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INTEGRAL

INTEGRAL. INTEGRAL TAK TENTU INTEGRAL TERTENTU. INTEGRAL TAK TENTU. 1. ∫ k dx = kx + c. CONTOH : ∫ 3 dx = 3x + c ∫ 5 dt = 5t + c ∫ 8 dQ = 8Q + c ∫ 5 6 du = 5 6 u + c. 2. ∫ ax b dx = a x b+1 + c b+1. CONTOH : ∫ 4X 3 dx = 4 x 4 + c = x 4 + c

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INTEGRAL

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  1. INTEGRAL • INTEGRAL TAK TENTU • INTEGRAL TERTENTU

  2. INTEGRAL TAK TENTU 1. ∫ k dx = kx + c • CONTOH : • ∫ 3 dx = 3x + c • ∫ 5 dt = 5t + c • ∫ 8 dQ = 8Q + c • ∫ 56 du = 56 u + c

  3. 2. ∫ ax b dx = a x b+1 + c b+1 • CONTOH : • ∫ 4X3 dx = 4 x 4 + c = x4 + c • 4 • 2. ∫ 3x8 dx = 3 x 9 + c =1/3X9 + C • 9

  4. 3. ∫ aUb dU = a U b+1 + c b+1 U=f(x) • CONTOH : • ∫(2X+ 1)dx = … 2. ∫(4X + 4) dX= … • X2 + X (4X2+8X+6)3 • Jawab : jawab : • Misal : U = X2 + X Misal : U =4X2+8X+6 • dU =( 2X + 1)dX dU =(8X+8)dX • ∫(2X + 1)dx = ∫dU dU =2(4X+4)dX • X2 + X U dU =(4X+4)dX • = Ln U + C 2 • = Ln ( X2 + X ) + C ∫dU= ∫ ½ U -3 dU • 2U3 • =½.1/-2 .U-2+ C • = - ¼(4X2+8X+6) -2 + C -1 4 (4x2+8x+6)2

  5. 4.∫UdV = U.V - ∫VdU RUMUS DI ATAS ADALAH RUMUS INTEGRAL PARSIAL CONTOH : ∫X.eX dx = …. Misal : U = X du = dx dv = eX dx V=∫eX dX = eX + C ∫X.eX dx = U.V - ∫V dU = X.eX - ∫ eX dx = X.eX - eX + C

  6. 5.∫ ex dx = ex + c 6.∫[f(x) + g(x)] dx =∫ f(x)dx+∫g(x)dx 7.∫n.f(x)dx = n∫f(x)dx

  7. SOAL SELESAIKANLAH ! • ∫ X3 dX = … 6. ∫ √ 2 + 5X dX = … • ∫X -4 dX = … 7.∫ (X2 + 3X + 4)3(2X + 3)dx =… • ∫9X2 dX = … 8. ∫ X2 + 3X – 2 dX = … • ∫5/X dX = … X • ∫(X2 -√X + 4) dX = … 9. ∫X.ex²dX = …

  8. INTEGRAL TERTENTU UNTUK a < c < b,berlaku b b b b 1.∫ f(x) dx = [F(X)] = F(b)- F(a) 4. ∫ k f(x) dx =k ∫ f(x) dx a a a a a b b b 2.∫ f(x) dx = 0 5. ∫ [f(x) + g(x)]dx = ∫f(x)dx + ∫g(x)dx a a a a b a c b b 3.∫ f(x) dx = - ∫ f(x) dx 6. ∫f(x)dx + ∫f(x)dx = ∫ f(x)dx a b a c a

  9. SOAL 6 0 1.∫ X dX = …. 5. ∫ (X2 – 2X + 3) dX = … 4 3 3 3 2. ∫ (X2 – 2X + 3 ) dX = … 6. ∫ (2X + 1)(3 – X) dX = … 0 1 1 4 3. ∫ (2X + 5) dX = … 7. ∫ ( √ X – X )2 dX = … -1 1 -4 8 4. ∫ (3X2 + 2X) dX = … 8. ∫ (X1/3 – X-1/3) dX = ….. -6 1 2 2a 9. ∫ (X + 9X3) dX = … 10. ∫ (a + X ) dX = … 1 a

  10. SURPLUS KONSUMEN DAN SURPLUS PRODUSEN BY AMIRULSYAH,MSi

  11. SURPLUS KONSUMEN SK Fungsi demand Fungsi demand SK SK P1 Q Q O Q1 O P P

  12. SURPLUS PRODUSEN P P Fungsi supply SP P1 SP Fungsi supply P1 Q Q O Q1 O Q1

  13. SURPLUS KONSUMEN DAN SURPLUS PRODUSEN P P Fungsi demand SK SK Fungsi supply P1 P1 SP SP Q O Q1 O Q 0 Q1

  14. PENGETAHUAN DASAR CARA I : L= axt 2 L= 4x3 2 L= 6 satuan luas CARA II : Integral 4 L= ∫(5-3/4x)dx – 2x4 0 4 = (5X – ¾.1/2X ²)] - 8 0 = (5.4 – 3/8.16) – (5.0-1/4.0) – 8 = (20 – 6) – 0 – 8 = 14 - 8 = 6 satuan luas Y 5 LUAS DAERAH LUAS = …? 2 X O 4 CARA III: INTEGRAL 5 L = ∫ ( ) dy 2 Y= 5-3/4x X= 20/3 – 4y 5 L = ∫ (20/3 – 4/3Y)dy 2 L= 6 satuan luas

  15. P CARA I: INTEGRAL 5 L = ∫ ( 6 – 3/25Q²)dQ – 3x5 0 5 L = (6Q – 3/25.1/3Q³)] – 15 0 L = 10 satuan luas 6 P= 6 – 3/25 Q² LUAS 3 LUAS DAERAH Q 0 5 CARA II: INTEGRAL 6 L = ∫ (50 – 25/3P)1/2dP 3 6 L = { 2/3(50 – 25/3P)3/2.(-3/25)} ] 3 L = { - 2/5 (50 – 25/3P)3/2 L = 10 satuan luas

  16. P LUAS= …? CARA II : INTEGRAL 6 L = 6X6 - ∫(2 + 2/3Q)dQ 0 6 L = 36 – {2Q + 2/3.1/2Q² } ] 0 L = 36 – 24 = 12 satuan luas 6 2 Q 0 6 CARA I : RUMUS L = axt 2 L = 4 x 6 2 L = 12 satuan luas CARA III : integral 6 L = ∫( 3/2 P – 3 ) dP 2 6 L = ( 3/4P – 3P ) ] = 9 + 3 = 12 satuan luas 2

  17. LUAS DAERAH P P = 2 + 1/5Q² 7 CARA I : INTEGRAL 5 L = 7x5 - ∫( 2 + 1/5Q²)dQ 0 5 L = 35 - (2Q + 1/5.1/3Q³)] 0 L = 35 - 10 - 8 1/3 L = 16 ⅔ satuan luas LUAS 2 Q 0 5 CARA II : INTEGRAL 7 L = ∫(5P - 10)1/2 dP 2 7 L = { 2/3(5P - 10)3/2. ⅕ }] 2 L = 2/15.{ 25 } 3/2 L = 16 ⅔ satuan luas

  18. SOAL P P = 5 + 1/12Q2 2. 12 1.Fungsi pendapatan dari suatu pabrik diberikan sebagai berikut : R = 6 + 350Q – 2Q2 Fungsi produksinya : Q = 3L Jika jumlah tenaga kerja yang ada 25 orang,berapakah MPRL dan jelaskan artinya . LUAS I 8 LUAS II P = 12 - 1/9Q2 5 Q 0 6

  19. JAWABAN P 6 P = 5 + 1/12Q2 2. 12 0 6 Luas I = ∫(12 - 1/9Q2)dQ - 8X6 = ( 12Q + 1/9.1/3Q3) ] - 48 = (12.6 + 1/27.63 – (12.0 + 1/27.03) - 48 = (72 + 1/27.216 – 0) - 48 = (72 + 8 – 0) - 48 = 80 – 48 = 32 LUAS I 0 8 LUAS II P = 12 - 1/9Q2 5 Q 0 6

  20. JAWABAN P 6 P = 12 - 1/9Q2 2. 12 0 6 Luas II = 6X8 - ∫(5 + 1/12Q2)dQ = 48 – ( 5Q + 1/12.1/3Q3) ] = 48 – (5.6 + 1/36.63 – (5.0 + 1/36.03) = 48 – (30 + 1/36.216 – 0) = 48 - (30 + 6 - 0) = 48 – 36 = 12 LUAS I 0 8 LUAS II P = 5 + 1/12Q2 5 Q 0 6

  21. JAWABAN 1.Fungsi pendapatan dari suatu pabrik diberikan sebagai berikut : R = 6 + 350Q – 2Q2 Fungsi produksinya : Q = 3L Jika jumlah tenaga kerja yang ada 25 orang,berapakah MPRL dan jelaskan artinya . Jawab : R = 6 + 350Q - 2Q² Q = 3L dR = 350 – 4Q dQ = 3 dQ dL MPRL = dR = dR . dQ dL dQ dL = (250 – 4Q).3 L = 25 Q =3L = 75 dR = (350 – 300).4 = 200 dL Artinya: Untuk setiap penambahan Tenaga Kerja sebanyak 25 orang akan menyebabkan penambahan pendapatan sebanyak 200 ,dan sebaliknya

  22. SOAL 1.Seorang anak mempunyai uang Rp 1000.Ia akan membeli permen susu (Y) dan permen coklat (X).Harga permen susu Rp100 dan permen coklat Rp 100. Fungsi nilai guna adalah U=XY.Berapa jumlah permen susu dan coklat yang dikomsumsi anak tersebut ? 2.Jika harga permen coklat meningkat menjadi Rp 200 .berapa jumlah permen coklat dan permen susu yang dikonsumsi anak tersebut ? 3.Jika preferensi untuk coklat meningkat menjadi U = X2Y, berapa konsumsi permen coklat dan permen susu ?

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