Latihan 7 18101105031.docx

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REGRESI X1

X2

X3

12

3

10

11

1

15

9

5

13

10

2

18

8

4

16

𝑏1 =

(βˆ‘ π‘₯2 2 )(βˆ‘ π‘₯1 𝑦)βˆ’(βˆ‘ π‘₯1 π‘₯2 ) βˆ‘ π‘₯2 𝑦 (βˆ‘ π‘₯1 2 )(βˆ‘ π‘₯2 2 )βˆ’(βˆ‘ π‘₯1 π‘₯2 )2

𝑏2 =

(βˆ‘ π‘₯1 2 )(βˆ‘ π‘₯2 𝑦)βˆ’(βˆ‘ π‘₯1 π‘₯2 ) βˆ‘ π‘₯1 𝑦 (βˆ‘ π‘₯1 2 )(βˆ‘ π‘₯2 2 )βˆ’(βˆ‘ π‘₯1 π‘₯2 )2

𝑏0 =

βˆ‘π‘Œ 𝑛

βˆ‘ 𝑋1

βˆ’ 𝑏1 (

𝑛

βˆ‘ 𝑋2

) βˆ’ 𝑏2 (

𝑛

)

Contoh Misalkan 𝑓 ∢ ℝ β†’ ℝ adalah fungsi yang didefinisikan sebagai : 1 βˆ’ π‘₯, π‘₯ ≀ 1 𝑓(π‘₯) = { 2π‘₯, π‘₯ > 1 1 𝑓(π‘₯) = { 2π‘₯

, ,

π‘₯≀1 π‘₯>1

Maka, lim 𝑓(π‘₯) = limβˆ’ 1 βˆ’ π‘₯ = 0 π‘‘π‘Žπ‘› lim+ 𝑓(π‘₯) = lim+ 2π‘₯ = 2

π‘₯β†’1βˆ’

π‘₯β†’1

π‘₯β†’1

π‘₯β†’1

INTEGRAL Misalkan 𝑓 π‘‘π‘Žπ‘› 𝑔 kontinu pada [π‘Ž, 𝑏]. Dengan 𝑓(π‘₯) = 𝑒 π‘₯ + sin π‘₯ π‘‘π‘Žπ‘› 𝑔(π‘₯) = 𝑒 π‘₯ + cos π‘₯ π‘‘π‘’π‘›π‘”π‘Žπ‘› π‘Ž = 2

𝑏

𝑏

πœ‹ 2

π‘‘π‘Žπ‘› 𝑏 = 2πœ‹

𝑏

[βˆ«π‘Ž 𝑓(π‘₯) 𝑔(π‘₯)𝑑π‘₯] = βˆ«π‘Ž 𝑓(π‘₯)2 𝑑π‘₯. βˆ«π‘Ž 𝑔(π‘₯)2 𝑑π‘₯ Ganti 𝑓(π‘₯) π‘‘π‘Žπ‘› 𝑔(π‘₯) dan batas integral sesuai dengan yang diberikan!!!!! 2πœ‹

2

2πœ‹

2πœ‹

2

2

[βˆ«πœ‹ 𝑒 π‘₯ + sin π‘₯ 𝑒 π‘₯ + cos π‘₯ 𝑑π‘₯] = βˆ«πœ‹ (𝑒 π‘₯ + sin π‘₯)2 𝑑π‘₯. βˆ«πœ‹ (𝑒 π‘₯ + cos π‘₯)2 𝑑π‘₯ 2

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