Page 52 - TUTORIAL_MAT_IPA_K11-2

INTEGRAL FUNGSI ALJABAR

4 4
2
48. ∫ + = + | = 8 + 4 = 0 → = −2
0 2 0
2 4
= − ∫ + + ∫ + = 8
0 2
2 4
2
2
− − | + { + | } = 8 → −2 − 2 + {(8 + 4 ) − (2 + 2 )} = 8 →
2 0 2 2
4 = 8 → = 2 ; = −2.2 = −4 → ( ) = 2 − 4

2
2
2
49. = 4 − ( ) = 4 − ; = 4 − ( − 4) = 4 − ( − 4) = −12 + 8 − ; = 4

2 4
2
2
= ∫ 4 − (4 − ) + ∫ 4 − (−12 + 8 − ) =

2
0
2 4 1 2 1 4
3
2
2
3
2
∫ + ∫ 16 − 8 + = | + 16 − 4 + | =
0 2 3 0 3 2
2 3 2 1 3 2 1 3 16
+ {(16.4 − 4. 4 + . 4 ) − (16.2 − 4. 2 + . 2 )} = sat luas
3 3 3 3

2
2
50. Menentukan absis titik potong antara kurva dan garis → = (2 − 1) → −
(2 − 1) = 0 → ( − (2 − 1)) = 0 → = 0 , = 2 − 1. Di sini (2m – 1) punya dua
kemungkinan nilai, positif atau negatif.
a. Jika (2 − 1) bernilai positif.
2 −1 2 −1 1 2 − 1 9
3
2
2
= ∫ (2 − 1) − = − | =
0 2 3 0 2
2 −1 1 1 9
3
3
2
→ . (2 − 1) − (2 − 1) = (2 − 1) =
2 3 6 2
3
→ (2 − 1) = 27
3
3
→ (2 − 1) = 3
→ 2 − 1 = 3 → = 2
b. Jika (2 − 1) bernilai negatif maka 2 − 1 = −3 → = −1

51.

‘LEARNING IS FUN’ 51