Download - INTEGRAL
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INTEGRALINTEGRAL TAK TENTU INTEGRAL TERTENTU
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INTEGRAL TAK TENTU
1. ∫ k dx = kx + cCONTOH :
1.∫ 3 dx = 3x + c2.∫ 5 dt = 5t + c
3.∫ 8 dQ = 8Q + c4.∫ 56 du = 56 u + c
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2. ∫ ax b dx = a x b+1 + c b+1
CONTOH :1.∫ 4X3 dx = 4 x 4 + c = x4 + c 4
2. ∫ 3x8 dx = 3 x 9 + c =1/3X9 + C 9
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3. ∫ aUb dU = a U b+1 + c b+1 U=f(x)
CONTOH :1. ∫ (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
-14 (4x2+8x+6)2
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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
4.∫UdV = U.V - ∫VdU
RUMUS DI ATAS ADALAH RUMUS INTEGRAL PARSIAL
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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
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SOAL
SELESAIKANLAH !
1. ∫ X3 dX = … 6. ∫ √ 2 + 5X dX = …2. ∫X -4 dX = … 7.∫ (X2 + 3X + 4)3(2X + 3)dx =…
3. ∫9X2 dX = … 8. ∫ X2 + 3X – 2 dX = …
4. ∫5/X dX = … X
5. ∫(X2 -√X + 4) dX = … 9. ∫X.ex² dX = …
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INTEGRAL TERTENTU
UNTUK a < c < b,berlaku b b b b1.∫ 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 b3.∫ f(x) dx = - ∫ f(x) dx 6. ∫f(x)dx + ∫f(x)dx = ∫ f(x)dx a b a c a
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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
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BY
AMIRULSYAH,MSi
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SURPLUS KONSUMEN
SK
Q
P
O
P
QO
SK
Q1
P1
Fungsi demand Fungsi demand
SK
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SURPLUS PRODUSEN
OQ
P
P1
SP
Fungsi supply
O
P
Q
SP
Q1 Q1
P1 Fungsi supply
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SURPLUS KONSUMEN DAN SURPLUS PRODUSEN
SK
SP
Fungsi supply
Fungsi demand
OQ
P
Q1
P1
O
Q
P
0
P1
Q1
SK
SP
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PENGETAHUAN DASAR
LUAS DAERAH
4
2
O
5
LUAS = …?
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
CARA III: INTEGRAL 5L = ∫ ( ) dy 2Y= 5-3/4xX= 20/3 – 4y 5L = ∫ (20/3 – 4/3Y)dy 2L= 6 satuan luas
X
Y
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LUAS DAERAH
3
6
50
P= 6 – 3/25 Q²
P
Q
LUAS
CARA I: INTEGRAL 5
L = ∫ ( 6 – 3/25Q²)dQ – 3x5 0 5
L = (6Q – 3/25.1/3Q³)] – 15 0L = 10 satuan luas
CARA II: INTEGRAL 6
L = ∫ (50 – 25/3P)1/2 dP 3 6
L = { 2/3(50 – 25/3P)3/2.(-3/25)} ] 3L = { - 2/5 (50 – 25/3P)3/2
L = 10 satuan luas
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LUAS= …?
0
2
6
CARA I : RUMUSL = axt 2L = 4 x 6 2L = 12 satuan luas
6
CARA II : INTEGRAL 6
L = 6X6 - ∫(2 + 2/3Q)dQ 0 6
L = 36 – {2Q + 2/3.1/2Q² } ] 0L = 36 – 24 = 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
Q
P
QP3
22
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LUAS
2
7
50
P = 2 + 1/5Q²
Q
P
CARA I : INTEGRAL 5
L = 7x5 - ∫( 2 + 1/5Q²)dQ 0 5
L = 35 - (2Q + 1/5.1/3Q³)] 0L = 35 - 10 - 8 1/3
L = 16 ⅔ satuan luas
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
LUAS DAERAH
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1.Fungsi pendapatan 1.Fungsi pendapatan dari suatu pabrik dari suatu pabrik diberikan sebagai diberikan sebagai berikut :berikut :
R = 6 + 350Q – 2QR = 6 + 350Q – 2Q22
Fungsi produksinya : Fungsi produksinya : Q = 3LQ = 3L
Jika jumlah tenaga Jika jumlah tenaga kerja yang ada 25 kerja yang ada 25 orang,berapakah orang,berapakah MPRL dan jelaskan MPRL dan jelaskan artinya .artinya .
2.
5
0
P
Q
12
8
6
LUAS I
LUAS II
P = 5 + 1/12Q2
P = 12 - 1/9Q2
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Luas I = ∫(12 - 1/9QLuas I = ∫(12 - 1/9Q22)dQ - 8X6)dQ - 8X6
= ( 12Q + 1/9.1/3Q= ( 12Q + 1/9.1/3Q33) ] - 48) ] - 48
= (12.6 + 1/27.6= (12.6 + 1/27.633 – (12.0 + – (12.0 + 1/27.01/27.033) - 48) - 48
= (72 + 1/27.216 – 0) - 48= (72 + 1/27.216 – 0) - 48
= (72 + 8 – 0) - 48= (72 + 8 – 0) - 48
= 80 – 48= 80 – 48= 32= 32
2.
5
0
P
Q
12
8
6
LUAS I
LUAS II
P = 5 + 1/12Q2
P = 12 - 1/9Q2
0
6
6
0
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Luas II = 6X8 - ∫(5 + 1/12QLuas II = 6X8 - ∫(5 + 1/12Q22)dQ )dQ
= 48 – ( 5Q + 1/12.1/3Q= 48 – ( 5Q + 1/12.1/3Q33) ] ) ]
= 48 – (5.6 + 1/36.6= 48 – (5.6 + 1/36.633 – (5.0 + – (5.0 + 1/36.01/36.033) )
= 48 – (30 + 1/36.216 – 0)= 48 – (30 + 1/36.216 – 0)
= 48 - (30 + 6 - 0)= 48 - (30 + 6 - 0)
= 48 – 36= 48 – 36= 12= 12
2.
5
0
P
Q
12
8
6
LUAS I
LUAS IIP = 5 + 1/12Q2
P = 12 - 1/9Q2
0
6
6
0
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1.Fungsi pendapatan dari suatu pabrik diberikan sebagai berikut :1.Fungsi pendapatan dari suatu pabrik diberikan sebagai berikut :
R = 6 + 350Q – 2QR = 6 + 350Q – 2Q22
Fungsi produksinya : Q = 3LFungsi produksinya : Q = 3L
Jika jumlah tenaga kerja yang ada 25 orang,berapakah MPRL dan Jika jumlah tenaga kerja yang ada 25 orang,berapakah MPRL dan jelaskan artinya .jelaskan artinya .
Jawab : Jawab :
R = 6 + 350Q - 2Q² Q = 3L R = 6 + 350Q - 2Q² Q = 3L
dR dR = 350 – 4Q = 350 – 4Q dQ dQ = 3 = 3
dQ dLdQ dL
MPRL = MPRL = dRdR = = dRdR . . dQdQ
dL dQ dLdL dQ dL
= (250 – 4Q).3= (250 – 4Q).3
L = 25 Q =3L = 75L = 25 Q =3L = 75
dR dR = (350 – 300).4 = 200 = (350 – 300).4 = 200
dLdL
Artinya: Untuk setiap penambahan Tenaga Kerja sebanyak 25 orang akan Artinya: Untuk setiap penambahan Tenaga Kerja sebanyak 25 orang akan menyebabkan penambahan pendapatan sebanyak 200 ,dan sebaliknyamenyebabkan penambahan pendapatan sebanyak 200 ,dan sebaliknya
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SOAL1.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 ?