Question

F(x)=∫te^(-t)dx上限为x^2下限为0求F‘(x)=?

Answer 1

令x²=u

F(x)=∫te^(-t)dt上限为u下限为0

F'(x)=[ue^(-u)]u'

=[x²e^(-x²)](2x)

=2x³e^(-x²)