设f(x)与g(x)在[a,b]上连续,证明:(1)若在[a,b]上f(x)>=0,且∫ f(x) dx=0,则在[a,b]上f(x)恒等于0(2)若在[a,b]上f(x)>=g(x),且∫ f(x) dx=∫g(x) dx,则在[a,b]上f(x)恒等于g(x)注:∫ 右上标为b,下标为a
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![设f(x)与g(x)在[a,b]上连续,证明:(1)若在[a,b]上f(x)>=0,且∫ f(x) dx=0,则在[a,b]上f(x)恒等于0(2)若在[a,b]上f(x)>=g(x),且∫ f(x) dx=∫g(x) dx,则在[a,b]上f(x)恒等于g(x)注:∫ 右上标为b,下标为a](/uploads/image/z/5561216-8-6.jpg?t=%E8%AE%BEf%28x%29%E4%B8%8Eg%28x%29%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E8%AF%81%E6%98%8E%EF%BC%9A%EF%BC%881%EF%BC%89%E8%8B%A5%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8Af%28x%29%3E%3D0%2C%E4%B8%94%E2%88%AB+f%28x%29+dx%3D0%2C%E5%88%99%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8Af%28x%29%E6%81%92%E7%AD%89%E4%BA%8E0%EF%BC%882%EF%BC%89%E8%8B%A5%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8Af%28x%29%3E%3Dg%28x%29%2C%E4%B8%94%E2%88%AB+f%28x%29+dx%3D%E2%88%ABg%28x%29+dx%2C%E5%88%99%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8Af%28x%29%E6%81%92%E7%AD%89%E4%BA%8Eg%28x%29%E6%B3%A8%EF%BC%9A%E2%88%AB+%E5%8F%B3%E4%B8%8A%E6%A0%87%E4%B8%BAb%2C%E4%B8%8B%E6%A0%87%E4%B8%BAa)
设f(x)与g(x)在[a,b]上连续,证明:(1)若在[a,b]上f(x)>=0,且∫ f(x) dx=0,则在[a,b]上f(x)恒等于0(2)若在[a,b]上f(x)>=g(x),且∫ f(x) dx=∫g(x) dx,则在[a,b]上f(x)恒等于g(x)注:∫ 右上标为b,下标为a
设f(x)与g(x)在[a,b]上连续,证明:
(1)若在[a,b]上f(x)>=0,且∫ f(x) dx=0,则在[a,b]上f(x)恒等于0
(2)若在[a,b]上f(x)>=g(x),且∫ f(x) dx=∫g(x) dx,则在[a,b]上f(x)恒等于g(x)
注:∫ 右上标为b,下标为a
设f(x)与g(x)在[a,b]上连续,证明:(1)若在[a,b]上f(x)>=0,且∫ f(x) dx=0,则在[a,b]上f(x)恒等于0(2)若在[a,b]上f(x)>=g(x),且∫ f(x) dx=∫g(x) dx,则在[a,b]上f(x)恒等于g(x)注:∫ 右上标为b,下标为a
(1)用反证法
不妨设存在一点p,使f(p)>0,那么连续函数由保号性,存在p一个领域(p-c,p+c),
当x∈(p-c,p+c)时,f(x)>0
∫ f(x) dx =∫ f(x) dx + ∫ f(x) dx +∫ f(x) dx
>= ∫ f(x) dx >0
与∫ f(x) dx = 0 矛盾.
所以f(x)=0
(2)f(x)>=g(x),则f(x)-g(x)>=0,
∫ f(x) dx=∫g(x)dx,则∫ f(x) dx - ∫g(x)dx = ∫ (f(x) -g(x))dx =0
由(1)结论有f(x)-g(x)=0,
证毕
∫ f(x) dx [a,b]
= area under the curve y= f(x) , from a->b
if f(x))>=0 and ∫ f(x) dx=0
=> f(x) = 0
if f(x)>=g(x),
let h(x)= f(x) - g(x)
then h(x) >=0
∫ h(x) dx = 0 [a,b]
=> h(x) =0
=> f(x) = g(x)