4/2 就不化简了 ? 到底是求导, 还是积分 ?
若是积分 : f'(x) = √(4-x^2), 则
f(x) = ∫√(4-x^2)dx = x√(4-x^2) - ∫[-x^2/√(4-x^2)]dx
= x√(4-x^2) - ∫[(4-x^2-4)/√(4-x^2)]dx
= x√(4-x^2) - ∫√(4-x^2)dx + ∫[4/√(4-x^2)]dx
= x√(4-x^2) - f(x) + ∫[4/√(1-x^2/4)]d(x/2)
2f(x) = x√(4-x^2) + 4arcsin(x/2) + 2C
f(x) = (x/2)√(4-x^2) + 2arcsin(x/2) + C
若是求导 :应为 f(x) = √(4-x^2) , 则 f'(x) = -x/√(4-x^2)
追问是积分,已知求导后的函数,把这个函数还原成求导之前的