Prove that：∫[0->a] x³ f(x²) dx = (1/2)∫(0->a²) x f(x) dx
Let u = x² and x = √u,then dx = 1/(2√u) du
x = 0,u = 0；x = a,u = a²
RHS = ∫[0->a²] (√u)³ * f(u) * 1/(2√u) * du
= (1/2)∫[0->a²] u^(3/2 - 1/2) * f(u) du
= (1/2)∫[0->a²] u f(u) du
= (1/2)∫[0->a²] x f(x) dx
= LHS很感谢你的答案谢谢，能再帮我解答以这道题吗？我主要是不知道[f(x)]'等于什么前面我已经解出来了，谢谢了y={f(sinx)+f(x^2)+[f(x)]}则y'=cosxf'(sinx)+2xf'(x^2)+?[f(x)]' = f'(x) * x' = f'(x)，有啥问题？