Page 46 - TUTORIAL_MAT_IPA_K11-2

INTEGRAL FUNGSI ALJABAR

1
2
25. = ∫ + 1 = + +
2
1 1
∴ (1, 0) → 0 = + 1 + → = − − 1
2 2
1 3 4
∴ (2, −1) → −1 = 2 + 2 + → −1 = 2 + 2 − − 1 → −2 = → = −
2 2 3
4 1 4 1
∴ = − → = − . − − 1 = −
3 2 3 3
2 2 1
→ = − + −
3 3
1 1
titik potong dengan sumbu Y → = 0 → = − → (0, − )
3 3

1
1
3
4
26. ( ) = ∫ + −3 = − −2 +
4 2
1 4 1 −2 11 1 11 6 3
(1) = 1 − 1 + = − → − + = − → = − = −
4 2 20 4 20 20 10
1 1 3
4
( ) = − −2 −
4 2 10

1
2
27. ∫ √ + 1 =
0
′
Misal : = + 1 → = 1 → = 1 → = ; = − 1

= 0 → = 0 + 1 = 1 ; = 1 → = 2
1 2
2
2
∫ √ + 1 = ∫ ( − 1) √
0 1

1/2 √
28. ∫ =
0 √1−
1 2
Misal : √ = sin → = cos → = 2√ cos ; =
2√
∴ = 0 → √0 = sin → = 0
1 1
∴ = → √ = sin → =
2 2 4

1/2 √ sin sin sin
∫ = ∫ 4 = ∫ 4 = ∫ 4 = ∫ tan
4
0 √1− 0 √1− 0 √ 0 cos 0
2
2

2 1 1 2 1 1 1 1 −2+1 2 1 −1 2
2
3
2
29. ∫ − = ∫ − −2 = . 2+1 − | = + |
1 2 2 1 2 2 1+2 1+(−2) 1 6 1
1 3 −1 1 3 −1 8 1 1 1 11 7 4 2
= ( . 2 + 2 ) − ( . 1 + 1 ) = ( + ) − ( + ) = − = =
6 6 6 2 6 1 6 6 6 3

1 ⁄ 3 1
30. ∫ 1 ⁄ 8 3 2 √1 + =

1 ′ 1 1 2
Misal : = 1 + → = − → = − → = −
2 2
1
1 ⁄ 3 1 1 ⁄ 3 1 ⁄ 1 3 ⁄ 3
2
∫ 1 ⁄ 8 3 2 √1 + = ∫ 1 ⁄ 8 3 2 √ . − = − ∫ 1 ⁄ 8 3 3 2 = −2 2| 1 ⁄

3 1 ⁄ 3 3 8
1 2 3 1 2 1 2 3 3
2
2
= −2 (1 + ) | = −2 (1 + ) − (−2 (1 + ) ) = −2(2 )2 + 2(3 )2 = 38
1 ⁄ 8 1 ⁄ 3 1 ⁄ 8

‘LEARNING IS FUN’ 45