Page 45 - TUTORIAL_MAT_IPS_K11-2

INTEGRAL FUNGSI ALJABAR

1 1 2 +2 2 2
20. ( ) = + → ( ) = + = → + 2 = + 2 → = → = 1
2 2 2
1 3
( ) = 6 → . 1. + = 6 → = 6 → = 4
2 2
1
( ) = + 4
2

6−
21. ∴ ( ) = 2 + → ( ) = 2 + = 6 → =
2
2
6−
2
( ) = 2 + = 2 ( ) +
2
+2
∴ , ( ), 2 → barisan aritmatika, maka ( ) = → 2 ( ) = + 2
2
2 2 2
6− 6− 6− 6− 6− 6−
→ 2. {2 ( ) + } = + 2 → 4 ( ) + 2 = + 2 → 4 ( ) =
2 2 2 2 2 2
6− 2 6− 6− 6−
→ 4 ( ) − = 0 → (4 − 1) = 0
2 2 2 2
6− 6−6 0
→ = 0 → = 6 → = = = 0 {tidak memenuhi}
2 2.6 12
6− 6− 6−4 1
→ 4 − 1 = 0 → 4 = 1 → 24 − 4 = 2 → 24 = 6 → = 4 → = =
2 2 2.4 4
1
. = . 4 = 1
4

1
8
1
3
2
3
22. ( ) = ∫ = + → (2) = . 2 + = − 19 → + = − 19 → = − 9
3 3 3 3 3
1 3
→ ( ) = − 9
3
1
3
Titik potong dengan sumbu X → = 0 → 0 = − 9
3
1
3
→ 9 = → = 3
3
Koordinat titik potong dengan sumbu X adalah (3, 0).

2
23. = ∫ 2 − 5 = − 5 + .
2
∴ (4, 7) → 7 = 4 − 5.4 + → 7 = −4 + → = 11
2
→ = − 5 + 11
2
Titik potong dengan sumbu Y → = 0 → = 0 − 5.0 + 11 = 11 → (0, 11)

3
2
2
24. = ∫ 3 + 4 + 6 = + 2 + 6 +
2
3
∴ (1, 14) → 14 = 1 + 2. 1 + 4.1 + 6 + = 13 + → = 1
3
2
→ = + 2 + 6 + 1
Titik potong dengan sumbu Y → = 0 → = 1 → (0, 1)

‘LEARNING IS FUN’ 44