(7)设函数f(x)在[0,1]上连续,则 (int )_(0)^1[ (int )_(x)^1[ f(t)+f(x)] dt] dx= ()-|||-(A) [f(x)dx. (B)|xf(x)dx. (C) (int )_(0)^1(1-x)f(x)dx. (D) (int )_(0)^1[ 1-xf(x)] dx.