Problem (7) : Integral


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  1. \int_0^2 (x – 2)2 dx = …

    Penyelesaian :

    Misal u = x – 2 \Rightarrow du = dx, kemudian substitusi

    \int_0^2 u2 du = \frac{1}{3} u3 \mid_{0}^{2}

    Substitusi lagi u = x – 2

    = \frac{1}{3} (x – 2)3 \mid_{0}^{2}

    = [\frac{1}{3} (2 – 2)3] – [\frac{1}{3} (0 – 2)3]

    = 0 – [-\frac{8}{3}]

    = \frac{8}{3}

  2. \int_1^4 (\frac{1}{\sqrt{x}}\sqrt{x}) dx = …

    Penyelesaian :

    \int_1^4 (\frac{1}{\sqrt{x}}\sqrt{x}) dx = \int_1^4 (x-1/2 – x1/2) dx

    = \frac{1}{2}x1/2\frac{2}{3}x3/2 \mid_{1}^{4}

    = [\frac{1}{2}41/2\frac{2}{3}43/2] – [\frac{1}{2}11/2\frac{2}{3}13/2]

    = [1 – \frac{16}{3}] – [\frac{1}{2}\frac{2}{3}]

    = \frac{1}{2}\frac{14}{3}

    = –\frac{25}{6}

  3. \int x\sqrt{x^2-3} dx = …

    Penyelesaian :

    misal u = x2 – 3 \Rightarrow du = 2x du, substitusi sehingga diperoleh

    \int x\sqrt{x^2-3} dx = \int x\sqrt{u} (du/2x)

    = \frac{1}{2} \int u1/2 du

    = \frac{1}{2} \frac{2}{3} u3/2 + C

    = \frac{1}{3} u3/2 + C

    Substitusi kembali u = x2 – 3

    = \frac{1}{3} (x2 – 3)3/2 + C

  4. \int cos2 x sin x dx = …

    Penyelesaian :

    misal u = cos x \Rightarrow du = -sin x dx, kemudian substitusi

    \int u2 (-du) = \frac{1}{3}u3 + C

    Substitusi kembali u = cos x, diperoleh

    = \frac{1}{3} cos3 x + C

  5. \int_{-1}^b (6x + 2) dx = 55

    tentukan nilai b = …

    Penyelesaian :

    3x2 + 2x \mid_{-1}^{b} = 55

    [3(b)2 + 2(b)] – [3(-1)2 + 2(-1)] = 55

    [3b2 + 2b] – [3 – 2] = 55

    3b2 + 2b – 56 = 0

    (3b + 14)(b – 4) = 0

    b = –\frac{14}{3} atau b = 4

  6. \int_{-\pi}^{\frac{\pi}{2}} (4x3 + cos x) dx = …

    Penyelesaian :

    \int_{-\pi}^{\frac{\pi}{2}} (4x3 + cos x) dx = x4 + sin x \mid_{-\pi}^{\frac{\pi}{2}}

    = [(\frac{\pi}{2})4 + sin \frac{\pi}{2}] – [(-\frac{\pi}{2})4 + sin -\frac{\pi}{2}]

    = (\frac{\pi}{2})4 + 1 -(\frac{\pi}{2})4 + 1

    = 2

  7. Tentukan luas daerah yang dibatasi oleh y = x2 + 1 (kurva 1) dan y = 3 – x (kurva 2)

    Penyelesaian :

    Photobucket

    cari batas-batasnya terlebih dahulu :

    x2 + 1 = 3 – x

    x2 + x – 2 = 0

    (x + 2)(x – 1) = 0

    x1 = -2 atau x2 = 1

    Luas Kurva = \int_{x_1}^{x_2} (kurva1 – kurva2) dx

    = \int_{-2}^{1} (x2 + x – 2) dx

    = \frac{1}{3}x3\frac{1}{2}x2 – 2x \mid_{-2}^{1}

    = [\frac{1}{3}13\frac{1}{2}12 – 2(1)] – [\frac{1}{3}(-2)3\frac{1}{2}(-2)2 – 2(-2)]

    = [\frac{1}{3}\frac{1}{2} – 2] – [-\frac{8}{3} – 2 + 4]

    = \frac{9}{2}

  8. Tentukan luas daerah yang dibatasi oleh y = sin 2x, sumbu x, garis x2 = –\frac{\pi}{3} dan x1 = \frac{\pi}{3}

    Penyelesaian :

    Photobucket

    misal u = 2x \Rightarrow du = 2 dx

    Volume kurva = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} sin 2x dx

    = \int_{-\frac{\pi}{3}}^{0} sin 2x dx + \int_{0}^{\frac{\pi}{3}} sin 2x dx

    = \int_{-\frac{\pi}{3}}^{0} sin u (du/2) + \int_{0}^{\frac{\pi}{3}} sin u (du/2)

    = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} sin u (du/2)

    = –\frac{1}{2} cos u \mid_{-\frac{\pi}{3}}^{0} + –\frac{1}{2} cos u \mid_{0}^{\frac{\pi}{3}}

    substitusi u = 2x, diperoleh

    = (-\frac{1}{2} [cos 2x \mid_{-\frac{\pi}{3}}^{0} ]) + (-\frac{1}{2} [cos 2x \mid_{0}^{\frac{\pi}{3}} ])

    = (-\frac{1}{2} [cos \frac{2\pi}{3} – cos 0]) + (-\frac{1}{2} [cos 0 – cos \frac{-2\pi}{3})])

    = (-\frac{1}{2} [-\frac{1}{2} – 1]) + (-\frac{1}{2}[1-(-\frac{1}{2})])

    = \frac{3}{4} + (-\frac{3}{4})

    karena luas kurva tidak mungkin negative, maka |-\frac{3}{4}| = \frac{3}{4}

    = \frac{3}{4} + \frac{3}{4}

    = \frac{3}{2}

  9. Tentukan luas benda putar yang terjadi oleh kurva y = x2, sumbu x dan garis x = 2 diputar 3600 mengelilingi sumbu x

    Penyelesaian :

    Karena benda putar mengelilingi sumbu-x, maka kita harus ubah persamaan kedalam variable x dan batas-batasnya berada di sumbu-x

    batas-batas : x = 0 dan x = 2

    Volume benda putar = \pi \int_{0}^{2} y2 dx

    = \pi \int_{0}^{2} x4 dx

    = \pi \frac{1}{5}x5 \mid_{0}^{2}

    = \pi \frac{1}{5}(25 – 05)

    = \frac{32}{5} \pi

  10. Hitunglah volume benda putar yang terjadi oleh kurva y = x2 – 2, sumbu x diputar 3600 mengililingi sumbu y

    Penyelesaian :

    Photobucket

    Karena benda putar mengelilingi sumbu-y, maka kita harus ubah persamaan kedalam variable y dan batas-batasnya berada di sumbu-y

    y = x2 – 2

    \sqrt{y+2} = x

    batas-batas : y = -2 dan y = 0

    Volume benda putar = \pi \int_{-2}^{0} x2 dx

    = \pi \int_{-2}^{0} (y + 2) dx

    = \pi \frac{1}{2}y2 + 2y \mid_{-2}^{0}

    = \pi([\frac{1}{2}02 + 2.(-2)) – [\frac{1}{2}(-2)2 + 2(-2)])

    = 2\pi

  11. Jika \int_0^b (2x – 3) dx = 4 dan \int_0^a \frac{1}{2} \quad \sqrt[3]{x^2} dx = \frac{3}{10}

    Jika a, b > 0 maka tentukan nilai a2 – 2ab + b2 = …

    Penyelesaian :

    \int_0^b (2x – 3) dx = 4

    x2 – 3x |0b = 4

    (b2 – 3b) – (02 – 3(0)) = 4

    b2 – 3b – 4 = 0

    (b – 4)(b + 1) = 0

    b = 4 atau b = -1

    \int_0^a \frac{1}{2} \quad \sqrt[3]{x^2} dx = \frac{3}{10}

    \int_0^a \frac{1}{2} x2/3 dx = \frac{3}{10}

    \frac{3}{10} x5/3 |0a = \frac{3}{10}

    \frac{3}{10} a5/3\frac{3}{10} 05/3 = \frac{3}{10}

    a5/3 = 1

    a = 1

    karena a, b > 0 maka yang memenuhi adalah a = 1 dan b = 4

    a2 – 2ab + b2 = (a – b)2

    = (1 – 4)2

    = 9

NOTE : kurva dibuat (plot) menggunakan software Mathematica6

2 comments on “Problem (7) : Integral

  1. ini gmn kk
    Tentukan luas permukaan benda putar apabila kurva y = akar x dengan interval 0 s/d 4
    Diputar mengelilingi sumbu y?

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