∫(a,b)√ [(b-x)(x-a)]dx=(b-a)^2∫(0,1)√[(1-x)x]dx令x=(sint)^2 dx=2sintcostdt √[(1-x)x]=sintcost原式=(b-a)^2∫(0,1)√[(1-x)x]dx=2(b-a)^2∫(0,π/2)[(sint)^2-(sint)^4]dt =2(b-a)^2[(1/2)(π/2)-(3/4)(1/2)(π/2)]=(π/8)(b-a)^2