Question

设 f'(x)在[0,1]上连续,试求∫[1+x f'(x)]e^ f'(x)dx ( 范围是0到1)

Answer 1

∫[1+x f'(x)]e^ f(x)dx

=∫e^ f(x)dx+∫xf'(x)e^ f(x)dx ‘先将括号打开,拆为两个积分

=∫e^ f(x)dx+∫xe^ f(x)df(x)

=∫e^ f(x)dx+∫xde^ f(x)

=∫e^ f(x)dx+xe^ f(x)-∫e^ f(x)dx '分部积分法则

=xe^ f(x)

最后结果1*e^ f(1)-0*e^ f(0)=e^ f(1)