设f(x)在[a,b]上连续,且f(x)不恒等于零,证明∫(a,b)[f(x)]²dx>0
举一反三
- 设f(X)及g(X)在[a,b]上连续(a<b),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)
- 设函数f(x)在对称区间【-a,a】上连续,证明∫(-a,a)f(x)dx=∫(0,a)[f(x)+f(-x)]dx
- 设f(x)及g(x)在[a,b]上连续,证明:若在[a,b]上,f(x)≥0,且。
- 设ab>0,f(x)在[a,b]上连续,在(a,b)内可导,证明:存在ε∈(a,b),使得设f(x)在[a,b]上连续,在(a,b)设f(x)在[a,b]上连续,在(a,b)内连续可导,x。∈(a,b)是f(x)的唯一驻点.若f(x。)是极小值,证明:x∈(a,x。)时,fˊ(x)<0;x∈(x。,b)时,fˊ(x)>0
- 设函数在[a,b]上可微且f`连续,f(a)=0.求证:∫[f(x)]^2dx