设函数f(x)在给定的积分区间上连续,已知:∫(a,b)f(x)dx=(b-a)∫(0,1)f(a+(b-a)x)dx.

2024-12-04 16:18:07
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回答1:

∫(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