A periodic function with period 1 is integrable over any finite interval. Also, for two real numbers `a ,b` and two unequal non-zero positive integers `ma n dn` `int_a^(a+n)f(x)dx=int_b^(b+m)f(x)dxdotC a l c ul a t et h ev a l u eofint_m^nf(x)dx`

Explanation

Given `f(1+x)=f(x)`
`:. int_(a)^(a+n)f(x)dx=n int_(0)^(1)f(x)dx` [ `:' f(x)` is periodic]
similarly `int_(b)^(b_m)f(x)dx="m" int_(0)^(1)f(x)dx`
Given `int_(a)^(a+n)f(x)dx=int_(b)^(b+m)f(x)dx`
or `n int_(0)^(1)f(x)dx=m int_(0)^(1)f(x)dx`
or `(n-m)int_(0)^(1)f(x)dx=0`
or `int_(0)^(1)f(x)dx=0 ( :'n!=m)`................1
`:. int_(m)^(n)f(x)dx=int_(0)^(n-m)f(m=x)dt=int_(0)^(n-m)f(x)dx ( :' f` is periodic )
`=(n-m)int_(0)^(1)f(x)dx` (Assume `ngtm`)
`=0` [From equation 1]