 28 h - ^ 9 - 21 hA MES 2 = = 3 - 2 G + 7 - 12 + 12 A = - 1 22 ÖRNEK 1 ÖRNEK 3 F (x) = * x2 x < 0 PMNBLÐ[FSF F_ x i = * 6x2 + 1 , x < 1 ise 2x + 1 x $ 0 4x - 1 , x $ 1 ise CJ¿JNJOEFUBONM' Y GPOLTJZPOVWFSJMJZPS 3 2 # F (x) dx JOUFHSBMJOJOTPOVDVLBÀUS #VOBHÌSF # F_ x i dx JOUFHSBMJOJOTPOVDVLBÀUS –2 –1 3 03 21 2 2 # # #F^ x h dx = a 6x2 + 1 k dx + ^ 4x - 1 hdx x dx + # # #F^ x h dx = ^ 2x + 1 h dx –2 –2 0 –1 –1 1 3 0 2 3 1 2 x = + x + x & 2x3 + x + 2 - x 3 2x –2 0 –1 1 = = 0 - d - 8 nG + 7^ 9 + 3 h - 0 A = 8 + 12 = 44 = 7^ 2 + 1 h-^ -2 - 1 hA+7^ 8 - 2 h-^ 2 - 1 hA 3 33 = (3 + 3) + (6 - 1) = 11 44 32 1 3. 11 1. 2. - 2 3
www.aydinyayinlari.com.tr ÷/5&(3\"- 7. MODÜL ·/÷7&34÷5&:&)\";*3-*, .VUMBL%FôFS'POLTJZPOVOVO#FMJSMJ÷OUFHSBMJ ÖRNEK 3 TANIM 3 .VUMBL EFóFS GPOLTJZPOVOVO CFMJSMJ JOUFHSBMJOJ # a x - 2 + x + 1 k dx JOUFHSBMJOJOTPOVDVLBÀUS CVMNBL J¿JO NVUMBL EFóFSJO J¿JOJ ZBQBO EF- óFSMFSFHËSFJOUFHSBMJOTOSMBSQBS¿BMBOS -2 33 # #x - 2 dx + x + 1 dx –2 –2 2 3 –1 3 # # # #= 2 - x dx + x - 2 dx + - x - 1dx + x + 1dx –2 2 –2 –1 ÖRNEK 1 2 2 2 3 2 –1 2 3 x x - 2x +f - x -xp x = 2x - + + +x –2 2 22 –2 2 3 2 –1 # x dx JOUFHSBMJOJOTPOVDVLBÀUS 9 19 1 = 8 + - 6 + 2 + + + 3 + = 17 -2 2 22 2 3 03 # # #x dx = - x dx + x dx –2 –2 0 2 02 3 0 -x x & + –2 2 2 = 7 0 - ^ - 2 hA + = 9 - 0 G = 2 + 9 = 13 2 22 ÖRNEK 2 ÖRNEK 4 2 4 # x2 - x dx JOUFHSBMJOJOTPOVDVLBÀUS # x + x - 2 dx JOUFHSBMJOJOTPOVDVLBÀUS –1 -3 2 01 2 4 24 # # #x + x - 2 dx = x - x + 2 dx + 2x - 2 dx –3 –3 2 a x2 - x k dx + 2 2 # # # #2 a x - x k dx + a x - x k dx 24 x - x dx = # #= 2 dx + 2x - 2 dx –1 –1 0 1 –3 2 32 0 23 13 2 2 xx xx xx +- +- 11 2 4 &- –1 2 3 03 2 = = 2x 2 + - 2x 2 16 x 32 –3 = 10 + (16 - 8) - (4 - 4) = 10 + 8 = 18 13 11 33 3. 17 4. 18 1. 2. 2 6
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr ÖRNEK 5 ÖRNEK 8 2 2x + 3 , x < 0 F_ x i = * 4x + 3 , x $ 0 # x + 1 dx integralinin sonucu LBÀUS fonksiyonu veriliyor. -3 1 2 –1 2 #VOBHÌSF # Fa x k dx integralinin eöJUJLBÀUS # # #x + 1 dx = - x - 1 dx + x + 1dx –2 –3 –3 –1 2 –1 2 2 -x x = -x + +x 1 01 2 –3 2 –1 # # #Fa x k dx = F^ - x h dx + F^ x h dx =d - 1 +1 n-d -9 +3 n+4-d 1 -1 n –2 –2 0 22 2 0 13 1 13 # F^ - x h dx u = -Y EV= -dx = + +5- = –2 22 22 022 # # #- F^ u h du = F^ u h du = ^ 4x + 3 h dx ÖRNEK 6 200 2 21 # x - 1 . dx JGBEFTJOJOTPOVDVLBÀUS # #= ^ 4x + 3 h dx + ^ 4x + 3 h dx –2 00 2x2 + 3x 2 + 2x2 + 3x 1 00 2 12 14 + 2 + 3 = 19 # # #x - 1 dx = 1 - x dx + x - 1dx –2 –2 1 2 12 2 x x =x- + -x –2 2 2 1 =d 1- 1 n-^-2-2h+0-d 1 -1 n 22 11 = +4+ =5 22 ÖRNEK 7 ÷OUFHSBMWF5ÑSFW÷MJöLJTJ %m/*m 2 d # F_ x i dx = F_ x i # x2 - x dx JöMFNJOJOTPOVDVLBÀUS dx 0 # dF_ x i = F_ x i+ c 21 2 # f d F_ x i p dx = F_ x i + c # # #2 2 dx x )BUSMBUNB#FMJSMJJOUFHSBMMFSEFDZB[MNB[ - x dx = ^ - x2 + x h dx + ^ x - x h dx 34 8. 19 00 1 32 13 2 2 xx 1 -x x +- =+ 03 2 32 =d -1 + 1 n+d 8 -2 n-d 1 - 1 n 32 3 32 118 11 =- + + -2- + = 1 323 32 13 5. 6. 5 7. 1 2
www.aydinyayinlari.com.tr ÷/5&(3\"- 7. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 1 ÖRNEK 4 F (x) = d 9 # ^ x2 - 1 hdxC f : R Z3 ' -3 ) = -1 ve F ( 2 ) = 2 dx 2 3 PMEVôVOBHÌSF # F2_ x i.d _ F_ x i i EFôFSJLBÀUS PMEVôVOBHÌSF # F (x) dx LBÀUS –3 1 #F^ x h = d ^ x2 - 1 h dx 2 dx # F2 ^ x h.d F^ x h u =' Y EV= F'(x) dx –3 3 3 3 ^ h 2 –3 #& F^ h 2 - ^ 2 - h #& 2 u F x x = 1& 1 dx du = = x x u 1 33 33 = ^ 9 - 3 h - d 1 - 1 n = 20 = 8 -d -1 n= 3 x 33 & -x 31 33 ÖRNEK 2 ÖRNEK 5 f (x) = 3x - 1 3 F_ x i = x2 PMEVóVOBHËSF 2x + 1 x+1 PMEVôVOBHÌSF # da f-1^ x h kLBÀUS 2 1 # d_ F_ x i i integralinin sonucu LBÀUS #3 –1 –1 3 1 df ^ x h = f ^ x h 1 22 1 # d F^ x h = F^ x h & –1 ^ x h = -x - 1 11 f 2x - 3 22 -x - 1 3 -4 -d -2 n= - 10 & x 41 5 = -1 = =-= 2x - 3 1 3 3 x+1 1 3 2 6 ÖRNEK 3 ÖRNEK 6 F_ x i = # f x2 + 4 + x2 - 1 p dx d J 3 _ x2 + 1 i3.2xdx N integralinin sonucu LBÀUS dx K O x+3 KK # OO PMEVôVOBHÌSF 'h LBÀUS P L2 #F'^ x h = d f x2 + 4 + 2 + 1 p dx 3 dx x + 3 x #d f ^ x2 + 1 h3.2xdx p = ? & F' (x) = x2 + 4 + 2 - 1 dx x+3 x 2 55 ÷OUFHSBMJOTPOVDVCJSTBZÀLBS F' (1) = + 0 = d ^ say› h = 0 olur. dx 44 20 10 5 35 4. 3 5. 5 6. 0 1. 2. - 3. 6 3 3 4
TEST - 16 .VUMBL%FôFSWF1BSÀBM'POLTJZPOVO÷OUFHSBMJ 1. F_ x i = * x+2, x < 1 5. ôFLJMEFZ=' Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõUJS 2x + 2, x $ 1 fonksiyonu veriliyor. y 4 3 #VOB HÌSF # F_ x - 1 i dx integralinin sonucu 0 LBÀUS \" # $ % & 24 x O –3 y = F(x) 2. F ( x ) = 2x + 6 fonksiyonu veriliyor. 4 2 #VOB HÌSF # F_ x i .F'_ x i dx integralinin so- #VOB HÌSF # Fa x k dx integralinin sonucu 0 OVDVLBÀUS LBÀUS –2 A) - 27 C) - 25 2 2 #) –13 \" m # m $ % & E) - 23 2 % m 3 3 3. # x - 1 - 2 dx 6. # x2 - 4x + 4 dx 0 1 integralinin sonuDVLBÀUS integralinin sonucu BöBôEBLJMFSEFO IBOHJTJ- EJS A) 5 # $ 7 % & 9 2 2 2 \" # $ % & 3 % 46 56 5 3 E) 4. # x2 - 4 dx 7. # a x - 4 + x k dx 3 –3 3 JOUFHSBMJOJOTPOVDVLBÀUS integralinin sonucu kaÀUS \" # $ \" # $ % & 1. D 2. & 3. C 4. D 36 5. C 6. B 7. C
.VUMBL%FôFSWF1BSÀBM'POLTJZPOVO÷OUFHSBMJ TEST - 17 1. F_ x i = * 4x - 2 , x < 2 4 2x , x $ 2 fonksiyonu veriliyor. 4. # x - 2 dx 6 –2 #VOBHÌSF # F_ x - 3 i dx integralinin sonucu integralJOJO TPOVDV BöBôEBLJMFSEFO IBOHJTJ- EJS 0 BöBôEBLJMFSEFOIBOHJTJEJS \" # $ % & A) - # -12 C) - % -6 E) -3 2. ôFLJMEFZ=' Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõUJS 5. F_ x i = * 3x2 + 4 , x # 1 fonksiyonu veriliyor. y g_ x i, x > 1 –3 O x 44 –2 4 # F_ x i dx = 6 # g_ x i dx y = F(x) –1 1 –2 4 1 # F_ x i dx = - 4 , # F_ x i dx = 6 PMEVôVOBHÌSF # g_ x i dx integralinin sonucu –3 –2 4 PMEVóVOBHËSF BöBôEBLJMFSEFO hBOHJTJEJS A) - # - $ % & 44 6. ôFLJMEFZ=' Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõUJS # F_ x i dx - # F_ x i dx y –3 –3 4 y = F(x) integralinin sonucu aöBôEBkilerden hangisi- EJS x O 35 A) - # - $ % & –2 3 5 3. # 3x2 - 3 dx #VOBHÌSF # F_ x i .F'_ x i dx integralinin so- –2 0 integralinJO TPOVDV BöBôEBLJMFSEFO IBOHJTJ- OVDVBöBôEBLJMFSEFOIBOHJTJEJS EJS \" # $ % & \" # $ % & 1. \" 2. \" 3. D 37 4. C 5. B 6. &
TEST - 18 .VUMBL%FôFSWF1BSÀBM'POLTJZPOVO÷OUFHSBMJ 1. ôFLJMEFZ=' Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõUJS 4. F_ x i = * x - 2 , x < 2 y 4x + m , x $ 2 y = F(x) 3 1234 # F_ x i dx = 12 1 –3 x –2 O 5 PMEVôVOBHÌSF NLBÀUS A) 1 # $ 3 % & 5 2 2 2 #VOBHÌSF 2 F'_ x i 5 F'_ x i dx # dx + # 0 F'_ x i 4 F'_ x i JOUFHSBMJOJO TPOVDV BöBôEBLJlerden hangisi- EJS A) - # - $ % & 2x - 1, x < 2 5. F_ x i = * 3x2, fonksiyonu veriliyor. x$2 2 3 2. # a x2 - 2x + 1 - x + 1 kdx #VOB HÌSF # F_ - x i dx integralinin sonucu –1 –2 integraliOJO TPOVDV BöBôEBLJMFSEFO IBOHJTJ- BöBôEBLJMFSEFOIBOHJTJEJS EJS A) - # -18 C) - % -12 E) -10 \" # $ % & Z ] x-5 , x$4 3. F_ x i = [ ] x-3, x<4 PMEVóVOBHËSF 2, x#1 F_ x i = * - 1, \\ 6. x > PMEVóVOBHËSF 1 6 3 # F_ x i dx # x .F_ x i dx 2 –2 JOUFHSBMJOJO TPOVDV BöBôEBLJMFSEFO IBOHJTJ- EJS integralinin sonucu aöBôEBLJMFSEFO IBOHJTJ- EJS \" C $ % & A) - # $ % & 1. B 2. C 3. \" 38 4. C 5. & 6. C
.VUMBL%FôFSWF1BSÀBM'POLTJZPOVO÷OUFHSBMJ TEST - 19 1. F (x) = 1 PMEVóVOBHËSF 4. ôFLJMEF Z = F ( x - GPOLTJZPOVOVO HSBGJóJ WFSJM- x-1 NJõUJS 3 y # d a F–1_ x i k y = F(x–1) 6 2 5 JOUFHSBMJOJOTPOVDVLBÀUS 4 A ) - 1 # - 1 C ) - 1 % 1 E) 0 3 2 63 –2 O 3 x 2. y 1 y = F(x) #VOBHÌSF # F'_ x + 1 i dx integralinin sonu- 5 –2 DVBöBôEBLJMFSEFOIBOHJTJEJS 3 \" # $ % & 2 –3 3 x –2 O 5. ôFLJMEF Z = ' Y GPOLTJZPOVOVO HSBGJóJ WFSJMNJõ- :VLBSEBLJõFLJMEFZ=' Y GPOLTJZPOVOVOHSBGJóJ WFSJMNJõUJS tir. y 03 6 #VOB HÌSF # F'_ x i dx + # F'_ x i dx integrali- 4 –3 –2 y = F(x) niOTPOVDVBöBôEBLJMFSEFOIBOHJTJEJS \" # $ % & x –1 O 2 3. y 2 –2 O 135° x # x.F'_ x i dx = 10PMEVóVOBHËSF –2 3 –1 y = F(x) 2 # F_ x i dx –1 JOUFHSBMJOJO TPOVDV BöBôEBLJMFSEFO IBOHJTJ- EJS \" # $ % & :VLBSEBLJ õFLJMEF Z = F ( x ) fonksiyonunun grafi- 6. F_ x i = # x2 - 6 dx fonksiyonu veriliyor. óJWFSJMNJõUJS' Y GPOLTJZPOVOBY=BQTJTMJOPL- UBTOEBO¿J[JMFOUFóFUJYFLTFOJJMFMJLB¿ZBQ- 2x - 3 NBLUBES F ( x ) fonksiyonunun x =OPLUBTOEBLJUFôFUJ- OJOFôJNJBöBôEBLJMFSEFOIBOHJTJEJS 3 A) - # -2 C) - % & #VOB HÌSF # 7F' '_ x i + F'_ x i.F' '_ x iA dx in- –2 tegSBMJOJOTPOVDVBöBôEBLJMFSEFOIBOHJTJEJS A) - 1 # $ 1 % & 5 2 2 2 1. C 2. C 3. \" 39 4. \" 5. D 6. &
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr 3*&.\"//501-\".* TANIM ÖRNEK 1 B C` R ve a <CPMNBLÐ[FSF ôFLJMEFLJHSBGJLUF' Y = x2GPOLTJZPOVOVOHSBGJóJWFSJM- B Y0 < x1 < x2 < ... < xk–1 < xk < ... < xn C NJõUJS[ ] FõJUBMUBSBMóBBZSMZPS õFLMJOEFLJ TBZMBSO PMVõUVSEVóV LÐNFZF y y = F(x) <B C> BSBMóOO CJS iQBSÀBMBONBTu veya iCÌMÑOUÑTÑuEFOJS 1 \\Y0 Y1 Y2 Yk–1 Yn ^ õFLMJOEFLJ 1 O2 x CËMÐOUÐTÐOEF r <Yk–1 Yk >BSBMLMBSOBLBQBMBMUBSBML #VOBHÌSF IFTBQMBOBO a) 3JFNBOOBMUUPQMBNOCVMVOV[ r Dxk Yk – xk–1 TBZTOB CV BMU BSBMóO b) 3JFNBOOÐTUUPQMBNOCVMVOV[ c) )FSBMUBSBMóOPSUBOPLUBTOBHËSFIFTBQMBOBO V[VOMVóVWFZBCPZV b–a n 3JFNBOOUPQMBNOCVMVOV[ r )FSL`\\ O^J¿JO Dxk = ise 1CËlüntüsüne EÑ[HÑOCÌMÑOUÑEFOJS 1 \\x0 Y1 Y2 Yk–1 Yk Yn ^EÐ[HÐOCË- ¦Ì[ÑN lünUÐTÐOEF a) x1 – x0 Y2 – x1 Y3 – x2 Yn – xn–1 y EJS 4 y = F(x) 9 4 a = x0 C= xnPMNBLÐ[FSF [B C]BSBMóiOu 1 FõJUQBS¿BZBCËMÐOTÐO#VEVSVNEB 1 x 4 x =B Y=CEPóSVMBS YFLTFOJWF[B C] ara- O MóOEB Z = G Y GPOLTJZPOVOVO HSBGJóJ BMUOEB 1 1 32 22 LBMBOBMBOOZBLMBõLEFóFSJBõBóEBLJUPQMBN- MBSJMFIFTBQMBOBCJMJS n < > BSBMóO FõJU QBS¿BZB CËMFMJN &ó- / \"MUUPQMBN An = f^ xk–1 h.Dxk SJ BMUOEB LBMBO BMBO UBSBM EJLEËSUHFOMFSJO k=1 BMBOMBS UPQMBNOEBO CÐZÐLUÐS #V EJLEËSU- n HFOMFSJOBMBOMBSUPQMBNO\"4 BMUUPQMBN JMF HËTUFSFMJN ôFLJMEFLJ BMU EJLEËSUHFOMFSJO UB- / ¶TUUPQMBNÜn = f^ xk h.Dxk CBOV[VOMVLMBS 1 CJSJNEJS k=1 2 tk = xk–1 + xk PMNBLÐ[FSF 1 1 1 1 3 2 2 2 2 2 2 A4 = ·Ff p+ ·F^ 1 h + ·Ff p n A4 = 1·1 + 1 + 1·9 = 7 br2 olur. /Rn = f ^ tk h.Dxk ifBEFTJOF3JFNBOOUPQ- 24 2 24 4 k=1 MBNEFOJS #VUPQMBNMBSBSBTOEB\"n < Rn < ÜnFõJUTJ[- 3JFNBOOBMUUPQMBN 7 br2EJS MJóJWBSES 4 40
www.aydinyayinlari.com.tr ÷/5&(3\"- 7. MODÜL ·/÷7&34÷5&:&)\";*3-*, b) y F(x) = x2 NOT [ ]BSBMóOEBIB¿PLFõJUQBS¿BZBBZSEóN[- 4 EBHFS¿FLBMBOBEBIBZBLOCJSEFóFSCVMVSV[ 9 4 1 SONUÇ 1 y 4 y = F(x) O 1132 x 22 < > BSBMóO FõJU QBS¿BZB CËMFMJN &ó- SJ BMUOEB LBMBO BMBO UBSBM EJLEËSUHFOMFSJO BMBOMBS UPQMBNOEBO LпÐLUÐS #V EJLEËSU- HFOMFSJO BMBOMBS UPQMBNO ¶4 ÐTU UPQMBN x JMF HËTUFSFMJN ôFLJMEFLJ EJLEËSUHFOMFSJO UB- O a=x0 c1 x1 c2 x2 xn–1cn b=xn CBOV[VOMVLMBS 1 CJSJNEJS 2 Ü4 = 1 ·Fd 1 n+ 1 ·F^ 1 h + 1 ·Fd 3 n+ 1 F^ 2 h 2 2 2 2 2 2 = 1 · 1 + 1 ·1 + 1 · 9 + 1 ·4 j Ü = 15 br2 y =' Y GPOLTJZPOVOEB[B C] OBEFUFõJUBMU 24 2 24 2 44 BSBMóBCËMÐOTÐO olur. 3JFNBOOÐTUUPQMBN 15 br2EJS 4 D1 ` [ x0 x1] D2 ` [ x1 Y2 ] Dn ` [ xn-1 Yn ] D y F(x) = x2 ve Dx = b - a PMNBLÐ[FSFIFTBQMBOBO3JF- n 49 16 NBOOUPQMBN\" ise y =' Y FóSJTJOJOBMUOEB LBMBOBMBOZBLMBõLPMBSBL 25 16 9 A = DY' D1 ) + DY' D2) + ... + DY' Dn) 16 /n 1 A = TxF_ ck iPMBSBLJGBEFFEJMJS 16 1 1315 3 7 2 x O k=1 4 24 424 \"MUWFÐTUUPQMBNMBSCVMVSLFO <0 > aralóO [B C] EB ' Y GPOLTJZPOVOVO BMUOEB kalan P = * 0, 1 , 1, 3 , 2 4LÐNFTJOJOFMFNBOMBS BMBO [B C]OJOEBIBGB[MBBMUBSBMóBCËMÐONFTJ 22 EVSVNVOEB EBIB ZBLO EFóFSMFS WFSFDFóJ J¿JO JMF FõJU QBS¿BZB CËMÐQ IFTBQMBNBMBS ZBQ- UL 1 LÐNFTJOEF BSEõL FMFNBOMBSO BSJU- 3JFNBOOUPQMBNOJOTPOTV[BZBLMBõNBTEV- NFUJL PSUBMBNBMBSO CVMBSBL ZFOJ CJS 1h LÐ- NFTJFMEFFPE'F=MJN* 1 , 3 , 5 , 7 4 SVNVOEBZ=' Y JMFYFLTFOJBSBTOEBLBMBO 44 4 4 BMBOWFSFDFóJOEFOCVBMBO ôFLJMEFLJ EJLEËSUHFOMFSJO UBCBO V[VOMVLMBS n 1 bJSJNEJSôFLJMEFLJEJLEËSUHFOMFSJOBMBO- /lim 2 Dx.F^ ck h MJNJUJJMFIFTBQMBOS MBSUPQMBNO34 3JFNBOOUPQMBN JMFHËT- n\"3 UFSFMJN k=1 R = 1 ·Fd 1 n + 1 Fd 3 n + 1 Fd 5 n + 1 Fd 7 n #VSBEB' Y >PMEVóVOBEJLLBUFEJOJ[ 42 4 2 4 2 4 2 4 F ( x ) < 0 olNBT EVSVNVOEB ZVLBSEBLJ MJNJU EFóFSJ OFHBUJG PMBDBLUS #V EVSVNEB CV MJNJU EFóFSJGPOLTJZPOVOHSBGJóJJMFYFLTFOJBSBTOEB LBMBOBMBOOOFHBUJGEFóFSJOFFõJUPMBDBLUS 1 1 1 9 1 25 1 49 NOT R= · + · + · + · 4 2 16 2 16 2 16 2 16 R4 = 84 = 21 nb 2.16 8 /lim Dx.F^ c h = # f^ x h dx EJS :VLBSEBLJIFTBQMBNBMBSTPOVDVOEB n\"3 k k=1 a A4 < R4 < Ü4PMEVóVHËSÐMÐS 41
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr ÖRNEK 2 ÖRNEK 4 F : [ ] Z < >PMNBLÐ[FSF ôFLJMEFLJHSBGJLUF' Y = x2 +GPOLTJZPOVWFSJMNJõUJS y F(x) = x2 + 1 F ( x ) =YGPOLTJZPOVFöJUBMUBSBMôBCÌMFOEÑ[HÑO 1QBSÀBMBONBTOBBJUÑTUUPQMBN BMUUPQMBNEBOLBÀ CJSJNLBSFGB[MBES O1 x y ·TUUPQMBN 6 1.2 = 1.4 + 1.6 = 12 5 4 \"MUUPQMBN 2 1.2 + 1.4 = 6 [ ] BSBMô FöJU BMU BSBMôB BZSMSTB 3JFNBOO BMU 12 - 6 = 6 birimkare O 123 UPQMBNLBÀCJSJNLBSFPMVS x y \"4 = 1F(1)+ 1.F(2)+1.F(3)+1.F(4) \"4 = 1.2 + 1.5 + 1.10 + 1.17 = 2 + 5 + 10 + 17 = 34 x O 12345 ÖRNEK 3 y :BOEBLJHSBGJLUF ÖRNEK 5 13 x F ( x ) = -x2 + 1 fonksiyo- 3 F : [ ] Z < >PMNBLÐ[FSF O OVWFSJMNJõUJS F ( x ) = -x2 + 12x +GPOLTJZPOVOVFöJUBMUBSB- #VFôSJJMFYFLTFOJBSB- MôB CÌMFO EÑ[HÑO 1 QBSÀBMBONBTOB BJU BMU UPQMBN LBÀCJSJNLBSFEJS TOEB LBMBO CÌMHFOJO BMBOO [ ] O FöJU F(x) = –x2 + 1 BMU BSBMôB CÌMFSFL 3JF- NBOOÑTUUPQMBNZBSE- NZMBZBLMBöLPMBSBLIFTBQMBZO[ y y 43 1.28 + 35 + 1.40 + 1.43 3 5 7 9 11 13 x 40 = 146 birimkare 35 O Tx = 13 - 3 =2 28 5 x O 23456 \"= Dx.F(3) + Dx.F(5) + Dx.F(7) + Dx.F(9) + DxF(11) \"= 2.(-8) + 2.(-24) + 2.(-48) + 2.(-80) + 2(-120) \"= -560 y = ' Y GPOLTJZPOVOVO HSBGJôJ Y FLTFOJOJO BMUOEB LBM- EôOEBO 3JFNBOO ÑTU UPQMBN OFHBUJG PMVS #VOB HÌSF BMBOZBLMBöLPMBSBLCS2 dir. 2. 34 3. 560 42 4. 6 5. 146
3JFNBOO5PQMBN TEST - 20 1. \"öBôEBLJ QBSÀBMBONBMBSEBO IBOHJTJ EÑ[HÑO 5. [ ] BSBMóOEB UBONM CJSFCJS WF ËSUFO CJS G Y CJSQBSÀBMBONBMES 5 A) P1 = * 0, 1 , 1, 3 ,34 fonksiyonu için # f^ x h dx = 7 olarak verJMNJõUJS 2 2 2 # P = \" 1, 2, 3, 4, 6 , [ ]BSBMôOiOuFöJUQBSÀBZBCÌMFOEÑ[HÑOCJS 2 1 QBSÀBMBONBTOB BJU 3JFNBOO BMU UPQMBNOO C) P = * 2, 7 , 8 ,3, 10 4 BMBCJMFDFôJ FO CÑZÑL UBN TBZ EFôFSJ iBu 3JF- 3 3 3 NBOO ÑTU UPQMBNOO BMBCJMFDFôJ FO LÑÀÑL UBN 3 TBZ EFôFSJ iCu PMEVôVOB HÌSF B + C UPQMBN LBÀUS % P = * 0, 1 , 3 , 1 4 4 55 \" # $ % & E) P = * 1, 5 , 3 ,24 4 2 5 2. f : [ ] Z3UBONMG Y = 3x fonksiyonu veriliyor. 6. f : [ ] Z [ ]PMNBLÐ[FSF f ( x ) = x2GPOLTJZPOVOVOUBONMPMEVóVBSBMóJLJ FõJUQBS¿BZBCËMFOEÐ[HÐOCJS1QBS¿BMBONBTZB- QMZPS #VOBHÌSF BSBMôFöJUQBSÀBZBCÌMFOEÑ[HÑO #VOBHÌSF PMVöBO3JFNBOOBMUUPQMBNOO 3JF- CJS1QBSÀBMBONBTOBBJUBMUUPQMBNLBÀCJSJNLB- NBOOÑTUUPQMBNOBPSBOLBÀUS SFEJS A) 7 # 13 C) 13 % 16 25 25 25 E) A) 5 $ 9 % & 13 29 29 29 2 2 2 # 7. 3JFNBOOUPQMBN /n 1 1 16 1 81 1 1 F^ zK h DxK = 4 . n + 4 · n + 4 . n + . . . + 1· n k=1 n n n 3. f : [ ] Z3UBONMG Y = x2 + 1 fonksiyonu ve- PMBOCJSGPOLTJZPOVOVOJOUFHSBMJOJOEFôFSJLBÀ- riliyor. US #VOB HÌSF [ ] BSBMôO FöJU QBSÀBZB CÌMFO A) 1 # 1 C) 1 % 1 1 EÑ[HÑOCJS1QBSÀBMBONBTOBBJUÑTUUPQMBNLBÀ 6 5 43 E) CJSJNLBSFEJS 2 \" # $ % & 4. f ( x ) = x3 fonksiyonu x ekseni ve x =EPôSVTV m BSBTOEBLBMBOCÌMHFOJO[ ]BSBMôOEBiOuBMU 8. # _ m + k i dx = 24 [L N] BSBMó FõJU QBS¿BZB BSBMôBCÌMÑOEÑôÑOEFOZßJÀJO3JFNBOOUPQ- MBNLBÀUS k \" # $ % & BZSMEóOEB CV CËMÐOUÐOün HFOJõMJóJ CJSJN PMEV- k óVOBHËSF # 2x dx JOUFHSBMJOJOTPOVDVBöBô- 0 EBLJMFSEFOIBOHJTJEJS \" # 3 C) 5 % 25 25 2 E) 4 16 4 1. C 2. C 3. \" 4. \" 43 5. C 6. B 7. B 8. &
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr #&-÷3-÷÷/5&(3\"-÷-&\"-\"/)&4\"#* bx A2 %m/*m %m/*m F : [B C] Z3 Z= F ( x ) fonksiyonu integralle- y = F(x) y OFCJMFOCJSGPOLTJZPOPMNBLÐ[FSF A1 y =' Y GPOLTJZPOVOVOHSBGJóJ Y= a ve x = b Oa c EPóSVMBSJMFYFLTFOJBSBTOEBLBMBOTOSMCËM- HFOJOBMBO b A = # F_ x i dx integrBMJJMFIFTBQMBOS a y y = F ( x ) fonksiyonu [B C]OEBQP[JUJGWFOFHB- y = F(x) UJGEFóFSMFSJOIFSJLJTJOJEFBMZPSTB F_ x i = * - F_ x i , a # x # c ise F_ x i , c < x # b ise A Oa b x PMEVóVOEBO' Y GPOLTJZPOVOVn grBGJóJ x = a ve x =CEPóSVMBSJMFYFLTFOJBSBTOEB LBMBOCËMHFOJOBMBO y = F ( x ) fonksiyonu [B C]OEBQP[JUJGEFóFS- cb MJJTFZBOJ' Y GPOLTJZPOVOVOHSBGJóJ Y= a ve A = A1 + A2 = # F_ x i dx - # F_ x i dx x = C EPóSVMBS JMF Y FLTFOJ UBSBGOEBO TOSMB- OBO CËMHFOJO BMBO Y FLTFOJOJO Ð[FSJOEF LBM- ac | |yosa F ( x ) =' Y PMBDBóOEBOCVCËMHFOJO integrBMJJMFIFTBQMBOS BMBO UYARI b y A = # F_ x i dx integrBMJJMFIFTBQMBOS a y A2 a bx a A1 b O x O c y = F(x) A :VLBSEBZ=' Y GPOLTJZPOVOVOHSBGJóJWFFL- y = F(x) TFOMFSBSBTOEBLBMBOTOSMCËMHFMFSJOBMBOMBS y = F ( x ) fonksiyonu [B C]OEBOFHBUJGEFóFS- A1 ve A2PMNBLÐ[FSF MJJTFZBOJ' Y GPOLTJZPOVOVOHSBGJóJY= a ve x =CEPóSVMBSJMFYFLTFOJUBSBGOEBOTOSMB- c OBOCËMHFOJOBMBOYFLTFOJOJOBMUOEBLBMZPS- # F_ x i dx integraMJOJOEFóFSJ | |sa F ( x ) = -' Y PMBDBóOEBO CV CËMHFOJO a olur. BMBO c b # F_ x i dx = - A1 + A2 A = - # F_ x i dx integrBMJJMFIFTBQMBOS a a [B D]OEBF ( x )GPOLTJZPOVOVOHSBGJóJJMFYFL- TFOJBSBTOEBLBMBOTOSMCËMHFOJOBMBO c olur. # F_ x i dx = A1 + A2 a 44
www.aydinyayinlari.com.tr ÷/5&(3\"- 7. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 1 ÖRNEK 4 15 br2 y (SBGJLUF WFSJMFOMFre y = -x2 +YQBSBCPMÐ –3 y = F(x) x = - Y= -1 HÌSF 5 EPôSVMBS WF Y FLTFOJ JMF TOSM CÌMHFOJO BMBO LBÀ 2 5 br2 5 br2EJS x # F (x) dx LBÀUS –3 5 y x 2 # F^ x h dx = 15 - 5 = 10 –2 –1 O –3 y =–x2 + 2x –1 3 –1 x #A = - ^ - 2 + 2x h dx = 2 x 3 -x ÖRNEK 2 –2 –2 y = d - 1 - 1 n - d - 8 - 4 n = - 4 + 20 = 16 y = F(x) 3 3 333 S1 S3 ÖRNEK 5 3 S2 x y = -x2 + 1 7 QBSBCPMÑWFYFLTFOJJMFTOSMCÌMHFOJOBMBOLBÀ 7 br2EJS # F (x) dx = 15 br2 S1 = 7 br2 S2 = 6br2 -3 PME VôVOBHÌSF 43LBÀCS2EJS y1 -x 31 #1 ^ - 2 h = +x A= + 1 dx x –1 3 –1 7 –1 1 x =d -1 +1 n-d 1 -1 n # F^ x h dx = S - S + S = 15 O 33 123 –3 22 4 =+= = 7 - 6 +43 = 15 j43 = 14 33 3 ÖRNEK 3 F(x) = 2x2 (SBGJLUFLJ UBSBM CÌM ÖRNEK 6 x HFOJO BMBOO CV y lunuz. y = x3 - 4x 3 FôSJTJ WF Y FLTFOJ JMF TOSM CÌMHFOJO BMBO LBÀ CS2 –1 EJS y 0 #A = 2 ^ x3 - 4x h dx x–24 0 3 = 2f - 2 p 2x x 4 –2 #5BSBMBMBO= 2 –2 O 2 2x dx –1 = 2(0 - (4 - 8)) = 8 3 3 2x = 2 ^ 27 - ^ - 1 h h = 56 & –1 3 3 3 1. 10 56 45 16 4 6. 8 2. 14 3. 4. 5. 33 3
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- ÖRNEK 9 www.aydinyayinlari.com.tr y y = F(x) ÖRNEK 7 y = x3 -FôSJTJY= -1 ve x =EPôSVMBSJMF0YFL- TFOJOJOPMVöUVSEVôVBMBOLBÀCJSJNLBSFEJS y y = x3–1 –3 A1 O A3 x –1 45 A2 –1 O x 12 –1 12 :VLBSEBLJöFLJMEF\"1 =CJSJNLBSF \"2 = 9 birimka- SFWF\"3 =CJSJNLBSFPMEVôVOBHÌSF BöBôEBLJJO- # #A = ^ - x3 + 1 hdx + ^ x3 - 1 h dx UFHSBMMFSJOEFôFSMFSJOJCVMVOV[ –1 1 4 5 - x4 14 2 a) # f_ x i dx b) # F_ x i dx x –3 –3 = +x + -x 44 4 5 –1 1 D # F_ x i dx E # F_ x i dx =d - 1 +1 n-d - 1 -1 n+^4-2h-d 1 -1 n –1 –3 44 4 35 3 19 4 45 = + +2+ = 44 44 e) # F_ x i dx f) # F_ x i dx - # F_ x i dx 5 –1 4 a) 3 - 9 = -6 b) 3 - 9 + 4 = -2 c) -9 d) 3 + 9 + 4 = 16 e) -4 f) -9 - 4 = -13 ÖRNEK 8 :BOEBLJõFLJMEFZ= F ( x ) ÖRNEK 9 EPóSVTBM GPOLTJZPOVOVO y HSBGJóJWFSJMNJõUJS y 4 –1 O :BOEBLJõFLJMEF 6x y = F(x) F_ x i = x y = F(x) O2 _ x2 + 1 i2 5BSBMCÌMHFOJOBMBOOJOUFHSBMZBSENJMFCVMVOV[ 2 x fonksiyonunun gra- GJóJWFSJMNJõUJS xy #VOBHÌSF UBSBMCÌMHFOJOBMBOLBÀCJSJNLBSFEJS %PôSVEFOLMFNJ 6 + 4 = 1 02 x ^2h ^3h x dx dx + 12 - 2x # #A = - 2x + 3y = 12 j y = –1 ^ x2 + 1 h2 ^ 2 h + 1 3 x 6 0 #A = d 12 - 2x n dx u = x2+ EV= 2x dx 3 2 15 # #1 du 1 du 13 6 -+ = 1 ^ 12x - x2 h = 1 16 2 2 2 2 20 & ^ ^ 72 - 36 h - ^ 24 - 4 h h = 2 u 1 u 3 23 3 19 16 46 9. B m C m D m E F m G m 13 7. 8. 10. 20 4 3
#JS'POLTJZPOVO(SBGJôJJMFY<FOJ\"SBTOEB,BMBO#ÌMHFOJO\"MBO TEST - 21 1. ôFLJMEF41 42 43CVMVOEVLMBSCËMHFMFSJOBMBOMBS- 4. ôFLJMEF G Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõtir. ES y y –1 S1 S3 x S1 y = f(x) –3 2 S2 4 x 4 S2 7 10 S1 ve S2J¿JOEFCVMVOEVóVCËMHFMFSJOBMBOMBSO f(x) HËTUFSNFLÐ[FSF S1 = 21 br2 42 = 16 br2 ve 4 10 # f (x) dx = 6 ve S1 = 9 br2 # f(x) dx = 16 -3 –1 PMEVôVOBHÌSF 42 kBÀCS2EJS PMEVôVOBHÌSF 43BMBOLBÀbr2EJS \" # $ % & \" # $ % & 2. ôFLJMEFZ=G Y JOHSBGJóJWFSJMNJõUJS y y = f(x) A 12 C x –3 B 57 6 5. y YFLT FOJOJO \"#ZBZJMFTOSMBEóCËMHFOJOBMBO S2 5 x 12 br2 #$ZBZJMFTOSMBE óCËMHFOJOBMBO5 br2 –6 S1 1 y = f(x) 6 ôFLJMEF S = 12 br2 42 = 6 br2 ise PMEVôVOBHÌSF # f (x) dx LBÀUS 1 –3 5 \" # $ % & # af^xh + f^xh kdx -6 integrBMJOJOEFôFSJLBÀUS A ) - # - $ % & 3. y ôFLJMEF \"1 \"2 CVMVOEVLMBS CËM- –3 –2 A1 O 4x HFOJO BMBOO HËTUFSNFLUFEJS A2 A1 = 3 br2 \"2 = 8 br2 ve 6. y = x2 –2 4 QBSBCPMÑJMFZ= Y= 1 ve x =EPôSVMBSBSB- TOEBLBMBOBMBOLBÀCS2EJS # f(x) dx = - 5 ise # f_ x i dx \" # $ % & –3 –3 E ) 16 JOUFHSBMJOJOEFôFSJLBÀUS A ) - # -8 C ) -5 % 1. \" 2. & 3. \" 47 4. & 5. D 6. C
·/÷7&34÷5&:&)\";*3-*, 7. MODÜL ÷/5&(3\"- www.aydinyayinlari.com.tr ÷,÷'0/,4÷:0/6/(3\"'÷ó÷\"3\"4*/%\",\"-\"/#²-(&/÷/\"-\"/* %m/*m UYARI y y = f(x) y y = g(x) A1 b A1 y = g(x) x a Ob A2 A2 aO y = f(x) x c :VLBSEBHËTUFSJMFOG Y WFH Y GPOLTJZPOMBS- &óFSG Y WFH Y GPOLTJZPOMBSOOHSBGJLMFSJZV- OOHSBGJLMFSJJMFY= a ve x =CEPóSVMBSBSBTO- LBSEBLJ HJCJ WFSJMJSTF CPZBM CËMHFMFSJO UPQMBN EBLBMBOTOSMCËMHFOJOBMBO\"1 H Y GPOLTJ- BMBO \"1 ve A2BMBOMBSOOBZSBZSIFTBQMBOB- ZPOVOVOHSBGJóJY= a ve x =CEPóSVMBSJMFY SBLUPQMBONBTJMFCVMVOVS FLTFOJBSBTOEBLBMBOTOSMCËMHFOJOBMBO\"2 bc PMNBLÐ[FSF # #A = A + A = ^ f^ x h - g^ x h h dx + ^ g^ x h - f^ x h hdx b x i dx = A1 + A2 _ 12 b ab # f_ b bb a ` bb # g_ x i dx = A2 b bb a a bb & A1 = # f_ x idx - # g_ x idx aa b ÖRNEK 1 & A1 = # f_ x i - g_ x i dx y 3 a Sonuç olarak [B C]OEBJLJGPOLTJZPOVn grafJóJ –1 O x 3 BSBTOEB LBMBO BMBO CVMNBL J¿JO ÐTUUFLJ GPOL- TJZPOVOVO EFOLMFNJOEFO BMUUBLJ GPOLTJZPOVO EFOLMFNJ¿LBSMBSBLJOUFHSBMJBMOS :VLBSEBLJHSBGJLUFQBSBCPMJMFEPôSVBSBTOEBLBMBO CÌMHFOJOBMBOLBÀCJSJNLBSFEJS 1BSBCPMEFOLMFNJZ= -1 (x + 1) (x - 3) %PôSVEFOLMFNJZ= -x + 3 3 #5BSBMBMBO= 7 -^ x + 1 h^ x - 3 h - ^ - x + 3 hA dx 0 3 32 3 a - x2 + 3x k dx = - x 3x #= + = d - 9 + 27 n = 9 32 0 22 0 48 9 1. 2
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