设f(x)在【a,b】有连续的二阶导数,又f(a)等于f'(a)等于0,及定积分∫上线是b下线是a f(x)dx等于2,...
问题描述：
设f(x)在【a,b】有连续的二阶导数,又f(a)等于f'(a)等于0,及定积分∫上线是b下线是a f(x)dx等于2,...
设f(x)在【a,b】有连续的二阶导数,又f(a)等于f'(a)等于0,及定积分∫上线是b下线是a f(x)dx等于2,求∫上线b下线a f''(x)(x-b)∧2dx
答
∫上线b下线a f''(x)(x-b)∧2dx
=∫上线b下线a (x-b)∧2df'(x)=f'(x)(x-b)^2|(a,b)-∫上线b下线a 2f'(x)(x-b)dx=-∫上线b下线a 2(x-b)df(x)=-2f(x)(x-b)|(a,b)+2∫上线b下线af(x)dx=4.