It is known that `f(x)` is an odd function and has a period `p`. Prove that `int_(a)^(x)f(1)dt` is also periodic function with the same period.

Explanation

Let `F(x)=int_(a)^(x)f(t)dt`
`:.F(x+p)=int_(a)^(x+p)f(t)dt=int_(a)^(x)f(t)+int_(x)^(x+p)f(t)dt`
`=F(x)+int_(x)^(x+p)f(t)dt`……………1
Obviously, now we have to prove that `int_(x)^(x+p)f(t)dt` is zero.
Given that `f(x)` has period `p`. Then `int_(x)^(x+p)f(t)dt` is independent of `x`.
Let `x=-p//2`. Then `int_(x)^(x+p)f(t)dt=int_(-p//2)^(p//2)f(t)dt=0`
[As given `f(x)` is an odd function]
`:.F(x+p)=F(x)` ltbr Thus, `F(x)` is periodic with period `P`.