Bank Soal

10 Responses to Bank Soal

  1. Arrin Nugraha XII IPA 1 mengatakan:

    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

  2. Arrin Nugraha XII IPA 1 mengatakan:

    Maaf Pak, punya saya Soalny tipe A

  3. aseh ega sari mengatakan:

    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)

  4. Demi Rangga XII IPA 2 mengatakan:

    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

  5. vivin puspasari mengatakan:

    pak ini tugas saya vivin puspasari xii ipa2

  6. nurul mengatakan:

    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

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

  8. Hiro mengatakan:

    apa ada soal lomba OSN

Terima Kasih Atas Komentar Anda

Isikan data di bawah atau klik salah satu ikon untuk log in:

Logo WordPress.com

You are commenting using your WordPress.com account. Logout / Ubah )

Gambar Twitter

You are commenting using your Twitter account. Logout / Ubah )

Foto Facebook

You are commenting using your Facebook account. Logout / Ubah )

Foto Google+

You are commenting using your Google+ account. Logout / Ubah )

Connecting to %s

%d blogger menyukai ini: