Integral, Turunan Dan Limit.docx

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Integral 1. Hasil ∫ 6π‘₯ √3π‘₯ 2 + 2 dx Jawab : Misal : u=3x2 𝑑𝑒 𝑑π‘₯

= 6π‘₯

𝑑𝑒= 6x 𝑑π‘₯ 𝑑𝑒

𝑑π‘₯ = 6π‘₯

1

= ∫ 6π‘₯ (3π‘₯ 2 + 5)2 Γ—

∫ 6π‘₯ √3π‘₯ 2 + 2 dx

𝑑𝑒 6π‘₯

1 2

= ∫ π‘ˆ Γ— 𝑑𝑒 = =

1

1

Γ— π‘ˆ 2+1 + 𝐢

1 2

1+ 1 3 2

3

Γ— π‘ˆ2 + 𝐢 3

2

= 3 (3π‘₯ 2 + 5)2 + C 6π‘₯ 2

2. Hasil dari ∫ √π‘₯ 3

βˆ’4

𝑑π‘₯

Jawab : Misal : u = x3 - 4 𝑑𝑒 𝑑π‘₯

= 3π‘₯ 2

𝑑𝑒= 3π‘₯ 2 𝑑π‘₯ 𝑑𝑒

𝑑π‘₯ = 3π‘₯ 2 6π‘₯ 2

∫ √π‘₯ 3 βˆ’4 𝑑π‘₯

1

= ∫ 6π‘₯ 2 (π‘₯ 3 βˆ’ 4)2 Γ— 𝑑π‘₯ 1

= ∫ 6π‘₯ 2 . π‘ˆ 2 Γ—

𝑑𝑒 3π‘₯ 2

1 2

= 2 ∫ π‘ˆ . 𝑑𝑒 =2 =2

1 1 βˆ’ +1 2

1 1 2

1

Γ— π‘ˆ βˆ’2+1 + 𝐢 1

Γ— π‘ˆ2 + 𝐢 1

= 4 π‘ˆ2+ C 1

= 4 (π‘₯ 3 βˆ’ 4)2 + C = 4 √π‘₯ 3 βˆ’ 4 + 𝐢

2

3. Hasil dari ∫1 (4π‘₯ 2 βˆ’ π‘₯ + 5)𝑑π‘₯ Jawab : 2

2 4

1

∫1 (4π‘₯ 2 βˆ’ π‘₯ + 5)𝑑π‘₯ = ∫1 (3 π‘₯ 3 βˆ’ 2 π‘₯ 2 + 5π‘₯) 𝑑π‘₯ 4

1

= 3 π‘₯ 3 βˆ’ 2 π‘₯ 2 + 5π‘₯

2 1

4

= (3 (2)3 Γ— 56

1

4

1

π‘₯ 2 + 5 (2)) βˆ’ (3 (1)1 βˆ’ 2 (1)1 + 5(1)) 2

35

=(3)βˆ’(6) =

77 6

= 12

5 6

4. Hasil dari ∫ π‘π‘œπ‘  4 2π‘₯ sin 2π‘₯ 𝑑π‘₯ Jawab : ∫ π‘π‘œπ‘  4 2π‘₯ sin 2π‘₯ 𝑑π‘₯

1

= ∫ βˆ’ 2(4+1) π‘π‘œπ‘  (4+1) . 2π‘₯ + 𝐢 1

= βˆ’ 10 π‘π‘œπ‘  5 . 2π‘₯ + 𝐢

Turunan 2π‘₯βˆ’5

1. Diketahui 𝑓(π‘₯) = 3π‘₯βˆ’4 dari 𝑓 β€² (π‘₯) adalah turunan pertama dan f(x). Nilai 𝑓 β€² (1). . ? Jawab : 2π‘₯ βˆ’ 5 3π‘₯ βˆ’ 4 𝑒′ 𝑣 βˆ’ 𝑒𝑣 β€² 𝑓 β€² (π‘₯) = 𝑣2 2 (3π‘₯βˆ’4)βˆ’(2π‘₯βˆ’5)3 = (3π‘₯βˆ’4)2 𝑓(π‘₯) =

=

6π‘₯βˆ’8βˆ’6π‘₯+15 (3π‘₯βˆ’4)2 7

=

(3π‘₯βˆ’4)2 7

𝑓 β€² (1) = (3(1)βˆ’4)2 =

7 (βˆ’1)2

=7 2. Jika 𝑓(π‘₯) =

sin π‘₯βˆ’cos π‘₯ sin π‘₯

Jawab : sin π‘₯ βˆ’ cos π‘₯ sin π‘₯ β€² 𝑒 𝑣 βˆ’ 𝑒𝑣 β€² 𝑓 β€² (π‘₯) = 𝑣2 𝑓(π‘₯) =

1

maka nilai 𝑓 β€² (3 πœ‹) ?

= = =

((π‘π‘œπ‘  π‘₯+𝑠𝑖𝑛 π‘₯) sin π‘₯)+((𝑠𝑖𝑛π‘₯βˆ’cos π‘₯) cos π‘₯) (sin π‘₯) 2 (π‘π‘œπ‘  π‘₯.sin π‘₯+ 𝑠𝑖𝑛2 π‘₯)βˆ’(sin π‘₯.cos π‘₯+π‘π‘œπ‘  2 π‘₯) 𝑠𝑖𝑛2 π‘₯ cos π‘₯.sin π‘₯+𝑠𝑖𝑛2 π‘₯βˆ’sin π‘₯.cos π‘₯+π‘π‘œπ‘  2 π‘₯

= = 𝑓 β€² (1) =

= = =

𝑠𝑖𝑛2 π‘₯ 𝑠𝑖𝑛2 π‘₯+ π‘π‘œπ‘  2 π‘₯ 𝑠𝑖𝑛2 π‘₯ 1 𝑠𝑖𝑛2 π‘₯

1 1 𝑠𝑖𝑛2 (3 (180Β°)) 1 𝑠𝑖𝑛2 60 1 3 4

4 3

3. Pembangunan sebuah gedung akan diselesaikan dalam x hari. Biaya pembangunan gedung per hari (210 βˆ’

2500 π‘₯

βˆ’ 3π‘₯) juta rupiah. Biaya minimum

pembangunan gedung sebesar . . . . juta rupiah. Jawab : Biaya/hari = (210 βˆ’

2500 π‘₯

βˆ’ 3π‘₯) 𝑗𝑑

𝐡(π‘₯) = 210π‘₯ βˆ’2500 - 3π‘₯ 2 𝐡 β€² (π‘₯) = 210 βˆ’ 6π‘₯ Jika : 𝐡 β€² (π‘₯) = 0 210 βˆ’ 6π‘₯ = 0 210 = 6π‘₯ π‘₯ =

210 6

π‘₯ = 35 𝐡(35) = 210π‘₯ βˆ’ 2500 βˆ’ 3π‘₯ 2 = 210 (35) βˆ’ 2500 βˆ’ 3(35)2 = 7350 βˆ’ 2500 βˆ’ 3675 = 1175 π‘—π‘’π‘‘π‘Ž π‘Ÿπ‘’π‘π‘–π‘Žβ„Ž 4. Hasil turunan ketiga dari 𝑓(π‘₯) = 3π‘₯ 4 + 5π‘₯ 3 βˆ’ 2π‘₯ + 1 Jawab : 𝑓(π‘₯) = 3π‘₯ 4 + 5π‘₯ 3 βˆ’ 2π‘₯ + 1 𝑓 β€² (π‘₯) = 12π‘₯ 3 + 15π‘₯ 2 βˆ’ 2 𝑓 β€²β€² (π‘₯) = 36π‘₯ 2 + 30π‘₯ 𝑓 β€²β€²β€² (π‘₯) = 72π‘₯ + 30

𝑓 β€²β€²β€² (1) = 72(1) + 30 = 102

Limit 1. Tentukan nilai dari

2. Nilai dari

3. Nilai dari Jawab:

4. Tentukan hasil dari soal limit berikut

adalah...

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