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Luas daerah bidang datar (kalkulus integral)
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Luas daerah bidang datar (kalkulus integral)
1.
Nama : Nurkhalifah
Anwar NIM : 1911041007 Kelas : A1 Tanggal : 21 April 2020 Tugas Kalkulus Integral Luasan Daerah 1. Luasan 𝑅 dibatasi oleh kurva𝑦 = 𝑥2 − 2 dan 𝑦 = 2𝑥2 + 𝑥 − 4 Jawab : Mencari batas atas dan bawahnya 𝑥2 − 2 = 2𝑥2 + 𝑥 − 4 𝑥2 + 𝑥 − 2 = 0 𝑥 = −2 atau 𝑥 = 1 Sehingga diperoleh ∆𝐴 ≈ [(𝑥2 − 2) − (2𝑥2 + 𝑥 − 4)]∆𝑥 ∆𝐴 ≈ (−𝑥2 − 𝑥 + 2)∆𝑥 𝐴(𝑅) = ∫ (−𝑥2 − 𝑥 + 2) 𝑑𝑥 1 −2 = [− 1 3 𝑥3 − 1 2 𝑥2 + 2𝑥] −2 1 = (− 1 3 (1)3 − 1 2 (1)2 + 2(1)) − (− 1 3 (−2)3 − 1 2 (−2)2 + 2(−2)) = (− 1 3 − 1 2 + 2) − ( 8 3 − 2 − 4) = 9 2 𝐴(𝑅) = 9 2 ≈ 𝟒.𝟓 satuan luas.
2.
2. Luasan 𝑅
dibatasi oleh kurva-kurva𝑦 = 𝑥,𝑦 = 2𝑥 dan 𝑦 = 5 − 𝑥 Jawab: Mencari batas atas dan bawahnya, Diperoleh titik potongnya 𝑎 = 0, 𝑏 = 5 3 , 𝑐 = 5 2 . Sehingga diperoleh 𝐴(𝑅) = ∫ (2𝑥 − 𝑥) 𝑑𝑥 5/3 0 + ∫ (5 − 𝑥 5/2 5/3 − 𝑥)𝑑𝑥 = [ 1 2 𝑥2] 0 5/3 + [5𝑥 − 𝑥2]5/3 5/2 = ( 1 2 ( 5 3 ) 2 ) − (0) + (5 ( 5 2 ) − ( 5 2 ) 2 ) − (5( 5 3 ) − ( 5 3 ) 2 ) = ( 25 18 ) + ( 25 36 ) = 25 12 𝐴(𝑅) = 25 12 ≈ 𝟐.𝟎𝟖𝟑̇ satuan luas. 2𝑥 = 5 − 𝑥 𝑥 = 5 3 𝑥 = 5 − 𝑥 𝑥 = 5 2
3.
3. Luasan 𝑅
dibatasi oleh kurva𝑦 = √𝑥 dan 𝑦 = −𝑥 + 6 Jawab: Mencari batas atas dan bawahnya (Titik-titikpotong yang diperoleh) √𝑥 = −𝑥 + 6 𝑥 = 𝑥2 − 12𝑥 + 36 𝑥2 − 13𝑥 + 36 = 0 (𝑥 − 9)(𝑥 − 4) 𝑥 = 9 (tidak memenuhi) atau 𝑥 = 4 Titik potong pada sumbu x, y=0 −𝑥 + 6 = 0 √𝑥 = 0 −𝑥 = −6 𝑥 = 0 𝑥 = 6 Didapatkan (0,0); (4,2);(6, 0) . Daerah pada grafik di atas terdapat dua daerah, yaitu dari 𝑥 = 0 sampai 𝑥 = 4 dan dari 𝑥 = 4 sampai 𝑥 = 6. Sehingga diperoleh 𝐴(𝑅)1 = ∫ √𝑥 4 0 𝑑𝑥 = [ 2 3 𝑥 3 2] 0 4 = ( 2 3 (4) 3 2) − (0) = 16 3 ≈ 5.3̇ satuan luas. 𝐴(𝑅)2 = ∫ (−𝑥 + 6) 6 4 𝑑𝑥 = [− 𝑥2 2 + 6𝑥] 4 6 = (− (6)2 2 + 6(6)) − (− (4)2 2 + 6(4)) = 2 satuan luas.
4.
Diperoleh : ∆𝐴 ≈
𝐴(𝑅)1 + 𝐴(𝑅)2 Jadi, luasnya adalah : 5. 3̇ + 2 ≈ 𝟕. 𝟑̇ satuan luas. 4. Luasan 𝑅 dibatasi oleh kurva-kurva𝑦 = 𝑥 + 6, 𝑦 = 𝑥3 dan 2𝑦 + 𝑥 = 0. Kemudian hitunglah luasnya. Jawab: Mencari batas atas dan bawahnya (Titik-titikpotong yang diperoleh) - Titik potong antara garis 𝑦 = 𝑥 + 6 dan 2𝑦 + 𝑥 = 0 atau 𝑦 = − 𝑥 2 . 𝑥 + 6 = − 𝑥 2 2𝑥 + 12 = −𝑥 3𝑥 = −12 𝑥 = −4 yaitu di titik (-4, 2). - Titik potong antara kurva 𝑦 = 𝑥3 dan garis 2𝑦 + 𝑥 = 0 atau 𝑦 = − 𝑥 2 . 𝑥3 = − 𝑥 2
5.
2𝑥3 = −𝑥 2𝑥3
+ 𝑥 = 0 𝑥(2𝑥2 + 1) = 0 𝑥 = 0 atau 2𝑥2 + 1 = 0 → 2𝑥2 = −1 → 𝑥2 = − 1 2 (tidak terdefenisi) di titik (0,0) - Titik potong antara garis 𝑦 = 𝑥 + 6 dan kurva 𝑦 = 𝑥3. Yaitu berpotongan di titik (2, 8) dengan 𝑥 = 2. Didapatkan (−4,2); (0,0);(2, 8). Daerah pada grafik di atas terdapat dua daerah, yaitu dari 𝑥 = −4 sampai 𝑥 = 0 dan dari 𝑥 = 0 sampai 𝑥 = 2. Sehingga diperoleh : 𝐴(𝑅)1 = ∫ (𝑥 + 6) − (− 𝑥 2 ) 0 −4 𝑑𝑥 = ∫ ( 3 2 𝑥 + 6)𝑑𝑥 0 −4 = [ 3 4 𝑥2 + 6𝑥] −4 0 = ( 3 4 (0)2 + 6(0)) − ( 3 4 (−4)2 + 6(−4)) = 0 − (−12) = 12 𝐴(𝑅)2 = ∫ (𝑥 + 6) − 𝑥3 0 2 𝑑𝑥 = [ 𝑥2 2 + 6𝑥 − 𝑥4 4 ] 2 0 =( 02 2 + 6(0) − 04 4 ) − ( 22 2 + 6(2) − 24 4 ) = −10 (abaikan tanda negatif) Diperoleh : ∆𝐴 ≈ 𝐴(𝑅)1 + 𝐴(𝑅)2 Jadi, luasnya adalah : 12 + 10 ≈ 𝟑𝟐 satuan luas.
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