Question

关于一元函数定积分的证明题已知f(x)在闭区间[a,b]连续,求证 在[a,b]存在一点c,使得f(x)从a到c的定积分

Answer 1

g(x)=∫(a~x)f(t)dt-∫(x~b)f(t)dt,显然g(x)在[ a,b ]连续

g(a)=-∫(a~b)f(t)dt,g(b)=∫(a~b)f(t)dt,

(1)若∫(a~b)f(t)dt=0,则可取c=a或c=b

(2)若∫(a~b)f(t)dt≠0,则g(a)g(b)