Question

求定积分：f(-a,a)(x^2-x)*根号下（a^2-x^2）dx

Answer 1

[-a,a] ∫ (x²-x)√(a²-x²) dx
=[-a,a] ∫ x²√(a²-x²) dx - [-a,a] ∫ x√(a²-x²) dx （后半部分是奇函数，在对称区间的定积分为零）
=[0,a] 2 ∫ x²√(a²-x²) dx
=[0,a] 2 ∫ (a²-a²+x²)√(a²-x²) dx
=[0,a] 2 ∫ a²√(a²-x²) - (a²-x²)^(3/2) dx
= a²x√(a²-x²)+ a^4 arctan[x/√(a²-x²)] - ¼ x(5a²-2x²)√(a²-x²) - ¾ a^4 arctan[x/√(a²-x²)] | [0,a]
=¼ (2x²-a²)x√(a²-x²)+¼ a^4 arctan[x/√(a²-x²)] | [0,a]
=(x→a)lim¼ a^4 arctan[x/√(a²-x²)]
=(u→π/2)lim¼ u a^4 （令 x=asinu，则arctan[x/√(a²-x²)]=u）
=(πa^4)/8

Answer 2

∫(-a→a) (x^2-x)√(a^2-x^2) dx
= ∫(-a→a) x^2√(a^2-x^2) dx - ∫(-a→a) x√(a^2-x^2) dx
= 2∫(0→a) x^2√(a^2-x^2) dx - 0
x=a*siny,dx=a*cosydy
= 2∫(0→π/2) a^2sin^2y*a^2cos^2y dy
= 2a^4∫(0→π/2) 1/4*(sin2y)^2 dy
= (a^4/2)∫(0→π/2) (1-cos4y) dy
= (a^4/2)(y-1/4*sin4y)
= (a^4/2)(π/2-0)
= a^4π/4