f’(x)=arcsin(x-1)^2
先求出f(x)=∫arcsin(x-1)^2dx,由于f(0)=0,f(x)=∫(0,x)arcsin(y-1)^2dy
然后對(duì)f(x)進(jìn)行積分：
∫(0,1)f(x)dx
=∫(0,1)dx∫(0,x)arcsin(y-1)^2dy (交換積分順序）
=∫(0,1)dy∫(y,1)arcsin(y-1)^2dx
=∫(0,1)(1-y)arcsin(y-1)^2dy
=(-1/2)∫(0,1)arcsin(y-1)^2d(y-1)^2 (用公式∫arcsinxdx)
=(-1/2)[(y-1)^2arcsin(y-1)^2+√(1-(y-1)^4)] | (0,1)
=(-1/2)(1-π/2)
=(π-2)/4