Question

求个定积分.∫(√(1-x^2)+x)dx 上限1 下限-1

Answer 1

解:
∫(- 1 -> 1) [√(1 - x²) + x] dx
= ∫(- 1 -> 1) √(1 - x²) dx + ∫(- 1 -> 1) x dx
= 偶函数 + 奇函数
= 2∫(0 -> 1) √(1 - x²) dx + 0,用几何方法解∫(0 -> 1) √(1 - x²) dx
= 2 * 1/4 * π * 1^2
= π/2

用第二换元法解∫(0 -> 1) √(1 - x²) dx:
令x = siny,dx = cosy dy
2∫(0 -> 1) √(1 - x²) dx
= 2∫(0 -> π/2) √(1 - sin²y) * cosy dy
= 2∫(0 -> π/2) cos²y dy
= 2∫(0 -> π/2) [1 + cos(2y)]/2 dy
= ∫(0 -> π/2) [1 + cos(2y)] dy
= [y + (1/2)sin(2y)] |(0 -> π/2)
= π/2