Question

∫(-1,1)(x^2-x)√(1-x^2)dx

-1是下限，1是上限
求详解

Answer 1

∫[-1，1] (x² - x)√(1 - x²) dx
= ∫[-1，1] x²√(1 - x²) dx - ∫[-1，1] x√(1 - x²) dx
= 2∫[0，1] x²√(1 - x²) dx - 0，第一个是偶函数，第二个是奇函数
令x = sinθ，dx = cosθ dθ
= 2∫[0，π/2] sin²θcos²θ dθ
= 2(1/4)∫[0，π/2] sin²2θ dθ
= (1/4)∫[0，π/2] (1 - cos4θ) dθ
= (1/4)(θ - 1/4 * sin4θ)_[0，π/2]
= π/8