A-maths(Integration)問題1

(a) For P, sin(x/2) = sin x
sin x - sin (x/2) = 0
2 cos (3x/4) sin (x/2) =0
cos (3x/4) = 0 or sin (x/2) =0
3x/4 = (n- 0.5)π or x/2 = nπ
x= (4n/3 +2)π or x=2nπ
For x in (-π, 0), the only possible solution is
x= (4n/3 +2)π with n=-2
x= -2π/3
y=sin (-2π/3) = -√3 /2
Hence P( -2π/3, -√3 /2)

(b) Area required
= | ∫[-π ~ -2π/3] sin x dx + ∫[-2π/3 ~ 0] sin (x/2) dx |
= | [-cos x]| [-π ~ -2π/3] + [-2cos (x/2)]| [-2π/3 ~ 0] |
= | (0.5 - 1) + (-2 - (-1)) |
= | -0.5 - 1| = 3/2

2011-05-19 12:53:54 補充：
∫[-π ~ -2π/3] sin x dx.是指∫sin x dx，由-π 積到 -2π/3，因為打不出上下標所以以此代替