Pembahasan Latihan Soal Integral (1) UN SMA


  1. Diketahui \int_a^3 (3x2 + 2x + 1) dx = 25 Nilai \frac{1}{2} a = …

    A. – 4

    B. – 2

    C. – 1

    D. 1

    E. 2

    PEMBAHASAN :

    \int_a^3 (3x2 + 2x + 1) dx = x3 + x2 + x \mid_a^3

    25 = (33 + 32 + 3) – (a3 + a2 + a)

    a3 + a2 + a = 27 + 9 + 3 – 25

    a3 + a2 + a – 14 = 0

    (a – 2)(a2 + a + 7) = 0

    a = 2 atau a2 + a + 7 = 0

    jadi \frac{1}{2} a = 1

    JAWABAN : D

  2. Nilai \int_0^\pi sin 2x cos x dx = …

    A. -4/3

    B. -1/3

    C. 1/3

    D. 2/3

    E. 4/3

    PEMBAHASAN :

    \int_0^\pi sin 2x cos x dx = \int_0^\pi 2 sin x cos x cos x dx

    = \int_0^\pi 2 sin x cos2 x dx

    misal u = cos x \Rightarrow du = -sin x dx

    = \int_0^\pi 2 u2 (-du)

    = -\frac{1}{3} u3 \mid_0^\pi

    Substitusi u = cos x

    = -\frac{1}{3} cos3 x \mid_0^\pi

    = -\frac{1}{3} cos3 (\pi) + \frac{1}{3} cos3 0

    = -\frac{1}{3} (-1)3 + \frac{1}{3} .13

    = \frac{1}{3} + \frac{1}{3}

    = \frac{2}{3}

    JAWABAN : D

  3. Hasil dari \int_0^1 3x\sqrt{3x^2+1} dx = …

    A. 7/2

    B. 8/3

    C. 7/3

    D. 4/3

    E. 2/3

    PEMBAHASAN :

    \int_0^1 3x\sqrt{3x^2+1} dx = …

    misal u = 3x2 + 1 \Rightarrow du = 6x dx

    = \int_0^1 \sqrt{u} \frac{du}{2}

    = \int_0^1 \frac{1}{2} u1/2 du

    = \frac{1}{2} .\frac{2}{3} u3/2 \mid_0^1

    substitusi u = 3x2 + 1, sehingga diperoleh

    = \frac{1}{3} (3x2 + 1)3/2 \mid_0^1

    = \frac{1}{3} (3.12 + 1)3/2\frac{1}{3} (3.02 + 1)3/2

    = \frac{1}{3} 8 – \frac{1}{3} .1

    = \frac{7}{3}

    JAWABAN : C

  4. Hasil dari \int cos5 x dx = …

    A. -\frac{1}{6} cos6 x sin x + C

    B. \frac{1}{6} cos6 x sin x + C

    C. –sin x + \frac{2}{3} sin3 x + \frac{1}{5} sin5 x + C

    D. sin x – \frac{2}{3} sin3 x + \frac{1}{5} sin5 x + C

    E. sin x + \frac{2}{3} sin3 x + \frac{1}{5} sin5 x + C

    PEMBAHASAN :

    \int cos5 x dx = \int cos x (cos2 x)2 dx

    = \int cos x (1 – sin2 x)2 dx

    = \int cos x (1 – 2 sin2 x + sin4 x) dx

    misal u = sin x \Rightarrow du = cos x

    = \int (1 – 2u2 + u4) du

    = u – \frac{2}{3} u3 + \frac{1}{5} u5 + C

    substitusi u = sin x,

    = sin x – \frac{2}{3} sin3 x + \frac{1}{5} sin5 x + C

    JAWABAN : D

  5. Hasil dari \int cos x (x2 + 1) dx = …

    A. x2 sin x + 2x cos x + C

    B. (x2 – 1)sin x + 2x cos x + C

    C. (x2 + 3)sin x – 2x cos x + C

    D. 2x2 cos x + 2x2 sin x + C

    E. 2x sin x – (x2 – 1)cos x + C

    PEMBAHASAN :

    dalam penyelesaian soal ini akan menggunakan Integral Parsial

    u = x2 + 1 \Rightarrow du = 2x dx

    dv = cos x dx \Rightarrow v = sin x

    \int u dv = uv – \int v du

    = sin x (x2 + 1) – \int sin x 2x dx

    parsial lagi

    m = 2x \Rightarrow dm = 2 dx

    dn = sin x dx \Rightarrow n = -cos x

    = sin x (x2 + 1) – (2x (-cos x) – \int -cos x 2 dx)

    = sin x (x2 + 1) – (-2x cos x + 2 sin x) + C

    = sin x (x2 + 1) + 2x cos x – 2 sin x + C

    = sin x (x2 – 1) + 2x cos x + C

    JAWABAN : B

  6. Diketahui \int_p^3 (3x2 – 2x + 2) dx = 40. Nilai \frac{1}{2} p = …

    A. 2

    B. 1

    C. – 1

    D. – 2

    E. – 4

    PEMBAHASAN :

    \int_p^3 (3x2 – 2x + 2) dx = x3 – x2 + 2x \mid_p^3

    40 = (33 – 32 + 6) – (p3 – p2 + 2p)

    p3 – p2 + 2p = 27 – 9 + 6 – 40

    p3 – p2 + 2p + 16 = 0

    (p + 2)(p2 + p + 7) = 0

    p = -2 atau p2 + p + 7 = 0

    jadi \frac{1}{2} p = -1

    JAWABAN : C

  7. Hasil dari \int_0^{\frac{\pi}{2}} sin 3x cos 5x dx = …

    A. -10/6

    B. -8/10

    C. -5/16

    D. -4/16

    E. 0

    PEMBAHASAN :

    \int_0^{\frac{\pi}{2}} sin 3x cos 5x dx = \int_0^{\frac{\pi}{2}} \frac{1}{2} [sin 8x + sin (-2x)] dx

    [Sifat Trigonometri]

    = \int_0^{\frac{\pi}{2}} \frac{1}{2} sin 8x dx – \int_0^{\frac{\pi}{2}} \frac{1}{2} sin 2x dx

    misal u = 8x \Rightarrow du = 8 dx

    v = 2x \Rightarrow dv = 2 dx

    = \int_0^{\frac{\pi}{2}} \frac{1}{2} sin u \frac{du}{8} \int_0^{\frac{\pi}{2}} \frac{1}{2} sin v \frac{dv}{2}

    = -\frac{1}{16} cos u \mid_0^{\frac{\pi}{2}} + \frac{1}{4} cos v \mid_0^{\frac{\pi}{2}}

    substitusi u = 8x dan v = 2x

    = -\frac{1}{16} cos 8x \mid_0^{\frac{\pi}{2}} + \frac{1}{4} cos 2x \mid_0^{\frac{\pi}{2}}

    = [-\frac{1}{16} (cos 8(\frac{\pi}{2}) - cos 8(0))] + [\frac{1}{4} (cos 2(\frac{\pi}{2}) – cos 2(0))]

    = [-\frac{1}{16} (1 – 1)] + [\frac{1}{4} (-1 – 1)]

    = -\frac{1}{2}

    JAWABAN :

  8. \int_0^\pi x sin x dx = …

    A. \frac{\pi}{4}

    B. \frac{\pi}{3}

    C. \frac{\pi}{2}

    D. \pi

    E. \frac{3\pi}{2}

    PEMBAHASAN :

    dalam penyelesaian soal ini akan menggunakan Integral Parsial

    u = x \Rightarrow du = dx

    dv = sin x dx \Rightarrow v = -cos x

    \int u dv = uv – \int v du

    = -x cos x – \int (-cos x) dx

    = [-x cos x + sin x] \mid_0^\pi

    = [-\pi cos (\pi) + sin (\pi)] – [-0 cos 0 + sin 0]

    = -\pi (-1)

    = \pi

    JAWABAN : D

  9. Nilai \int_0^{\frac{\pi}{2}} (2x + sin x) dx = …

    A. \frac{1}{4}\pi^2 – 1

    B. \frac{1}{4}\pi^2

    C. \frac{1}{4}\pi^2 + 1

    D. \frac{1}{2}\pi^2 – 1

    E. \frac{1}{2}\pi^2 + 1

    PEMBAHASAN :

    \int_0^{\frac{\pi}{2}} (2x + sin x) dx = x2 – cos x \mid_0^{\frac{\pi}{2}}

    = ((\frac{\pi}{2})2 – cos (\frac{\pi}{2})) – (02 – cos 0)

    = (\frac{\pi^2}{4} – 0) – (02 – 1)

    = \frac{\pi^2}{4} + 1

    JAWABAN : C

  10. Nilai \int x sin(x2 + 1) dx = …

    A. –cos (x2 + 1) + C

    B. cos (x2 + 1) + C

    C. –½ cos (x2 + 1) + C

    D. ½ cos (x2 + 1) + C

    E. –2cos (x2 + 1) + C

    PEMBAHASAN :

    misal u = x2 + 1 \Rightarrow du = 2x dx

    \int x sin(x2 + 1) dx = \int sin u \frac{du}{2}

    = -\frac{1}{2} cos u + C

    substitusi u = x2 + 1

    = -\frac{1}{2} cos (x2 + 1) + C

    JAWABAN : C

NOTE : silahkan dikoreksi dan berikan komentar jika ada kesalahan atau masih ada keambiguan dalam penyelesaian soal-soal ini.

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10 comments on “Pembahasan Latihan Soal Integral (1) UN SMA

    • waalaikumslam
      setau saya uji integral ini hanya ingin menguji deretnya konvergen atau divergen. Jadi dengan menunjukkan integralnya [dengan btas atas = \infty dan batas bawah = 1] konvergen maka dretnya juga scra bersama-sama konvergen ato jika integralnya divergen maka deretnya juga divergen.

      contoh :
      cek apakah deret \sum \frac{1}{n^p} konvergen atau divergen untuk 0 < p < 1
      penyelesain :
      \int_1^{\infty} \frac{1}{X^p} dx = lim_{M \to \infty} \int_1^{M} \frac{1}{X^p} dx
      = \frac{1}{1-p} lim_{M \to \infty} (X^{1-p}) \mid_1^M
      = \frac{1}{1-p} lim_{M \to \infty} (M^{1-p}-1)
      = \infty
      karena integralnya divergen, maka deretnya divergen

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