Anti-Derivative Question

我上次俾左詳解但問題被你 delete 左，所以你要答應唔 delete 條題目我才給予詳解。

1 b
2 b
3 d

2007-01-17 23:39:56 補充：
1x² cos(x³)dx = 1/3 cos(x³) d(x³) = d (1/3 sinx³)2f'(0)= (sub x=0) (1/(x+4+e^(-3x)) d(x+4+e^(-3x))/dx)= 1/(4+e^0) (1+0+0) = 1/53d/dx (anti-D from 0 to x^2) sin(t^3) dt)= dx^2/dx . d/dx^2 (anti-D from 0 to x^2) sin(t^3) dt)= 2x sin((x^2)^3) 所以係我做錯左，答案應該係 e 才對

2007-01-17 23:45:13 補充：
Brian Lui
寫了題對了。我的 (1+0+0) 好應該係 (1+0-3) 才對，這次太求奇了。

Q.1
∫x² cos(x³)dx
= ∫(1/3) cos(x³) d(x³)
= (1/3) sin(x³) + C.
Answer: B

Q.2
f(x) = ln [x + 4 + e^(-3x)].
f'(x)
= [1 + 0 + (-3) • e^(-3x)] / [x + 4 + e^(-3x)]
= [1 - 3 e^(-3x)] / [x + 4 + e^(-3x)]
f'(0)
= (1 - 3 e^0) / (0 + 4 + e^0)
= (1 - 3 • 1) / (0 + 4 + 1)
= -2/5
Answer: A

Q.3
d/dx [(∫ from 0 to x²) sin(t³) dt]
= { du / dx } • d/du [(∫ from 0 to u) sin(t³) dt] , by putting u = x²
= { d (x²) / dx } • { sin(u³) }
= 2x sin (x^6)
Answer: E