若函数f(x)具有连续的导数，则d⼀dx∫上限是x下限是0 (x-t)f ✀(t)dt=? 求详细的解答.

解法一：d/dx[∫<0,x>(x-t)f '(t)dt]=(dx/dx)(x-x)f '(x)+∫<0,x>[d(x-t)/dx]f '(t)dt
=∫<0,x>f '(t)dt
=f(t)│<0,x>
=f(x)-f(0)；
解法二：d/dx[∫<0,x>(x-t)f '(t)dt]=d/dx[x∫<0,x>f '(t)dt-∫<0,x>tf '(t)dt]
=d/dx[xf(t)│<0,x>-∫<0,x>td(f(t))]
=d/dx[x(f(x)-f(0))-(tf(t))│<0,x>+∫<0,x>f(t)dt] (应用分部积分法)
=d/dx[x(f(x)-f(0))-xf(x)+∫<0,x>f(t)dt]
=d/dx[-xf(0)+∫<0,x>f(t)dt]
=d(-xf(0))/dx+d(∫<0,x>f(t)dt)/dx
=-f(0)+f(x)
=f(x)-f(0)。

d/dx∫[0,x] (x-t)f '(t)dt
=d/dx{x∫[0,x] f '(t)dt-∫[0,x] tf '(t)dt}
=∫[0,x] f '(t)dt+xd/dx∫[0,x] f '(t)dt-d/dx∫[0,x] tf '(t)dt
=f(x)+xf'(x)-xf'(x)
=f(x)

F(x)=∫[0,x](x-t)f'(t)dt=∫[0,x](x-t)df(t)
=[f(x)*(x-x)-f(0)*x]-∫[0,x]f(t)d(x-t)
=-f(0)*x+∫[0,x]f(t)dt
设g(x)=∫f(x)dx ∫[0,x]f(x)dx=g(x)-g(0)
d∫[0,x]f(t)dt /dx=g'(x)=f(x)
F'(x)=-f(0)+f(x)