Model Pengintegrasian Nilai Pendidikan Karakter

1.1 Memahami konsep integral tak tentu dan integral tentu.

  3. Integral tak tentu dan integral tertentu fungsi trigonometri

  integral fungsi trigonometri Siswa mampu menentukan hasil integral dengan metode substitusi dan parsial Siswa mampu mengguna- kan integral untuk menentukan luas daerah dan volume benda putar

Siswa mampu menggunakan konsep integral

dalam pemecahan masalah

  Mendefinisikan konsep integral fungsi aljabar

  Model Pengintegrasian Nilai Pendidikan Karakter

Integral

  Cermat dalam menentukan batas- batas daerah yang akan dihitung luas atau volumenya.

  5. Luas daerah di bawah kurva dan volume benda putar Teliti Cermat Teliti dalam menentukan hasil pengintegralan.

  4. Integral substitusi dan integral parsial

  2. Integral tak tentu dan integral tertentu fungsi aljabar

  1

  1. Integral sebagai lawan dari turunan (antiderivatif)

  Pada bab ini akan dipelajari:

  1.3 Menggunakan integral untuk meng- hitung luas daerah di bawah kurva dan volume benda putar.

  1.2 Menghitung integral tak tentu dan integral tentu dari fungsi aljabar dan trigonometri yang sederhana.

  Standar Kompetensi Kompetensi Dasar Nilai Indikator

  1. Menggunakan konsep integral dalam pe- mecahan masalah.

• Menentukan hasil integral tak tentu fungsi aljabar
• Menentukan hasil integral tertentu fungsi aljabar
• Menentukan rumus fungsi jika diketahui turuna>Menentukan hasil integral tak tentu fungsi trigono- metri
• Menentukan hasil integral tertentu fungsi trigonom>Melakukan pengintegralan dengan metode substitusi
• Melakukan pengintegralan dengan metode par
• Menentukan luas daerah yang dibatasi kurva
• Menentukan volume benda putar Mendefinisikan konsep integral fungsi trigonometri Menentukan hasil integral dengan metode peng- integralan Menggunakan integral untuk menentukan luas daerah dan volume benda putar Siswa mampu menentukan integral fungsi aljabar Siswa mampu menentukan

  1

  2 A. Pilihan Ganda

  v(t) = 0 ⇔ 5t – t = 0

  2

  1

  1. Jawaban: a _{2 2} ⇔ t(10 – t) = 0

  2 − 3x 4x x 3x 4x x

  dx = ( ) dx

• – x x

  ∫ ∫

  ⇔ t = 0 atau t = 10

  x x x x ₁ Jadi, benda berhenti setelah 10 detik.

  − ₂

  = (3 x – 4x) dx

  ∫ ₁

  5. Jawaban: b

  − +

  1

  3

  4 ₂ 1 + 1 dy

  = – ₁ x x + c

  1 _{2 1} dx

  1 1 = 4x + 5 + − +

  3

  4

  y = (4x + 5) dx

  ∫ ₂

  2

  = x x + c – _{1 2 = 2x + 5x + c}

  2

  2 Kurva melalui titik (–3, –3).

  2

  = 6 x – 2x + c

  2

  y = 2x + 5x + c

  2

  2. Jawaban: b ⇔ –3 = 2(–3) + 5(–3) + c ⇔ –3 = 2(9) – 15 + c

  (2x – 3)(3x + 2) dx

  ∫

  ⇔ –3 = 18 – 15 + c

  2

  = (6x – 5x – 6) dx

  ∫

  ⇔ –3 = 3 + c

  6

  5 2 + 1 1 + 1 c = –6

  ⇔ = – x x – 6x + c

• 2 1 1 1

  2 Jadi, persamaan kurva tersebut y = 2x + 5x – 6.

  6

  5

  3

  2

  = x x – 6x + c –

  6. Jawaban: c

  3

  2

  3

  3

  3

  2 1 1 1

• 5

  3

  2

  2  

  2

  = 2x x – 6x + c

• – x x − x

  (3x + 2x – 1) dx =  

  2 2 1 1 1 + +

  ∫  

  1

  1

  3

  3

  2

  3. Jawaban: d

• =  

  x x x −  

  1

  2

  f ′(x) = 3x + 6x – 5 dan f(–1) = 8

  3

  2

  3

  2

  = (3 + 3 – 3) – (1 + 1 – 1) f(x) = f ′(x) dx

  ∫

  = (27 + 9 – 3) – (1 + 1 – 1)

  2

  = (3x + 6x – 5) dx

  ∫

  = 33 – 1

  1

  1

  3

  2

  = 3 · x + 6 · x – 5x + c = 32

  3

  2

  3

  2

  = x + 3x – 5x + c

  3

  2

  7. Jawaban: c f(–1) = 8 ⇒ (–1) + 3(–1) – 5(–1) + c = 8 ⇔ –1 + 3 + 5 + c = 8

  2

  2

  1

  2 2 –2

  ⇔ c = 1 (x ) dx = (x – x ) dx ₂

• – x

  ∫ ∫

  3

  2

  1

  1 Jadi, f(x) = x + 3x – 5x + 1.

  2

  1

  1  3 1  − x − x

  =

   

  4. Jawaban: c

  −  3 1 

  1

  2 Percepatan = a(t) = 5 – t

  1

  1  

  x dv(t)

• 3

  =

     3 x 

  1

  = a(t) ⇒ v(t) = a(t) dt = (5 – t) dt

  ∫ ∫ dt

  8

  1

  1

  1

  2

  = ( + ) – ( + 1) = 5t – t + c

  3

  2

  3

  2

  19

  4 Benda bergerak dari keadaan diam maka

  = –

  6

  3 v(0) = 0 ⇒ c = 0.

  11

  1

  =

  2

  6 Kecepatan benda dirumuskan v(t) = 5t – t .

2 Pada saat benda berhenti berarti kecepatannya 0.

  2

  8. Jawaban: e ³

  3 − ₂

  4

  b. dx = 3 x dx

  ∫ ∫ x x

  2

  (–x + 6x – 8) dx ₃

  ∫ 1 3 − +

  2 _{2 4}

  = x + c ₃

  − +

  1

  2  ₂

  1 

  = − x 3x − 8x

• 3

     3  ₂ ¹ −

  3 ₂

  = + c

  1

  1 ^{1 x}

  3

  2

  3

  2 −

  = (– (4) + 3(4) – 8(4)) – (– (2) + 3(2) – 8(2)) ₂

  3

  3

  6

  64

  8

  = – + c = (– + 48 – 32) – (– + 12 – 16)

  x

  3

  3

  16

  20

  4

  2

  = (– ) – (– ) =

  c. (3x + 2) dx

  ∫

  3

  3

  3

  2

  = (9x + 12x + 4) dx

  ∫

  9. Jawaban: d

  3

  1

  1

  3

  2

  2

  = 9 · x + 12 · x + 4x + c (3x + 2x + 1) dx = 25

  3

  2 ∫ a

  3

  2

  = 3x + 6x + 4x + c

  3

  3

  2  

• ⇔ + x x x = 25  

  a

  d. (2 x + 1)(3 x – 2) dx

  ∫

  3

  2

  ⇔ (27 + 9 + 3) – (a + a + a) = 25

  3

  2

  = (6x – x – 2) dx

  ∫

  ⇔ 39 – a – a – a = 25 ₁

  3

  2

  ⇔ a + a + a – 14 = 0 ₂ = (6x – – 2) dx

  x ∫

  2

  ⇔ (a – 2)(a + 3a + 7) = 0 ₃

  2

  2

  2 ²

  ⇔ a = 2 atau a + 3a + 7 = 0 = 3x – – 2x + c

  x

  3 Oleh karena tidak ada nilai x yang memenuhi

  2

  2

  2

  = 3x – 2x + c – x x persamaan x + 3x + 7 = 0 maka penyelesaiannya

  3 a = 2.

  5

  1

  1

  2. a. 2g(x) dx = 6 Jadi, a = × 2 = 1. ∫

  2

  2 −

  2

  5

  10. Jawaban: b ⇔ 2 g(x) dx = 6

  4 ∫

  −

  2 ∫ f(x) dx = 2

  5

  4

  4

  ⇔ g(x) dx = 3

  ∫ −

  2

  2f(x) dx = 2 f(x) dx = 2 ⇔ 2

  ∫ ∫

  2

  2

  5

  4

  b. (2f(x) – 3g(x)) dx

  ∫

  f(x) dx = 1 ⇔

  ∫ −

  2

  2

  5

  5

  4

  2

  4

  = 2 f(x) dx – 3 g(x) dx

  ∫ ∫

  f(x) dx = f(x) dx + f(x) dx

  ∫ ∫ ∫ − 2 −

  2

  2

  2

  2

  = 2(8) – 3(3) f(x) dx + 1 f(x) dx = 2 – 1 = 1 ⇔ 2 = ∫ ⇔ ∫

  = 7

  2 p Jadi, f(x) dx = 1.

  ∫

  3. a. (4x – 5) dx = –3

  ∫ p

  2 B. Uraian  

  ⇔ 2x 5x = –3

  − _{2 1}   x 2 − ₂

  2

  1. a. dx = x dx ⇔ (2p – 5p) – 0 = –3

  ∫ ∫ x ₃

  2 ₂ ⇔ 2p – 5p + 3 = 0

  = x dx ⇔ (2p – 3)(p – 1) = 0

  ∫ ₃

  3

• 1

  1 ₂ ⇔ p = atau p = 1

  = x + c

  2 ³ + ₂

  1 ₅

  1

  2 ₂

  2

  = x + c = x x + c _{5 2}

  5

  3

  ∫

  2

  −   − + +  

  1 x 2x 3x

  2

  3

  2

  =

  2

• – 4x + 5) dx = 20 ⇔

  1

  f(x) dx =

  ⇒ 4(3) – 3(3)

  2

  ⇔ c = 3 Jadi, f(x) = –3x

  2 + 4x + 3.

  b.

  2 1 − ∫

  2 1 − ∫

  (4 – 6x) dx = 4x – 3x

  f(x) dx =

  2 1 − ∫

  (–3x

  2

  cos 2x dx – 2

  ∫

  2

  ∫

  ∫

  3

  = 20 ⇔(

  8p

  3

  p

  3

  7p

  7p

  f ′(x) dx =

  3

  = 21 ⇔ p = 21 ×

  3

  7

  = 9 4. a. f

  ′(x) = 4 – 6x f(x) =

  ∫

  (cos 2x – 2 sin x) dx =

  1. Jawaban: c

  2

  9

  2 ⁶ ₃  

  = (

  9

  2

  9

  2

  2

   

  1

  2 Jadi, nilai

  6 ∫

  f(x) dx = –31

  1

  2 .

  A. Pilihlah jawaban yang tepat.

  2x – 2x

       

  = (–8 + 8 + 6) – (1 + 2 – 3) = 6 – 0 = 6 5.

  f(x) dx =

  6 ∫

  f(x) dx =

  3 ∫

  f(x) dx +

  6

  3 ∫

  3 ∫

  2 x 4x

  (x + 4) dx +

  6

  3 ∫

  (2 – 4x) dx =

  3

  2

  1

  − +    

  3 x 2x 5x  

  1 p

• – 8 + 10) – (
• – 2 + 5) = 20 ⇔
• – 1 = 20 ⇔
• 12) – (0 + 0)
• (12 – 72) – (6 – 18) = (
• c f(3) = –12

• 12) – 0 + (–60) – (–12) =
• c = –12 ⇔ 12 – 27 + c = –12
• – 36 = –31
• 4x + 3) dx

  ∫

  2

  · 2 sin x cos x + c = 3 sin x cos x + c

  3. Jawaban: b sin a cos b =

  1

  2

  (sin (a + b) + sin (a – b))

  ∫

  4 sin 5x cos 3x dx =

  4 ·

  1

  2

  (sin 8x + sin 2x) dx = 2

  ∫

  (sin 8x + sin 2x) dx = 2(–

  3

  2

  sin 2x + c =

  (3 – 6 sin

  sin 2x – 2(–cos x) + c =

  1

  2

  sin 2x + 2 cos x + c

  2. Jawaban: d

  ∫

  2

  1

  x) dx =

  ∫

  3(1 – 2 sin

  2

  x) dx = 3

  ∫

  cos 2x dx = 3 ·

  1

  8

  2

  1

  1

  2

  x – π) dx =

  1

  2 ∫

  sin (x – 2 π) dx = –

  2

  2

  cos (x – 2 π) + c b.

  sin x dx =

  ∫

  1 (px

  2

  3

  sin 2(

  1

  cos 8x –

  4. Jawaban: b

  1

  2

  cos 2x) + c = –

  1

  4

  cos 8x – cos 2x + c

  ∫

  ∫

  sin (

  1

  2

  x – π) cos (

  1

  2

  x – π) dx =

4 A. Pilihan Ganda

• –(cos
• 3 sin
• 0) = (
• – sin 0) = –
• 3
• 3
• 0) – (–
• 0) =

  =

  π ∫

  cos 2x dx = ³

  1

  2 sin 2x

  π     

  −

  = –

  1

  2

  (sin

  2

  3 π

  1

  2

  (

  1

  2 3 – 0)

  = –

  1

  4

  3 B. Uraian 1. a.

  ∫

  (cos x + 2 sin x) dx =

  ∫

  cos x dx + 2

  ∫

  sin x dx = sin x + 2(–cos x) + c = sin x – 2 cos x + c b.

  x) dx = – ³

  x – sin

  2

  1

  1

  3

  (sin (2 π – π) – sin (

  3

  2

  π – π)) =

  1

  3

  (sin π – sin

  1

  2

  π) =

  3

  2

  (0 – 1) = –

  1

  3

  10. Jawaban: e ₃

  π ∫

  (sin x + cos x)(sin x – cos x) dx = ³

  π ∫

  (sin

  2

  x – cos

  2

  x) dx = ³

  π ∫

  ∫

  sin (2x +

  2

  3

  π) dx = –

  1

  2

  cos (2x +

  2

  3

  π) + c c.

  ∫

  6 sec

  2

  3x dx = 6 ·

  1

  3

  tan 3x + c = 2 tan 3x + c d.

• π
• 1

  (2 sin

  2

  −

  dx =

  ∫

  (

  6 tan 3x cos 3x

  ) dx =

  ∫

  (6 tan 3x sec 3x – sec

  3x) dx = 6 ·

  sin 2x + c 2. a.

  1

  3

  sec 3x –

  1

  3

  tan 3x + c = 2 sec 3x –

  1

  3

  ∫ 6 tan 3x sec 3x cos 3x

  2

  1

  cos 2x dx = 2(–3 cos

  3

  x – 3 cos 2x) dx = 2

  ∫

  sin

  1

  3

  x dx – 3

  ∫

  1

  3

  3

  ∫

  1

  2

  sin 2x) + c = –6 cos

  1

  3

  x –

  x) – 3(

  − π

  π π      

  2

  · (–

  1

  2

  ) + 3(

  1

  2 3 ) – (–

  1

  2

  1

  4

  2 3 ) – (–

  1

  ) =

  1

  3

  4

  2 3 =

  3

  4

  (1 + 2

  3 )

  7. Jawaban: d

  π ∫

  (sin 3x + cos x) dx =

  1

  3 cos 3x sin x

  2

  cos 0 + 3 sin 0) = (–

  − +

  (sin 2x + 3 cos x) dx = ^{1 3}

  5

  5. Jawaban: d

  ∫ 1 cos 2x − dx

  =

  ∫

  2 2 sin x dx

  =

  ∫ 2 sin x dx

  = –

  2 cos x + c

  6. Jawaban: e _{1 3}

  π ∫

  1

  2

  2 cos 2x 3 sin x)

  π      

  − +

  = (–

  1

  2

  cos

  2

  3 π

  3 π

  ) – (–

  1

  π     

  = (–

  3 sin (3x )

  3 −

  2

  sin (–2 π +

  1

  3

  π) =

  1

  2

  sin π –

  1

  2

  sin

  5

  π = 0 –

  π) –

  1

  2

  (

  1

  2 3 ) = –

  1

  4

  3

  9. Jawaban: b _{2 3 1 2}

  π π ∫

  cos (3x – π) dx = ^{2 3} _{1 2}

  1

  1

  3

  1

  π −π ∫

  3

  cos 3 π + sin π) – (–

  1

  3

  cos 0 + sin 0) = (

  1

  3

  1

  3

  2

  3

  8. Jawaban: b _{1 3}

  cos (2x +

  3 π

  1

  3

  π) dx = ^{1 3}

  1

  1

  2

  3 sin (2x )

  π −π     

  =

  1

  2

  sin (

  2

• – sec 3x cos 3x

  tan 3x + c b

  2

  b. (sin 2x – cos 2x) dx

  ∫

  sin 2x dx

  ∫

  2

  2 a

  = (sin 2x – 2 sin 2x cos 2x + cos 2x) dx

  ∫ b

  2

  2

  1  

  = − cos 2x = (sin 2x + cos 2x – 2 sin 2x cos 2x) dx 

  ∫ 2 

   a b

  = (1 – sin 4x) dx 

  1  ∫

  2

  = − (1 2 sin x) −

  

  1 a

  2  

  = x + cos 4x + c

  b

  4 1 

  2 

  = − + sin x

  

  2   a ₂ π

  1

  1

  2

  2

  = (– + sin

  b) – (– + sin

  a) 3. a. (cos 2x + sin 3x) dx

  2

  2 ∫

  2

  2 π ₂ = sin b – sin a

  1

  1  

  = sin 2x − cos 3x

   

  = (sin b – sin a)(sin b + sin a)

  

  2 3 

  1

  1

  3

  1

  1 π

  = c (sin b + sin a) = ( sin π – cos ) – ( sin 0 – cos 0)

  2

  3

  2

  2

  3

  1

1 = c (sin a + sin b)

  = (0 – 0) – (0 – ) =

  3

  3 b

  π ₂ Terbukti bahwa sin 2x dx = c(sin a + sin b).

  ∫ π a

  b. 2 cos ( – x) dx

  ∫

  4 π

  5. a. f ′(x) = 12 cos 2x

  4 ₂ π 2 π   f(x) =

  ∫ 12 cos 2x dx = sin ( − x)

    π

  −

  1 4  ₄

  1

  = 12 · sin 2x + c

  2 π

  = –2 (sin (– ) – sin 0)

  4

  = 6 sin 2x + c

  1  π 

  = –2 (–

  2 – 0) =

  2

  2

  f   = 8

  12   ₃ π π  

    + c = 8

  ⇔ 6 sin 2

  c. 6 sin x cos x dx

  12   ∫

  π π

  6 sin + c = 8 ⇔ ₃

  6

  1

  = 3 sin 2x dx

  ∫

  6 · + c = 8 ⇔

  2 ₃ π ⇔ 3 + c = 8 

  3

  = − cos 2x ⇔ c = 5

   2   Diperoleh f(x) = 6 sin 2x + 5.

  3

  2 π

  = – (cos – cos 0)

  2 3   π   π

  b. f   = 6 sin 2   + 5

  4

  4    

  3

  1

  = – (– – 1)

  π

  2

  2

  = 6 sin + 5

  2

  3

  3

  = 6 · 1 + 5 = – (– )

  2

  2

  = 11

  9

  =

  4 b

  4. cos x dx = c

  ∫ a b

    sin x

  = c ⇔

      a

  ⇔ sin b – sin a = c

  6

• 4
• 8x + 1
• ·
^{2dx =}

• ) dx
• 9x – 1
• 1
• −
• 9x – 1)
^{1 2} · (2x + 3)
• 8x + 1)

  2

  = 6x + 9 = 3(2x + 3) ⇔ (2x + 3) dx =

  du

  3 ∫ ₂

  2x

  3 3x 9x

  1

  dx =

  ∫

  (3x

  u ^{1 2}

  ∫

  2

  −

  ·

  du

  3

  =

  1

  3 ∫

  u ^{1 2}

  −

  du dx

  3 2x 4 + + c

  4. Jawaban: c Misalkan u = 3x

  ∫ ^{1 2}

  3. Jawaban: c Misalkan u = 2x

  3

  du dx

  = 6x

  2

  ⇔ du = 6x

  2

  dx

  ∫ ² ₃ 3x 2x

  4

  3 (2x 4) ⁻

  1

  2

  dx =

  1

  2 ∫ ^{1 2} u ⁻

  du =

  1

  2

  × _{1 2}

  1 ¹ ₂ u + c

  = u + c =

  du =

• – 5

  5. Jawaban: c Misalkan u = 4x

• – 5)
⁵ 7<
• – 5)
⁵ 7<
• 1 −

  (x – 2)

  1

  4

  (x – 2)

  4

  (4 dx) = x(x – 2)

  4

  (x – 2)

  4

  d(x – 2) = x(x – 2)

  4

  5

  1

  5

  2 3x 9x − 1 + + c

  5

  (x – 2)

  4

  (5x – (x – 2)) + c =

  1

  5

  (4x + 2)(x – 2)

  7 (2x 5) − + c

  2

  3

  6

  4

  (x – 2)

  4

  1

  3

  2

  7

  3

  ⇒ du = 4 dx dv = (x – 2)

  3

  dx ⇒ v =

  ∫

  (x – 2)

  3

  dx =

  ∫

  (x – 2)

  3

  d(x – 2) =

  1

  4

  (x – 2)

  4 ∫

  u dv = uv –

  ∫

  v du

  ∫

  4x(x – 2)

  3

  dx = (4x) ·

  · 2u ^{1 2} + c =

• – ∫
• – ∫
• – 1
• c =
• – 5)
^{2 7} + c =

  6

  3

  1

  2

  ( ¹ ₂

  1 1 +

  u ^{1 2}

  ) + c =

  1

  2

  ( _{3 2}

  1

  u ^{3 2} ) + c =

  2

  =

  (

  2

  3

  u ^{3 2} ) + c =

  1

  3

  u ^{3 2} + c =

  1

  3

  u u + c =

  1

  1

  2 ∫ u du

  (2x

  2 2x + 8x + 1

  7 A. Pilihan Ganda

  1. Jawaban: c Misalkan u = 2x

  2

  du dx

  = 4x + 8 ⇔ du = (4x + 8) dx

  ∫

  (2x + 4)

  2 2x + 8x + 1

  dx =

  ∫

  (2x + 4) dx =

  1

  ∫

  2 2x + 8x + 1

  (

  4x

  8

  2

  =

  ∫ u

  1

  2

  du =

  3

  2

  (2x

  1 1 −

  3

  du =

  1

  3 ∫

  u ^{5 7}

  −

  du =

  1

  3

  ( _{5 7}

  u ^{5 7}

  −

  ) + c =

  1

  3

  (

  7

  2

  u ^{2 7} ) + c =

  7

  6

  u ^{2 7} + c =

  7

  1

  u ^{5 7}

  2 2x + 8x + 1 + c

  ∫

  2. Jawaban: d Misalkan u = 2x

  3

  du dx

  = 6x

  2

  ⇔ du = 6x

  2

  dx

  ∫ ² _{3 5 7} 2x (2x 5) −

  dx =

  ∫

  2x

  2

  (2x

  3

  −

  dx =

  ∫

  (2x

  3

  −

  (2x

  2

  dx) =

  4

• c
• 1)

  4

  v du

  5

  2x + c

  10. Jawaban: b Misalkan u = (x

  2

  ∫

  sin x dx = –cos x

  ∫

  u dv = uv –

  ∫

  ∫

  10

  (x

  2

  2

  ∫

  (–cos x) (2x dx) = –(x

  2

  ∫

  2x cos x dx = (2 – x ) cos x + 2

  ∫

  x d(sin x) = (2 – x

  cos

  1

  ) cos x + 2 (x sin x –

  ) = –

  2 ∫

  cos

  4

  2x sin 2x dx =

  ∫

  u

  4

  (–

  du

  2

  1

  5

  2 ∫

  u

  4

  du = –

  1

  2

  ·

  1

  5

  u

  2

  ∫

  = –2 sin 2x ⇔ sin 2x dx = –

  ∫

  ∫

  2x(x

  2

  3

  dx =

  ∫

  (x

  2

  3

  · 2x dx =

  u

  du dx

  3

  du =

  1

  4

  u

  4

  1

  4

  (x

  2

  = 2x ⇔ 2x dx = du

  2

  (sin x) dx) = (2 – x

  ∫

  2

  ) cos x + 2x sin x – 2 (–cos x) + c = (2 – x

  2

  ) cos x + 2x sin x + 2 cos x + c = (4 – x

  2

  ) cos x + 2x sin x + c

  B. Uraian

  1. a. Misalkan u = 5 – x

  du dx

  = –1 ⇔ dx = –du

  2 5 x −

  b. Misalkan u = x

  dx =

  ∫

  2 u

  (–du) = –2

  ∫

  u ^{1 2}

  −

  du = –2 · 2u ^{1 2} + c

  = –4

  5 x − + c

  du

  du dx

• 1)
• – (a
• 1)
• 1)

  1

  2

  (a

  = 7 ⇔

  3

  2

  ) = 14 ⇔ 8 – (a

  3

  2

  3

  = 14 ⇔ 2((1 + 1)

      

  3 a (x + 1)

  1

  9. Jawaban: b Misalkan u = cos 2x

  3

  2

  · 2x dx = 14 ⇔ 6 ·

  2

  2

  1 a ∫ (x

  6

  dx = 14 ⇔

  2

  2

  1 a ∫ 12x(x

  6. Jawaban: c

  3

• 1)
• c = –
• 1 = 1 ⇔ a
• – 2) ⇒ du = 2x dx dv = sin x dx ⇒ v =
• – 2) sin x dx = (x
• – 2) (–cos x) –
• – 2) cos x +

  π ∫

  1

  2

  2

  = 1 ⇔ a

  2

  x sin x dx = 2

  π ∫

  (cos x)

  2

  d(–cos x) = –2

  π ∫

  (cos x)

  2

  d(cos x) = –2 ·

  3

  3 (cos x)

  7. Jawaban: c Misalkan u = sin 2x

  0) = –

  8

  3

  4

  (–1 – 1) =

  3

  2

  3

  π  

  π – cos

  3

  (cos

  3

  2

  = –

  (2 sin x cos x) cos x dx = 2

  = 0 ⇔ a = 0

  du dx

  = 2 cos 2x ⇔

  π ∫

  sin 2x cos x dx =

  π ∫

  8. Jawaban: c

  sin 2x sin 2x + c

  3

  1

  u u + c =

  3

  1

  u ^{3 2} + c =

  3

  2

  ·

  2

  1

  (sin 2x) ^{1 2} (cos 2x dx) =

  1

  2

  du = cos 2x dx

  ∫

  cos 2x sin 2x dx =

  u ^{1 2} du =

  ∫

  ∫

  u ^{1 2} (

  1

  2

  cos

  1

  2 ∫

  du) =

• – 3
• – 3)
• – 3)
• c =
• – 3)
• c

  2

  =

  = ²

  2

  1

  2 u

  π    

  = ²

  2

  1

  2 (1 cos x)

  π   −  

  1

  (1 – cos x) sin x dx = ²

  2

  (1 – cos

  2 π

  )

  2

  2

  (1 – cos 0)

  2

  =

  1

  π ∫ u du

  π ∫

• –2
• –2
• – 1
• –2
• – 1

  1 −

  5

  1

  1 u − −

       

  =

  1

  2

  (

  1

  5 −

  (1 – 0)

  = sin x ⇔ du = sin x dx ₂

  1

  2

  ·

  4

  5

  =

  2

  5 b. ² π ∫

  (1 – cos x) sin x dx Misalkan u = 1 – cos x

  du dx

  ) =

  2

• –1
• c =

• c =
2<
• c =
2<
• c 2. a. Misalkan u = x
• – 4x – 1
• – ∫

• –2
>– 4x – 1)

  1

  2

  sin (2x – 1) + c b.

  ∫

  (3x + 2) cos (3x + 2) dx =

  ∫

  (3x + 2) d(

  1

  3

  sin (3x + 2)) = (3x + 2) ·

  sin (3x + 2)

  3

  cos (2x – 1) dx = –x cos (2x – 1) +

  1

  3

  sin (3x + 2) d(3x + 2) = (x +

  2

  3

  ) sin (3x + 2) +

  1

  3

  1

  2

  2

  2x sin (2x – 1) dx =

  (1 – 1)

  2

  1

  2

  1

  2

  =

  1

  2 3. a.

  ∫

  ∫

  cos (2x – 1)) d(2x) = –x cos (2x – 1) +

  2x d(–

  1

  2

  cos (2x – 1)) = 2x · (–

  1

  2

  cos (2x – 1))

  (–

  1

  2

  ∫

    −  

  1

  ∫

  (2x – 3)

  2 2x 3 − + c

  d. Misalkan u = 4 – 3x

  2 du dx

  = –6x ⇔ x dx =

  du

  6 − ∫ ^{2 2}

  3x (4 3x ) −

  dx = 3

  (4 – 3x

  2

  2

  )

  · x dx = 3

  ∫

  u

  ·

  du

  6 −

  =

  3

  5

  u ^{5 2} + c =

  u

  ∫

  9

  c. Misalkan u = 2x – 3

  du dx

  = 2 ⇔ dx =

  du

  2 ∫

  (4x – 6)

  2x 3 −

  dx =

  2(2x – 3)(2x – 3) ^{1 2} dx = 2

  5

  ∫

  (2x – 3) ^{3 2} dx = 2

  ∫

  u ^{3 2} ·

  du

  2

  =

  ∫

  u ^{3 2} du =

  2

  6 − ∫

  du = –

  =

  5

  1 u

  − − −

  ∫

  · (–

  1

  2

  ) du = –

  1

  2

  2

  5

  1 u

  − − −

  ∫

  du = –

  1

  2

  5

  1

  1 1u

  − − −

  2

  (2 – x) dx =

  1

  ∫

  2

  ·

  1

  1 −

  u

  1 2u

  1 2(4 3x ) −

  1 8 6x −

  2

  du dx

  = 2x – 4 ⇔ du = (2x – 4) dx ⇔ (2 – x) dx = –

  1

  2

  du x = 0 ⇒ u = 0 – 0 – 1 = –1 x = 2

  ⇒ u = 4 – 8 – 1 = –5

  2

  2

  2 2 x (x 4x 1)

  − − − ∫

  dx =

  2

  2 (2x

• – ∫
• – 1

  cos (3x + 2) + c

  4x π

  dx ₂

  ∫ 4 x −

  2 ^{1 1}

  4. sin 2x sin x dx

  ∫ _{2 2}

  = 4x(–2(4 – x) ) – (–2(4 – x) ) 4dx ₁ ∫ ₁

  π _{2 2}

  = –8x(4 – x) – 8 (4 – x) (–1) dx

  2 ∫

  2

  = (2 sin x cos x) sin x dx _{1 1}

  ∫ _{2 2}

  = –8x(4 – x) – 8 (4 – x) d(4 – x)

  ∫ π _{1 3}

  2 ₂

  2 ₂

  2

  2

  = –8x(4 – – 8 · (4 – + c

  x) x)

  =

  4 sin x cos x sin x dx

  3 ∫

  16 π

  3

  = –8x

• – 2

  4 x (4 x) + c − −

  3

  2

  2

  = – 4(1 – cos x) cos x (–sin x) dx

  ∫

  16

  3 4 x

• – Jadi, f(x) dx = –8x (4 x) + c.

  − − ∫

  3 π

  2