g(x)=f(tx^2)在0至sinx上对t进行积分,求g(x)的导数,并判断其连续性

Question

求导并判断导函数的连续性

Answer 1

当x=0时，F(0)=0
当x≠0时，令u=tx^2，则t=u/x^2，dt=du/x^2
F(x)=∫(0,sinx*x^2) f(u)du/x^2
=[∫(0,sinx*x^2) f(u)du]/x^2
F'(x)=[(cosx*x^2+sinx*2x)*f(sinx*x^2)*x^2-2x*∫(0,sinx*x^2) f(u)du]/x^4
=[(cosx*x+2sinx)*f(sinx*x^2)]/x-2*[∫(0,sinx*x^2) f(u)du]/x^3
=cosx*f(sinx*x^2)+2sinx*f(sinx*x^2)/x-[2∫(0,sinx*x^2) f(u)du]/x^3
lim(x->0)F'(x)=f(0)+2f(0)-lim(x->0)[2∫(0,sinx*x^2) f(u)du]/x^3
=3f(0)-2*lim(x->0)(cosx*x+2sinx)*f(sinx*x^2)/3x
=3f(0)-2f(0)*lim(x->0)(3cosx-xsinx)/3
=f(0)
F'(0)=lim(x->0)[F(x)-F(0)]/x
=lim(x->0)[∫(0,sinx*x^2) f(u)du]/x^3
=f(0)
因为lim(x->0)F'(x)=F'(0)，所以导函数在R上连续本回答被提问者和网友采纳