amaths . integration

1. ∫[0,3]1/(x^2+3) dx
=∫[0,pi/6] 1/(3sec^2 θ) (√3sec^2 θ)dθ
=∫[0,pi/6] dθ
=pi/6

2. volume of solid
=pi ∫[1,2](x^2)^2dx
=pi ∫[1,2] x^4 dx
=pi (2^5-1^5)
=31pi

2010-12-18 23:16:52 補充：
Correction of Q1:
x=√3 tanθ
dx=√3sec^2 θ dθ
when x=3, tanθ=√3, θ=pi/3
∫[0,3]1/(x^2+3) dx
=∫[0,pi/3] [1/(3sec^2 θ)] (√3sec^2 θ dθ)
=∫[0,pi/3] dθ
=pi/3

2010-12-18 23:32:10 補充：
Q1
x=√3 tanθ
dx=√3sec^2 θ dθ
when x=3, tanθ=√3, θ=pi/3
∫[0,3]1/(x^2+3) dx
=∫[0,pi/3] [1/(3sec^2 θ)] ( √3 sec^2 θ dθ)
=1/√3 ∫[0,pi/3] dθ
=√3pi/9

Q2
volume of solid
=pi ∫[1,2](x^2)^2dx
=pi ∫[1,2] x^4 dx
=pi (2^5-1^5)/5
=31pi/5

thx 自由自在~
I am so careless...

2010-12-19 16:40:27 補充：
是, pi = π

Q2
條公式係一開始就有..
你可以去呢度睇下
http://en.wikipedia.org/wiki/Solid_of_revolution

Q1
∫[0,π/3] dθ 的意思是IN果舊野係1, 1省略唔寫
∫[0,π/3] dθ= ∫[0,π/3] 1 dθ
IN 完好自然係π/3
再乘番1/√3
答案=√3π/9

(1) (√3 π)/9
(2) 31π/5