Self-inverse
Let f be a differentiable function such that f ( f ( x ) ) = x for x ∈ [ 0 , 1 ] . Suppose f ( 0 ) = 1 .
Determine the value of
∫ 0 1 ( x − f ( x ) ) 2 0 1 6 d x
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1 solution
Can you explain the second step in a detailed way?
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The first step (from integrating wrt x to integrating wrt y ) is just substitution. In the second stage, I just add and subtract a copy of ∫ 0 1 ( f ( y ) − y ) 2 0 1 6 d y .
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I understood the first part. But how does − ∫ 0 1 ( f ( y ) − y ) 2 0 1 6 d y become equals to − I ?
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@Dhruva V – ∫ 0 1 g ( y ) d y = ∫ 0 1 g ( x ) d x . Just change the dummy variable.
If we make the substitution y = f ( x ) , we note that x = f ( y ) , so that d x = f ′ ( y ) d y and hence I = ∫ 0 1 ( x − f ( x ) ) 2 0 1 6 d x = − ∫ 0 1 ( f ( y ) − y ) 2 0 1 6 f ′ ( y ) d y = − ∫ 0 1 ( f ( y ) − y ) 2 0 1 6 ( f ′ ( y ) − 1 ) d y − ∫ 0 1 ( f ( y ) − y ) 2 0 1 6 d y = − [ 2 0 1 7 1 ( f ( y ) − y ) 2 0 1 7 ] 0 1 − I = 2 0 1 7 2 − I which makes the integral I = 2 0 1 7 1 .
A direct calculation with f ( x ) = 1 − x confirms this.