1.

Consider f:[0,1]toR has a continuous derivative and int_(0)^(1)f(x)dx=0, then for ever alpha epsilon(0,1) |int_(0)^(1)f(x)dx|lex. "Max"_(0le x le 1)|f^(')(x)| then

Answer»

`[1/k]gt 5`
`1/8 [1/k]EPSILON[0,1]`
`kepsilon(0,1/5)`
`1/(2k)+1` is an odd number

Solution :`|int_(0)^(9)f(x)dx|=|a int_(0)^(1)f(at)dt|le 1/8 M`


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