函数f(x)(?)(x)和g(x)(?)(x)分别在[a,b]上连续,在(a,b)内(?)(x)≠0则必存在一点ξ∈(a,b)使得g(ξ)integral from n=1 to ab f(x)(?)(x)dx=f(ξ)integral from n=1 to b(a)g(x)(?)(x)dx成立.这个结论对于多个函数对f_i(x)(?)(x),i=1,2,...函数f(x)(?)(x)和g(x)(?)(x)分别在[a,b]上连续,在(a,b)内(?)(x)≠0则必存在一点ξ∈(a,b)使得g(ξ)integral from n=1 to ab f(x)(?)(x)dx=f(ξ)integral from n=1 to b(a)g(x)(?)(x)dx成立.这个结论对于多个函数对f_i(x)(?)(x),i=1,2,…,2n也成立.展开更多
文摘函数f(x)(?)(x)和g(x)(?)(x)分别在[a,b]上连续,在(a,b)内(?)(x)≠0则必存在一点ξ∈(a,b)使得g(ξ)integral from n=1 to ab f(x)(?)(x)dx=f(ξ)integral from n=1 to b(a)g(x)(?)(x)dx成立.这个结论对于多个函数对f_i(x)(?)(x),i=1,2,…,2n也成立.