Page 110 - BacII 2011-2017 by Lim Seyha

វិទយល័យសេម�ចឳ េខត�េស ម�ប 108

(x + 2)(x – 2)
◦ ប (c) កាត់អ័ក អាប់សុីស ល y = 0 ⇔ 0 = ⇔ x = –2; x = 2
(1 – x)
y

(d) :
6
(c)
4
y = –x – 1

2

x
• •
–8 –6 –4 –2 2 4 6 8
–2 •

–4 •

(d) : y = –x – 1 –6 x
x 0 –1 =
y –1 0 –8 1

IV. គណនាអាំង ល អនុគមន៍៖
∫ [ 3 2 ] 3 3 2 ( 3 2 )
3 ( ) x x 3 3 1 1
2
ក. I = 2x – 3x + 1 dx = 2 – 3 + x = 2 –3 +3– 2 – 3 + 1 =
1 3 2 1 3 2 3 2
27 2 3
18 – + 3 – + – 1
2 3 2
2 22 22
= 20 – 12 – = ដូច ះ I =
3 3 3
2x + 1 –1 3
ខ. f(x) = ; បងា ញថា f(x) = +
2
x – 5x + 4 x – 1 x – 4
–1 3 –(x – 4) + 3(x – 1) –x + 4 + 3x – 3 2x + 1
យ + = = = = f(x)
x – 1 x – 4 (x – 1)(x – 4) x – 5x + 4 x – 5x + 4
2
2
–1 3
ដូច ះ f(x) = +
x – 1 x – 4
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