Transcript manasa 14

0, −1 < 𝑥 < 0
1. 𝑓(𝑥) = {
𝑥, 0 < 𝑥 < 1
Penyelesaian :
1
∫ 𝑓(𝑥)𝑃𝑚 (𝑥)𝑑𝑥
= 𝐶𝑙
−1
0
2
2𝑚 + 1
1
∫ 0 . 𝑃0 (𝑥)𝑑𝑥 + ∫ 𝑥 . 𝑃0 (𝑥)𝑑𝑥 = 𝐶0
−1
0
2
2(0) + 1
1
0 + ∫ 𝑥 . 1 𝑑𝑥
= 𝐶0 . 2
0
1
∫ 𝑥 𝑑𝑥
= 𝐶0 . 2
0
1 2 1
𝑥 |
2
0
= 𝐶0 . 2
1
1
(1)2 − (0)2 = 𝐶0 .2
2
2
1
= 𝐶0 .2
2
𝐶0 =
1
4
1
∫ 𝑓(𝑥)𝑃𝑚 (𝑥)𝑑𝑥
= 𝐶𝑙
−1
0
2
2𝑚 + 1
1
∫ 0 . 𝑃1 (𝑥)𝑑𝑥 + ∫ 𝑥 . 𝑃1 (𝑥)𝑑𝑥 = 𝐶1
−1
0
2
2(1) + 1
1
0 + ∫ 𝑥 . 𝑥 𝑑𝑥
0
= 𝐶1 .
1
∫ 𝑥 2 𝑑𝑥
0
1 3 1
𝑥 |
3
0
= 𝐶1 .
= 𝐶1 .
2
3
2
3
2
3
1
1
2
(1)3 − (0)3 = 𝐶1 .
3
3
3
1
2
= 𝐶1 .
3
3
𝐶1 =
1
2
1
∫ 𝑓(𝑥)𝑃𝑚 (𝑥)𝑑𝑥
= 𝐶𝑙
−1
0
2
2𝑚 + 1
1
∫ 0 . 𝑃2 (𝑥)𝑑𝑥 + ∫ 𝑥 . 𝑃2 (𝑥)𝑑𝑥 = 𝐶2
−1
0
1
1
0 + ∫ 𝑥 . ( (3𝑥 2 − 1)) 𝑑𝑥
2
0
1
3
1
∫ ( 𝑥 3 − 𝑥) 𝑑𝑥
2
0 2
3 4 1 2 1
𝑥 − 𝑥 |
8
4
0
2
2(2) + 1
= 𝐶2 .
2
5
= 𝐶2 .
2
5
= 𝐶2 .
2
5
3
1
2
(1)4 − (0)2 = 𝐶2 .
8
4
5
3 1
2
−
= 𝐶2 .
8 4
5
𝐶2 =
𝑓(𝑥) =
5
16
1
1
5
𝑃0 (𝑥) + 𝑃1 (𝑥) +
𝑃 (𝑥) + … … …
4
2
16 2
𝑛
𝜋
∑ ∫ 𝑏𝑖 sin 𝑖𝑥 . cos 𝑥 𝑑𝑥 = 0
𝑖=1 −𝜋
𝑛
𝜋
1
∑ ∫ 𝑏𝑖 ( (sin(𝑖𝑥 + 𝑥) + sin(𝑖𝑥 − 𝑥)) ) 𝑑𝑥
2
−𝜋
𝑖=1
𝑛
𝜋
𝑏𝑖
(sin(𝑖 + 1)𝑥 + sin(𝑖 − 1)𝑥) 𝑑𝑥
−𝜋 2
∑∫
𝑖=1
𝑛
∑
𝑖=1
𝑛
∑
𝑖=1
𝑛
∑
𝑖=1
𝑛
∑
𝑖=1
𝑏𝑖 𝜋
∫ (sin(𝑖 + 1)𝑥 + sin(𝑖 − 1)𝑥) 𝑑𝑥
2 −𝜋
𝑏𝑖
1
1
𝜋
cos (𝑖 + 1) 𝑥 −
cos(𝑖 − 1) 𝑥)]
[(−
2
𝑖+1
𝑖−1
−𝜋
𝑏𝑖
1
1
cos (𝑖 + 1) 𝜋 +
cos(𝑖 − 1) 𝜋)
[(−
2
𝑖+1
𝑖−1
1
1
− (
cos (𝑖 + 1)(−𝜋) −
cos(𝑖 − 1) − 𝜋)]
𝑖+1
𝑖−1
𝑏𝑖
[0 − 0]
2
=0
𝜋
1
1
𝜋
𝑎0 cos 𝑥 𝑑𝑥 = [ 𝑎0 sin 𝑥]
2
−𝜋
−𝜋 2
1
1
= [ 𝑎0 sin 𝜋 − 𝑎0 sin(−𝜇) ]
2
2
=0+
=0+0
=0
∫