作业帮f(x)在[0,1]上连续,(0.1)内可导,f(0)=3∫(2/3~4)f(x)dx,证明在（0,1）内c存在,f(c)导数=0
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![f(x)在[0,1]上连续,(0.1)内可导,f(0)=3∫(2/3~4)f(x)dx,证明在（0,1）内c存在,f(c)导数=0](/uploads/image/z/2789573-5-3.jpg?t=f%28x%29%E5%9C%A8%5B0%2C1%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%280.1%29%E5%86%85%E5%8F%AF%E5%AF%BC%2Cf%280%29%3D3%E2%88%AB%282%2F3%7E4%29f%28x%29dx%2C%E8%AF%81%E6%98%8E%E5%9C%A8%EF%BC%880%2C1%EF%BC%89%E5%86%85c%E5%AD%98%E5%9C%A8%2Cf%28c%29%E5%AF%BC%E6%95%B0%3D0)
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f(x)在[0,1]上连续,(0.1)内可导,f(0)=3∫(2/3~4)f(x)dx,证明在（0,1）内c存在,f(c)导数=0
f(x)在[0,1]上连续,(0.1)内可导,f(0)=3∫(2/3~4)f(x)dx,证明在（0,1）内c存在,f(c)导数=0
f(x)在[0,1]上连续,(0.1)内可导,f(0)=3∫(2/3~4)f(x)dx,证明在（0,1）内c存在,f(c)导数=0
你写错了吧,积分上限是1.由积分中值定理,存在b位于（2/3 1)之间,使得积分值=3*(1-2/3)f(b),即f(0)=f(b).在[0 b]上用Rolle中值定理得结论.