ࡱ>  }G` n1bjbj \<( ;;;8;>o2? HM"jMjMM.TbT$T$h nحQTS|*TTTحjMM.)LzLzLzTljMMLzTLzLz* f,M? `M;rl J<?0o4w6vwxwdTTLzTTTTTححyjTTToTTTT2;; INTEGRAL INTEGRAL TAK TENTU Definisi : Fungsi F dikatakan anti turunan dari fungsi f pada selang I jika F(x) = f(x) untuk semua x di I. Notasi : F(x) =  EMBED Equation.3 f(x) dx Integral tak tentu adalah Anti/Invers/Kebalikan turunan. Contoh :  EMBED Equation.3   EMBED Equation.3  Integral tak tentu adalah operator linear, yaitu bersifat :  EMBED Equation.3  =  EMBED Equation.3   EMBED Equation.3  =  EMBED Equation.3  +  EMBED Equation.3  Rumus-rumus Dasar Integral Tak Tentu 1.  EMBED Equation.3 , n `" - 1 2.  EMBED Equation.3  3.  EMBED Equation.3  4.  EMBED Equation.3  5.  EMBED Equation.3  6.  EMBED Equation.3  7.  EMBED Equation.3  8.  EMBED Equation.3  9.  EMBED Equation.3  10.  EMBED Equation.3  11.  EMBED Equation.3  12.  EMBED Equation.3  13.  EMBED Equation.3  Contoh :  EMBED Equation.3  INTEGRAL TENTU Definisi : Misal f fungsi yang didefinisikan pada [a,b], f dikatakan terintegralkan pada [a,b] jika  EMBED Equation.3  ada, selanjutnya  EMBED Equation.3  disebut Integral Tentu (Integral Riemann) f dari a ke b, dan didefinisikan  EMBED Equation.3  =  EMBED Equation.3 . INCLUDEPICTURE "../Documents%20and%20Settings/PC%20Dosen-01/Local%20Settings/Temp/slsmetnum/slscalus/CALCULUS_files/Image406.gif" \* MERGEFORMAT   EMBED Equation.3  menyatakan luas daerah yang tercakup diantara kurva y = f(x) dan sumbu x dalam selang [a,b], jika  EMBED Equation.3 bertanda negatif maka menyatakan luas daerah yang berada dibawah sumbu x. Definisi :  EMBED Equation.3  = 0  EMBED Equation.3  = -  EMBED Equation.3 , a > b Teorema Dasar Kalkulus Teorema Dasar Kalkulus memberikan kemudahan untuk menghitung Integral Tentu, berikut teorema tersebut : Misal f kontinu pada [a,b] dan F sebarang anti turunan f, maka  EMBED Equation.3  = F(b) F(a) Selanjutnya ditulis F(b) F(a) =  EMBED Equation.3  Contoh : Perlihatkan bahwa jika r ( Q dan r ( -1, maka  EMBED Equation.3  Jawab : Karena F(x) =  EMBED Equation.3  suatu anti turunan dari f(x) = xr, maka menurut TDK,  EMBED Equation.3  Integral tentu sebagai operator linear, yaitu bersifat : Misal f dan g terintegralkan pada [a,b] dan k suatu konstanta, maka kf dan f + g terintegralkan, dengan  EMBED Equation.3  k   =  +  EMBED Equation.3  Contoh : Hitung  EMBED Equation.3  Jawab :  EMBED Equation.3  = 4 EMBED Equation.3  = 4 EMBED Equation.3  = ( 12 Sifat-Sifat Integral Tentu Sifat Penambahan Selang Teorema : Jika f terintegralkan pada suatu selang yang mengandung tiga titik a, b dan c, maka  EMBED Equation.3  =  EMBED Equation.3  +  EMBED Equation.3  bagaimanapun urutan a, b dan c. Contoh :  EMBED Equation.3  2.  EMBED Equation.3  EMBED Equation.3  3.  EMBED Equation.3  Sifat Simetri Teorema : Jika f fungsi genap [f(-x) = f(x)] , maka EMBED Equation.3  = 2  EMBED Equation.3  dan Jika f fungsi ganjil [f(-x) = - f(x)], maka  EMBED Equation.3  = 0. Contoh :  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  = 0 TEKNIK-TEKNIK PENGINTEGRALAN Teknik Subtitusi Biasa a. Subtitusi Dalam Integral Tak Tentu Teorema : Misal g fungsi yang terdiferensialkan dan F suatu anti turunan dari f, jika u = g(x) maka EMBED Equation.3 f(g(x))g(x) dx =  EMBED Equation.3 f(u) du = F(u) + c = F(g(x)) + c Prinsipnya: dimunculkan variabel baru menggantikan variabel lama, sehingga soal dibawa ke rumus yang tersedia. Ketika variabel baru muncul, variabel lama hilang dari semua, dan soal integral dapat diselesaikan menggunakan rumus yang ada. Contoh 1 : Hitunglah EMBED Equation.3 . Soal ini tidak dapat langsung dengan rumus (tidak ada rumus yang sesuai). Deangan pemisalan dimunculkan variabel baru sebagai sustitusi kompoen variabel lama, diperoleh integral yang dapat diselesaikan dengan rumus yang sesuai. Jawab : Misalkan u =  EMBED Equation.3  = x1/2 sehingga du =  EMBED Equation.3  dx maka  EMBED Equation.3  = 2 EMBED Equation.3 = 2 EMBED Equation.3  = 2cosu + c = 2cos EMBED Equation.3  + c Contoh 2. Hitunglah EMBED Equation.3  DImisalkan  EMBED Equation.3  ini disubstitusikan ke soal sehingga  EMBED Equation.3  Dapat diselesaikan dengan rumus pertama integral  EMBED Equation.3  Catatan. Setelah dipilih yang dimisalkan, diperoleh yang akan menggantikan dx Setelah substitusi, variable lama harus hilang semua, dan soal sesuai dengan salah satu rumus yang ada Jika setelah substitusi masih ada variable lama, berarti salah memilih yang dimisalkan, ulangi lagi dengan pemisalan komponen yang lain b. Subtitusi Dalam Integral Tentu. Teorema : Misal g mempunyai turunan kontinu pada [a,b] dan f kontinu pada daerah nilai g, maka EMBED Equation.3  Contoh : Hitung  EMBED Equation.3  Jawab : Misal u = x2+2x+6 sehingga du = 2x+2 dx = 2(x+1)dx perhatikan u = 6 jika x = 0 dan u = jika x = 1, jadi  EMBED Equation.3  =  EMBED Equation.3  =  EMBED Equation.3  =  EMBED Equation.3  Contoh lain : Tentukan integral dari :  EMBED Equation.3  Penyelesaian : a. Misal : EMBED Equation.3  Maka:  EMBED Equation.3  Sehingga :  EMBED Equation.3  b. Misal u = sin x  EMBED Equation.3  Sehingga :  EMBED Equation.3  2. Pengintegralan Parsial Pengintegralan parsial (sebagian) dapat dilakukan jika pengintegralan dengan teknik subtitusi tidak memberikan hasil, dan dengan catatan bagian sisa pengintegralan lebih sederhana dari integral mula-mula.  EMBED Equation.3  Contoh : 1.  EMBED Equation.3  EMBED Equation.3  Misalkan u = x, dv = ex dx maka du = dx , v = ex  EMBED Equation.3  =  EMBED Equation.3  = xex ex + c Contoh lain : Tentukan :  EMBED Equation.3  Penyelesaian : a. Misal 2x = u maka 2 dx = du Misal dv =  EMBED Equation.3   EMBED Equation.3  b. Misal x = u maka dx = du Misal dv = sin x dx maka v = -cos x  EMBED Equation.3  c. Perlu dimisalkan 2 kali Misal  EMBED Equation.3  ( EMBED Equation.3  Integral ruas kanan perlu dimisalkan lagi  EMBED Equation.3  ( EMBED Equation.3  SOAL GSLC (untuk kelas 02PDT, 02PIT, 02PLT) Kerjakan soal-soal berikut sebagai tugas kelompok 1 kelompok maksimum 5 mahasiswa Jangan lupa nomor absen Jangan terlambat, batas akhir menjelang f2f berikutnya I. Substitusi 1.  EMBED Equation.3  2.  EMBED Equation.3  3.  EMBED Equation.3  4.  EMBED Equation.3  5.  EMBED Equation.3  6.  EMBED Equation.3  7.  EMBED Equation.3  8.  EMBED Equation.3  9.  EMBED Equation.3  10.  EMBED Equation.3  II. Integral Parsial 1. EMBED Equation.3  2. EMBED Equation.3  3. EMBED Equation.3  4.  EMBED Equation.3  5.  EMBED Equation.3  3. Pengintegralan Bentuk-Bentuk Trigonometri a.  EMBED Equation.3 sin n x dx,  EMBED Equation.3 cos n x dx Jika n bilangan bulat positif ganjil, maka keluarkan faktor sin x atau cos x dan kemudian gunakan kesamaan sin 2 x + cos 2 x = 1. Jika n bilangan bulat positif genap, maka gunakan rumus setengah sudut sin 2 x =  EMBED Equation.3  , cos 2 x =  EMBED Equation.3  Contoh :  EMBED Equation.3 cos 4 x dx =  EMBED Equation.3  =  EMBED Equation.3 (1 + 2 cos 2x + cos 2 2x) dx =  EMBED Equation.3 dx +  EMBED Equation.3 cos 2x (2) dx +  EMBED Equation.3 (1 + cos 4x) dx =  EMBED Equation.3 x +  EMBED Equation.3 sin 2x +  EMBED Equation.3  sin 4x + c b.  EMBED Equation.3 sin m x cos n x dx Jika m atau n bilangan bulat positif ganjil dan eksponen lain sembarang, maka keluarkan faktor sin x atau cos x yang berpangkat ganjil tersebut kemudian gunakan kesamaan sin 2 x + cos 2 x = 1. Jika m dan n bilangan bulat positif genap, maka gunakan rumus setengah sudut. Contoh : Tentukan : 1. EMBED Equation.3 sin 3 x cos 4 x dx 2. EMBED Equation.3  sin 2 x cos 4 x dx c.  EMBED Equation.3 tg n x dx,  EMBED Equation.3 cotg n x dx. Keluarkan faktor tg 2 x = sec 2 x 1 dalam kasus tg atau faktor cotg 2 x = cosec 2 x 1 dalam kasus cotg. Contoh :  EMBED Equation.3 cotg 4 x dx =  EMBED Equation.3 cotg 2 x (cosec 2 x 1) dx =  EMBED Equation.3 cotg 2 x cosec 2 x dx  EMBED Equation.3 cotg 2 x dx = - EMBED Equation.3 cotg 2 x d(cotg x) -  EMBED Equation.3 (cosec 2 x 1) dx = - EMBED Equation.3 cotg 3x + cotg x + x + c d.  EMBED Equation.3 tg m x sec n x dx,  EMBED Equation.3 cotg m x cosec n x dx Jika n genap dan m sembarang, maka keluarkan faktor sec 2 x atau cosec 2 x. Jika m ganjil dan n sembarang, keluarkan faktor tg x.sec x. Contoh : Tentukan : 1.  EMBED Equation.3 tg 3/2 x sec 4 x dx 2.  EMBED Equation.3 tg 3 x sec 1/2 x dx e.  EMBED Equation.3 sin mx cos nx dx,  EMBED Equation.3 sin mx sin nx dx,  EMBED Equation.3 cos mx cos nx dx. Gunakan kesamaan : sin mx cos nx = [sin (m+n)x + sin (m n)x] sin mx sin nx = -[cos (m+n)x - cos (m n)x] cos mx cos nx = [cos (m+n)x + cos (m n)x] Contoh :  EMBED Equation.3 sin 2x cos 3x dx = 1/2 EMBED Equation.3 sin 5x + sin (-x) dx = 1/10  EMBED Equation.3 sin 5x d(5x)  EMBED Equation.3 sin x dx = - 1/10 cos 5x + cos x + c. 3. Pengintegralan Parsial Pengintegralan parsial (sebagian) dapat dilakukan jika pengintegralan dengan teknik subtitusi tidak memberikan hasil, dan dengan catatan bagian sisa pengintegralan lebih sederhana dari integral mula-mula.  EMBED Equation.3  Contoh : 1.  EMBED Equation.3  EMBED Equation.3  Misalkan u = x, dv = ex dx maka du = dx , v = ex  EMBED Equation.3  =  EMBED Equation.3  = xex ex + c 4. Integral Fungsi Akar (Subtitusi yang Merasionalkan). a. Fungsi Integran yang memuat bentuk  EMBED Equation.3  Penyelesaian dengan menggunakan subtitusi : u =  EMBED Equation.3  Contoh : Hitung  EMBED Equation.3  Jawab : Misalkan u =  EMBED Equation.3  maka  EMBED Equation.3  = x 4 dan 3 EMBED Equation.3 du = dx Shg  EMBED Equation.3 =  EMBED Equation.3  b. Integran yang memuat bentuk  EMBED Equation.3  Gunakan berturut-turut subtitusi : x = a sin t, x = a tg t dan x = a sec t. Contoh : Tentukan  EMBED Equation.3  Jawab : Jawab : Misalkan x = 2 sin t maka dx = 2 cos t dt dan  EMBED Equation.3 = 2 cos t , shg  EMBED Equation.3  =  EMBED Equation.3  = - ctg t t + c =  EMBED Equation.3  5. Integral Fungsi Rasional Fungsi Rasional merupakan fungsi hasil bagi dua fungsi Polinom yang ditulis :  EMBED Equation.3  , P(x) dan Q(x) fungsi  fungsi Polinom dengan Q(x) `" 0 Fungsi Rasional dibedakan atas : Fungsi Rasional Sejati yaitu fungsi rasional dimana derajat fungsi polinom pada pembilang lebih kecil dari pada derajat fungsi polinom pada penyebut. Fungsi Rasional Tak Sejati yaitu fungsi rasional dimana derajat fungsi polinom pada pembilang lebih besar dari atau sama dengan derajat fungsi polinom pada penyebut. Fungsi Rasional Tak Sejati dapat ditulis sebagai penjumlahan fungsi polinom dengan Fungsi Rasional Sejati dengan jalan membagi fungsi pembilang dengan fungsi penyebut. Permasalahan mengintegralkan fungsi rasional terletak pada bagaimana mengintegralkan fungsi rasional sejati. Suatu fakta, bahwa fungsi rasional sejati dapat ditulis sebagai jumlah dari fungsi rasional sejati yang lebih sederhana Contoh :  EMBED Equation.3  a. Penjabaran Fungsi Rasional atas Faktor Linear yang Berbeda Contoh : Tentukan  EMBED Equation.3  Jawab :  EMBED Equation.3  maka 5x + 3 = A(x+1)(x-3) + Bx(x-3) + Cx(x+1) dengan menyamakan koefisien pada kedua polinom diruas kiri dan ruas kanan maka diperoleh : A = -1 , B =  EMBED Equation.3  , dan C =  EMBED Equation.3  sehingga  EMBED Equation.3 =  EMBED Equation.3  = - ln  EMBED Equation.3  b. Penjabaran Fungsi Rasional atas Faktor Linear yang Berulang Contoh : Tentukan  EMBED Equation.3  Jawab :  EMBED Equation.3  maka x = A(x-3) + B dengan menyamakan koefisien pada kedua polinom diruas kiri dan ruas kanan diperoleh : A = 1 dan B = 3 sehingga  EMBED Equation.3  Yang perlu diperhatikan untuk tiap faktor  EMBED Equation.3  dalam penyebut, maka ada sebanyak k suku penjabarannya, yaitu :  EMBED Equation.3  c. Penjabaran Fungsi Rasional atas Faktor Kuadrat yang Berbeda Contoh : Tentukan  EMBED Equation.3  Jawab :  EMBED Equation.3  Selanjutnya tentukan A, B dan C seperti cara diatas dan kemudian hitung integral setiap sukunya. PENGGUNAAN INTEGRAL TERTENTU Luas Daerah Bidang Rata a. Daerah Antara Kurva dan Sumbu Koordinat. Perhatikan gambar daerah rata dibawah ini  Daerah R dibatasi oleh grafik-grafik y = f(x), x = a, x = b dan y = 0, luasnya A(R) ditentukan oleh : A(R) =  EMBED Equation.3  Jika gambar terletak dibawah sumbu X maka integral diatas bernilai negatif, karena luas daerah tidak mungkin bilangan negatif maka nilai integral tersebut dimutlakkan. Perhatikan pula gambar daerah rata berikut ini :  Daerah R dibatasi oleh grafik-grafik x = f(y), y = c, y = d dan x = 0, luasnya A(R) ditentukan oleh : A(R) =  EMBED Equation.3  Jika gambar terletak disebelah kiri sumbu Y maka integral diatas bernilai negatif, karena luas daerah tidak mungkin bilangan negatif maka nilai integral tersebut dimutlakkan. Contoh : Tentukan luas daerah yang dibatasi oleh fungsi :  INCLUDEPICTURE "http://archives.math.utk.edu/visual.calculus/5/area2curves.2/int2.gif" \* MERGEFORMATINET  INCLUDEPICTURE "http://archives.math.utk.edu/visual.calculus/5/area2curves.2/g2.gif" \* MERGEFORMATINET Untuk menghitung luas daerah rata ikuti pola berfikir sebagai berikut : Gambar daerah yang bersangkutan Potong daerah menjadi jalur-jalur dan beri nomor pada satu jalur tertentu Hampiri luas jalur tertentu tersebut dengan luas persegi panjang Jumlahkan luas jalur-jalur pada daerah tersebut Ambil limit dari jumlah diatas dengan lebar jalur menuju 0, maka diperoleh integral tertentu. b. Daerah antara 2 Kurva Perhatikan kurva-kurva y = f(x) dan y = g(x) dengan g(x) ( f(x) pada selang [a,b], sebagai gambar berikut :  EMBED Equation.3  A =  EMBED Equation.3   Kita gunakan cara : potong, aproksimasikan, dan integralkan. Tugas Terstruktur Tentukan : a.  EMBED Equation.3  b.  EMBED Equation.3  c.  EMBED Equation.3  d.  EMBED Equation.3  e.  EMBED Equation.3  f.  EMBED Equation.3  2. Hitung luas daerah yang dibatasi oleh kurva y =  EMBED Equation.3  dan y =  EMBED Equation.3  3. 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Equation Native !h_1170936883$F8`M8`MOle #L, [f(x)+g(x)]dx  +" FMicrosoft Equation 3.0 DS Equation Equation.39q0pt f(x)dxCompObj#%$fObjInfo&&Equation Native 'L_1170936886)F8`M8`M  +" FMicrosoft Equation 3.0 DS Equation Equation.39q0, g(x)dx  +"Ole )CompObj(**fObjInfo+,Equation Native -L_1170936873@.F8`M8`MOle /CompObj-/0fObjInfo02 FMicrosoft Equation 3.0 DS Equation Equation.39qpUt x n +" dx=1n+1x n+1 +c FMicrosoft Equation 3.0 DS EqEquation Native 3_11709368693F8`M8`MOle 6CompObj247fuation Equation.39qP(S, sinxdx="cosx+c  +" FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo59Equation Native :l_1170936865;18F8`M8`MOle <CompObj79=fObjInfo:?Equation Native @h_1170936860=F8`M8`MLt cosxdx=sinx+c  +" FMicrosoft Equation 3.0 DS Equation Equation.39qYp, 1xdx=lnx  +" +cOle BCompObj<>CfObjInfo?EEquation Native Fu_1170936856J6BF8`M8`MOle HCompObjACIfObjInfoDK FMicrosoft Equation 3.0 DS Equation Equation.39qOt e x +" dx=e x +c FMicrosoft Equation 3.0 DS EqEquation Native Lk_1170936848GF8`M8`MOle NCompObjFHOfuation Equation.39qdU, a x +" dx=a x lna+c FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfoIQEquation Native R_1170936844OELF8`M8`MOle TCompObjKMUfObjInfoNWEquation Native X_1170936840QF8`M8`MyPd^ dx 1"x 2 =sin "1 x+c  +" FMicrosoft Equation 3.0 DS Equation Equation.39qqAW dx1+Ole [CompObjPR\fObjInfoS^Equation Native _x 2 =tgn "1 x+c  +" FMicrosoft Equation 3.0 DS Equation Equation.39q dxx x 2 "1  =sec "1_1170936835VF8`M8`MOle bCompObjUWcfObjInfoXeEquation Native f_1170936826[F8`M8`MOle iCompObjZ\jf x+c  +" FMicrosoft Equation 3.0 DS Equation Equation.39qZLU sec 2 xdx=tgnx+c  +"ObjInfo]lEquation Native mv_1170936823cY`F8`M8`MOle oCompObj_apfObjInfobrEquation Native s~_1170936820eF`M`M FMicrosoft Equation 3.0 DS Equation Equation.39qb T cosec 2 xdx=ctgx+c  +" FMicrosoft Equation 3.0 DS Equation Equation.39qOle uCompObjdfvfObjInfogxEquation Native yx\d secxtgnxdx=secx+c  +" FMicrosoft Equation 3.0 DS Equation Equation.39ql_T cosecx_1170936812w^jF`M`MOle {CompObjik|fObjInfol~Equation Native _1170936803oF`M`MOle CompObjnpf      "!#%$&'(+),-.hg0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefijkmlnopqrtsuwvxyz|{}~ctgxdx=cosecx+c  +" FMicrosoft Equation 3.0 DS Equation Equation.39qT (2x 3 +" +5cosx)dx=1ObjInfoqEquation Native _1170936787tF`M`MOle 2x 4 +5sinx+c FMicrosoft Equation 3.0 DS Equation Equation.39qT 'lim P!0 f(x i )x ii=1n "CompObjsufObjInfovEquation Native _1170936792rmyF`M`MOle CompObjxzfObjInfo{Equation Native V FMicrosoft Equation 3.0 DS Equation Equation.39q: d f(x)dx ab +" FMicrosoft Equation 3.0 DS Eq_1170936779~F`M`MOle CompObj}fObjInfouation Equation.39q:AT f(x)dx ab +" FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native V_1170936782hF`M`MOle CompObjfObjInfoEquation Native _1170936776|F`M`MOle _d 'lim P!0 f(x i )x ii=1n " FMicrosoft Equation 3.0 DS Equation Equation.39q:d f(x)dxCompObjfObjInfoEquation Native V_1170936772F`M`M ab +" FMicrosoft Equation 3.0 DS Equation Equation.39q: T f(x)dx ab +"Ole CompObjfObjInfoEquation Native V_1170936769F`M`MOle CompObjfObjInfo FMicrosoft Equation 3.0 DS Equation Equation.39q:d f(x)dx aa +" FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native V_1170936763F`M`MOle CompObjfObjInfoEquation Native V_1170936766F`M`MOle :P< f(x)dx ab +" FMicrosoft Equation 3.0 DS Equation Equation.39q: f(x)dxCompObjfObjInfoEquation Native V_1170936690F`M`M ba +" FMicrosoft Equation 3.0 DS Equation Equation.39q:(: f(x)dx ab +"Ole CompObjfObjInfoEquation Native V_1170936686F`M`MOle CompObjfObjInfo FMicrosoft Equation 3.0 DS Equation Equation.39q3_T [F(x)] ab FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native O_1170936683F`M`MOle CompObjfObjInfoEquation Native _1170936676F`M`MOle  x rab +" dx=b r+1 r+1"a r+1 r+1 FMicrosoft Equation 3.0 DS Equation Equation.39qCompObjfObjInfoEquation Native T_1170936672F`M`M88 T x r+1 r+1 FMicrosoft Equation 3.0 DS Equation Equation.39qp x rab +" dx=F(b)"F(a)=Ole CompObjfObjInfoEquation Native b r+1 r+1"a r+1 r+1 FMicrosoft Equation 3.0 DS Equation Equation.39qB(S kf(x)dx= ab +"_1170936635F`M`MOle CompObjfObjInfoEquation Native ^_1171037283,DF`M`MOle CompObjf FMicrosoft Equation 3.0 DS Equation Equation.39q:$W g(x)dx ab +" FMicrosoft Equation 3.0 DS EqObjInfoEquation Native V_1170936610F`M`MOle CompObjfObjInfoEquation Native s_1170936603F`M`Muation Equation.39qWU (4x"6x 2"12 +" )dx FMicrosoft Equation 3.0 DS Equation Equation.39qOle CompObjfObjInfoEquation Native t (4x"6x 2"12 +" )dx=4x "12 +" dx"6x 2"12 +" dx FMicrosoft Equation 3.0 DS Equation Equation.39q_1170936599F`M`MOle CompObjfObjInfoEquation Native _1170936593F`M`MOle CompObjf  !"#&)*+.3678;>ADGJMPQRUXY\_bglorwz}T< x 2 2[] "12 "6x 3 3[] "12 FMicrosoft Equation 3.0 DS Equation Equation.39q{DO 42"ObjInfoEquation Native _1170936583F`M`MOle  12()"683+13() FMicrosoft Equation 3.0 DS Equation Equation.39q;_, f(x) ac +" dxCompObj fObjInfo Equation Native W_1170936586F`M`MLZb #, {QD[r4N346q'bY}o=E5fOTBDd h  s *A? 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Equation.39qK 6x 7x"9  dx  +" FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfoEquation Native g_1429756937Fd`Md`MOle CompObjfObjInfoEquation Native h_1429757020Fd`Md`ML0 3xcos(4x"5)dx  +" FMicrosoft Equation 3.0 DS Equation Equation.39qBp= xe 3xOle CompObjfObjInfoEquation Native ^"7 dx  +" FMicrosoft Equation 3.0 DS Equation Equation.39qH,   +" FMicrosoft Equation 3.0 DS Eq_1170936365"Fd`Md`MOle CompObj!#fObjInfo$Equation Native 4_1170936362/ 'Fd`Md`MOle CompObj&(fuation Equation.39q(   +" FMicrosoft Equation 3.0 DS Equation Equation.39q2Ut 1"cos2ObjInfo)Equation Native 4_1170936354\X,Fd`Md`MOle CompObj+-fObjInfo.Equation Native N_11709363571Fd`Md`Mx2 FMicrosoft Equation 3.0 DS Equation Equation.39q2XLU 1+cos2x2Ole CompObj02fObjInfo3Equation Native N_11709360236Fd`Md`MOle CompObj57fObjInfo8 FMicrosoft Equation 3.0 DS Equation Equation.39qx+   +" FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native 4_1170936018>R;Fd`Md`MOle CompObj:<fObjInfo=Equation Native 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