设函数f（x）连续,则定积分∫tf（x∧2-t∧2）dt上限是x下限是0的导是多少?

Question

如题所述

Answer 1

简单分析一下，答案如图所示

Answer 2

首先换元,再利用变上限积分求导

令u=x^2 -t^2.则t=√(x^2-u). dt=d√(x^2-u)=1/2•1/√(x^2-u)du
当t=0时,u=x^2.当t=x时,u=0

∴∫(0→x)tf(x^2 -t^2)dt
=∫(x^2→0)√(x^2-u) f(u)d√(x^2-u)
=∫(x^2→0)√(x^2-u) f(u)•(1/2•1/√(x^2-u))du
=∫(x^2→0) f(u)•1/2du
=1/2∫(x^2→0) f(u)du
d/dx•∫(0→x)tf(x^2 -t^2)dt
=d/dx•（1/2∫(x^2→0) f(u)du）
=(1/2∫(x^2→0) f(u)du)′
=1/2•f(x^2)(x^2)′
=1/2•f(x^2)•2x
=xf(x^2)