设f(x)在[a,b]上连续,则f(x)在[a,b]上的平均值是( )
A: f(a)+f(b)2
B: ∫baf(x)dx
C: 12∫baf(x)dx
D: 1b-a∫baf(x)dx
A: f(a)+f(b)2
B: ∫baf(x)dx
C: 12∫baf(x)dx
D: 1b-a∫baf(x)dx
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f为[a,b]上的非负可积函数,在x0∈[a,b]连续且f(x0)>0,证明:∫baf(x)dx>0.
- 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
- 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),g(x)在区间[a,b]上连续,且g(x) A: π∫ab[2m-f(x)+g(x)][f(x)-g(x)]dx B: π∫ab[2m-f(x)-g(x)][f(x)-g(x)]dx C: π∫ab[m-f(x)+g(x)][f(x)-g(x)]dx D: π∫ab[m-f(x)-g(x)][f(x)-g(x)]dx