Bank Soal

12 Responses to Bank Soal

  1. Arrin Nugraha XII IPA 1 says:
    17/09/2010 pukul 4:27 am

    Arrin Nugraha (XII IPA 1)

    1.Tentukan penyelesaian integral ∫▒〖3√x〗 – x√x dx
    Jawab :
    =∫▒〖( 3√x〗 – x√x )
    =∫▒〖( 3x〗1/2 – x.x1/2 )
    =∫▒〖(3x〗1/2 – x3/2 )
    = 3×3/2 / 3/2 – x5/2 / 5/2
    = 6/3 x3/2 – 2/5 x5/2
    = 2×3/2 – 2/5 x5/2 + C

    2.Selesaikan dengan subtitusi integral ∫▒(4x-2)/((x^2- x )) dx
    Jawab :
    = ∫▒(4x-2)/((x^2- x ))
    = ∫▒〖(4x-2〗) (x2 – x )- 4
    dx = du/(2x-1)
    = ∫▒〖2(2x-1) 〗. U- 4 . du/(2x-1)
    = ∫▒〖2 .〗 U- 4 . du
    = 2 ∫U- 4 . du
    = 2 (-1/3 U- 3 ) = -2/3 (x2 – x) + C

    3.Selesaikan dengan cara parsial ∫▒〖x sin⁡〖(x- π〗 〗) dx
    Jawab :
    ∫▒〖x sin⁡〖(x- π〗 〗)
    u = x dv = ∫▒sin⁡〖(x〗 – π)
    du/dx = 1 v = -1/x cos ( x – π )
    du = 1. dx

    = u.v – ∫▒〖v .du〗
    = x. (-1/x cos ( x – π )) – ∫▒〖-1/x cos ( x – π )〗 . dx
    = – cos ( x – π ) + 1/x2 sin ( x – π ) + C

    4.Hitunglah luas daerah yang dibatasi oleh kurva y = x2 – 9x + 15, dan yang dibatasi oleh kurva y = – x2 + 7x – 15
    Jawab :
    y1 = y2
    x2 – 9x + 15 = – x2 + 7x – 15
    x2 – 9x + 15 + x2 – 7x + 15 = 0
    2×2 – 16x + 30 = 0 : 2
    x2 – 8x +15 = 0
    (x – 5 ) (x – 3 )
    x= 5 x= 3

    L = ∫_a^b▒(f (x) – g(x) )
    = ∫_3^5▒( x2 – 9x + 15 ) – (- x2 + 7x – 15)
    = ∫_3^5▒ x2 – 9x + 15 + x2 – 7x + 15
    = ∫_3^5▒ 2×2 – 16x + 30
    = [ 2/3 x3 – 8×2 + 30x ]35
    = [ 2/3 (5)3 – 8(5)2 + 30(5)] – [ 2/3 (3)3 – 8(3)2 + 30(3)
    = [250/3 – 200 + 150] – [ 18 – 72 + 90 ]
    = [ 250/3 – 50 ] – [36]
    = [ 250/3 – 50 – 36 ]
    = [ 250/3 – 86 ]
    = [ 250/3 – 258/3 ]
    = 8/3 satuan luas

    5.Hitunglah volume benda putar yang dibatasi oleh kurva y = x2 – 1 , sumbu X, x = 1 dan x= -1 mengelilingi sumbu X sejauh 360⁰.
    Jawab :
    v = π ∫_a^b▒〖(y)〗2
    = π ∫_(-1)^1▒( x2 – 1 )2
    = π ∫_(-1)^1▒(x4 – 2×2 + 1)
    = π (1/5 x5 – 2/3 x3 + x )-11
    = π [ 1/5 – 2/3 + 1] – [ -1/5 + 2/3 – 1]
    = π [1/5 – 2/3 + 1 + 1/5 – 2/3 + 1]
    =π [ 2/5 – 4/3 + 2 ]
    = π [(6-20+30)/15]
    = 16/15 π satuan volume

    Balas
  2. Arrin Nugraha XII IPA 1 says:
    17/09/2010 pukul 6:03 am

    Maaf Pak, punya saya Soalny tipe A

    Balas
  3. aseh ega sari says:
    17/09/2010 pukul 6:26 am

    Bagian D
    1)

    2)

    3)

    4) Hitung luas daerah yang dibatasi oleh kurva . Sumbu x dan x=6

    5)

    1) Evaluasi Bab I

    2)

    3)

    4)

    Balas
  4. Demi Rangga XII IPA 2 says:
    17/09/2010 pukul 7:20 am

    1.  2 x . dx =  2 . x1/2
    = 4/3 x3/2

    2. 2 cos (4x + 5) dx = 2 ¼ sin (4x + 5)
    = ½ sin (4x + 5)

    3. a3 (3×2 + 2x + 1) dx = 25 Nilai ½ a
    = a3 [x3 + x2 + x] a3 = [(3) 3 + (3) 2 + (3) – [a3 + a2 + a]
    = 27 +9 + 3 – (a3 + a2 + a) = 25
    = – (a3 + a2 + a) = 25 – 39
    = a3 + a2+ a = 14
    = (a – 2) . (a2 + 3a + 7) = 0
    Jadi, nilai ½ a
    a – 2 = 0
    a = 2
    ½ a = ½ . 2 = 1

    4. 0 sin 2x. cos x.dx = 2 sin x. cos x. cos x
    2 sin x. cos2 x
    x = cos x
    dx = – sin x
    -dx = sin x
    = -dx sin x. dx
    = -2  x2 . dx = -2  x2 . dx
    = -2. 1/3 x3 + 1 = -2/3 cos3 x + 1
    = [2/3 cos ()3] – [2/3 .cos [0] 3
    = 2/3 cos 3 – 2/3 [1] 3 = 2/3 cos 3 – 2/3
    = cos 3

    5. 01 3 3×2 + 1 x dx
    Misal x = 3×2 + 1
    dx = 6x dx
    1/6 dx = x dx
    3. 1/6 3×2 + 1
    [1/2 . 2/3 + 3/2]
    1/3 [3x + 1] 3/2
    = 1/3 ([1] + [1]) 3/2 – 1/3 [3(0) + (1)]
    = 1/3 [4] 3/4 – 1/3

    6.  (cos2 x) . cos x dx
     (1 – sin2 x) 2 . cos x
     cos x dx – 2  sin2 x . cos x x +  sin4 x cos x dx
    Misal
    x = sin x
    dx = cos x dx = d sin x
     d (sin x) – 2 sin2 x d sin x + sin4 x d sin x
    Sin x – 2. 1/3 sin3 x + 1/5 sin5 x + c
    Sin x – 2/3 sin3 x + 1/5 sin5 x + c

    7. 0 x sin x . dx
    v = x dv = sin x dx
    dv = dx  dv =  sin x dx
    v = – cos x
     x sin x dx = -x cos x –  -cos x
    = – x cos x + sin x
    = – . Cos  + sin  + 0 cos o – sin 0
    = –  – 1 + 0 + 0. 1 – 0
    = 

    8.  x 9 – x2 . dx =  9 – x2 . x. dx
    Misal x = 9 – x2
    dx = -2x dx
    x dx = – ½ dx
    =  x – ½ dx
    = – ½  x 1/2
    = – ½ (1/3 x3/2) + c
    = 1/3 x. x1/2 + c
    = 1/3 (9 – x2) 9 – x2 + c

    9. x1 = x1
    X + X = 6 x = 6 – x
    X1 = x2
    X2 = 6 – x
    X + x – 6 = 0
    (x+3) (x-2)
    X = -3 x = 2
    =  x2 + x – 6 = 0
    = [1/3 x3 + ½ x2 – 6x]
    = 1/3 [-3] 3 + ½ (-3) 2 – 6 (-3) – 1/3 (2) 3 – ½ (2) 2 + 6 (2)
    = -9 + 9/2 + 18 – 8/3 – 4/2 + 12
    = 19 + 4/2 – 8/3 = 19 + 27-16 = 9 + 11/6
    6
    = 114 + 1 = 125 = 20 5/6
    6 6

    10. Daerah yang dibatasi oleh kurva y=x2 dan x+y-2=0  y=x-2
    x+x2-2=0 (x-1) (x+2)
    X= 1 x= -2
    =  1-2 (x-2) 2 – (x2) 2 x
    =  (x2 – 4x + 4 – x4)
    = [1/3×3 – 2x + 4x – 1/5x] 1-2
    = 1/3 – 2 + 4 – 1/5 – [1/3 (-2) 3 – 2 (-2) 2 + 4 (-2) – 1/5 (-2) 5]
    = 1/3 – 2 + 4 – 1/3 – [-8/3 – 8 – 8 + 32/5]
    = 1/3 – 2 + 4 – 1/5 + 8/3 + 8 + 8 – 32/5
    = 9/3 + 16 – 33/5 = 21 – 33/5
    = 105 – 33 = 72 = 14 2
    5 5 5

    11. A.  (x2 + 1) cos x dx =
    v = x2 + 1 dx = cos x
    dx = 2x dx  dx =  cos x  v = sin x
    = x – sin x –  sin x 2x V = 2x
    = x – sin x – 2x – cos x  – cos x 2 dv = 2 dx
    = x2 + 1 sin x + 2x . cos x – 2 sin x  dv = sin x
    B. = ½  0 2x + sin x = ½   2x +  sin x V = – cos x
    = ½  0 [x + cos x] ½  0
    = (1/2 )2 – cos ½  – [(0) – cos (0)]
    = ¼  – 0 – (-1)
    = ¼ 

    12. A.  x. sin (x2 + 1) dx
    Y = x2 + 1
    dx = 2x x dx = ½ dy
     sin ½ dy = ½  sin y dx
    – ½ cos (x2 + 1) + e = – ½ cos x + c
    B. ½ 0 (sin2 x – cos2 x) dx =  cos 2x = ½ sin 2x

    13. x = 3
    y = x2 – 4x + 5
    y = -x2 + 6x – 5
    X1 = y2
    X2 – 4x + 3 = – y2 + 6x – 5
    2×2 – 10 x + 8 : 2
    X2 – 5x +4
    (x – 4) (x – 1)
    X = 4 x = 1

    LI = 31 2×2 – 10x + 8
    = [2/3 x3 – 5×2 + 8x] = 18 – 45 + 24 – 2/3 – 5
    = [- 2/3] = 2/3

    LII = 43 2×2 – 10x + 8
    = 125/3 – 70 + 32 – 18 + 45 – 24
    = 128/3 – 25
    = 128 – 105 = 23/3 = 7 2/3
    3

    Ltot = LI + LII
    = 2/3 + 23/3
    = 25/3

    14. y = 9 – x2 x = 4 = x =  2
     = 9 – x2
    =  2-2 (9 – x2) 2 – (5) 2 . dx
    = 2-2 81 – 18 x2 + x4 – 25
    = 2-2 x4 – 18 x2 + 56
    = 1/5 x5 – 6×3 + 56x
    = 1/5 [2] 5 – 6 [2] 3 + 56 [2] – (1/5 [-2] 5 – 6 [-2] 3 + 56 [-2])
    = (32/5 – 42 + 112) – (-32/5 + 42 – 112)
    = 32/5 – 42 + 112 + 32/5 – 42 + 112
    = 32/5 + 140 = 32 + 700 = 732 = 146 2/5
    5 5

    15.  3 x – x x =  3x½ – x. x3/2
    = 2×3/2 – 2×2 . 2/3×3/2
    = 2×3/2 – 2×2 . 2/3×3/2

    16.  4x – 2
    (x2 – x) 4
    V = x2 – x
    dv = (2x + 1)
    dx

    4x – 2 dx =  v4 . dv
    (x2 – x) 4
    2 = 1/5 v2 + L
    = 1/5 (x2 – x) 5
    = 2/5 (x2 – x) 5 + L

    17.  x sin (x – )
    V = x  dv = sin (x – )
    dv = dx v = -cos x
    = -x cos x  – cos (x – )
    = -x cos x – sin (x – )

    18. y = x2 – 1 x = 1 dan x = -1
    (x + 1) (x – 1)
    X = -1 x = 1
    =  1-1 (x2 – 1) 2 dx
    =  1-1 (x – 1) 2 dx
    = (x2 – 2x + 1)
    = 1 – 2 + 1 – [(-1) 2 – 2 (-1) + 1]
    = 1 – 2 + 1 – 1 + 2 + 1
    = 2

    Balas
  5. vivin puspasari says:
    20/09/2010 pukul 3:20 am

    pak ini tugas saya vivin puspasari xii ipa2

    Balas
  6. nurul says:
    21/02/2013 pukul 10:22 pm

    pak kenapa UN sekarang 20 paket,, gimana cara menjawabnya pak?
    kami sebagai pelajar merasa tertekan dengan adanya 20 paket pak. kami mohon agar saran kami bisa di terima

    Balas
  7. Anisa Fernanda (@anisa_fernanda) says:
    18/07/2014 pukul 10:00 am

    ada kisi2 soal tes untuk kelas XI yang pakai kurikulum 2013 nggak pak?

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  8. Hiro says:
    06/01/2016 pukul 12:08 am

    apa ada soal lomba OSN

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  9. Anonim says:
    06/07/2017 pukul 2:22 pm

    Film 89f cinta 8.go . Dengan fb 3cj9 cinta. 5 90n0 foto gg gf?b9 bik, tolong td bu
    N
    Ih 6mcgykv5 erat dengan 9j8gjj0

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  10. Siti Mulyani says:
    11/12/2020 pukul 8:46 pm

    30.Luas daerah yang di batasi oleh kurvaf(x)=x2+2x+3dan g(x)=3–x adalah…
    satuan luas………….

    Balas

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