If f(x) is an odd pefiodc function defined on the interval [T/2,T/2], where T is rthe perod of f(x). Then `phi(x)=int_(a)^(x) f(t)dt,` is

Options

A. periodic with period T
B. non-periodic
C. periodic with period 2T
D. perodic with period aT

Answer

A

Explanation

We have,
`phi(x+T)=underset(a)overset(x+T)int f(t)dt=underset(a)overset(x)int f(t) dt+underset(x)overset(x+T)int f(t)dt`
`rArr phi(x+T)=phi(x)+underset(x)overset(x+T)int f(t)dt`
`rArr phi(x+T)=phi(x)+Psi(x),"where"Psi(x)=underset(x)overset(x+T)int f(t)dt`
Now,
`Psi(x)=underset(x)overset(x+T)int f(t)dy=underset(x)overset(0)int f(t)dt+underset(0)overset(T)int f(t)dt+underset(x)overset(x+T)int f(t)dt`
Putting `t-t=y` in the last integral, we have,
`underset(x)overset(x+T)int f(dt)=underset(0)overset(x)intf(y+T)dy=underset(0)overset(x)intf(y)dy`
`:. Psi(x)=underset(x)overset(x)intf(t)dt+underset(0)overset(T)intf(t)dt+underset(0)overset(x)intf(t)dt=underset(0)overset(T)intf(t)dt`
`rArr Psi(x)` is independent of x.
`rArr Psi(x)=Psi(-T//2)`
`rArr Psi(x)=underset(T//2)overset(T//2)intf(t)dt[ "Putting" x=-(T)/(2)"in" Psi=underset(x)overset(x+T)intf(t)dt]`
`rArr Phi(x)=0 " " [ :. f(t) "an odd function" ]`
`:. phi(x+T)phi(x)` for all x