IF f(x+f(y))=f(x)+y AA x, y in R and f(0...

IF `f(x+f(y))=f(x)+y AA x, y in R and f(0)=1`, then `int_(0)^(10)f(10-x)dx` is equal to

`underset(0)overset(1)int f(x) dx`

`10 underset(0)overset(1)int f(x)dx`

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The correct Answer is:

We have,
f(x+f(y))=f(x)+y for all `x,y in R`
Replacing y by 0, we obtain
`rArr f(x+f(0))=f(x)+0` for all ` xin R`
`rArr f(x+1)=f(x)` for all `x in R`
`rArr f(x)` is periodic with period 1.
Now,
`underset(0)overset(10)int f(10-x)dx=-underset(10)overset(0)int f(t) dt`, where t=10-x
`=underset(0)overset(10)int f(t) dt=10underset(0)overset(1)int f(t)dt[:' f(x)" is periodic with period "1]`

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