Page 29 - BacII 2011-2017 by Lim Seyha

វិទយល័យសេម�ចឳ េខត�េស ម�ប 27

2 x 2 x
∫ ( )
IV. ក. គណនាអាំង ល I = – + 3 dx
1 3 2
2 x 2 x x 3 x 2 8 1 1
∫ ( ) [ ] 2 ( )
I = – + 3 dx = – + 3x = – 1 + 6 – – + 3
1 3 2 3 × 3 2 × 2 1 9 9 4
8 1 1 7 1 28 + 72 + 9 109
= + 5 – + – 3 = + 2 + = =
9 9 4 9 4 36 36
109
ដូច ះ I =
36
2 – x 1 1
ខ. អនុគមន៍ f(x) = – បងា ញថា f(x) = – +
(x – 1) 2 (x – 1) 2 x – 1
(
1 1 –1 + x – 1 ) –1 + x – 1 2 – x
យ – + = = = = f(x) ពិត
(x – 1) 2 x – 1 ( x – 1 ) 2 ( x – 1 ) 2 ( x – 1 ) 2
1 1
ដូច ះ f(x) = – +
(x – 1) 2 x – 1
0 1 1 1 1
∫ ∫ ( ) [ ] 0
គណនា K = f(x)dx = – + dx = + ln |x – 1|
–1 –1 (x – 1) 2 x – 1 x – 1 –1
( )
1 1
= –1 + ln 1 – – + ln 2 = – – ln 2
2 2
1
ដូច ះ K = – – ln 2
2
− → − → −→ − → − → − → − → − → − → − → − → − →
V. ១. មានវុ ចទ័រ u = i – j + 2k, v = – i + 2 j + 2k, w = i + j – 2k បាន
( − → − → − → ) − → − → − → − →
− → −→
− →
− →
− →
u + v = j + 4k
ក. u + v = i – j + 2k + – i + 2 j + 2k = j + 4k − → − →

−→ −→ − →
i j k

−
− → → − → − → − →
ខ. w × u = 1 1 –2 = (2 – 2) i – (2 + 2) j + (–1 – 1)k

1 –1 2

− → − → − → − → → − → − → − →
−
= 0 i – 4 j – 2k w × u = 0 i – 4 j – 2k

−→ − → −→
i j k

− → − → − →
−
− → →
គ. w × v = 1 1 –2 = (2 + 4) i – (2 – 2) j + (2 + 1)k

–1 2 2

− → − → − → − → → − → − → − →
−
= 6 i – 0 j + 3k w × v = 6 i – 0 j + 3k
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