Let f (x) be an odd function defined on R with a period T. Then $F(x)=\int_0^xf(t)dt$ is
Answer & explanation
Correct answer: option 1
$F (x + T) = F(x) +\int\limits_x^{x+T}f(t)dt$ ..... (i)
Let $g(x)=\int\limits_x^{x+T}f(t)dt$
$g'(x)=f(x+T)-f(x)=0$, ∴ g(x) = constant
But $g(-\frac{T}{2})=\int\limits_{-T/2}^{T/2}f(t)dt=0$ {∵ f(t) is an odd function}
$∴g(x)=g(-\frac{T}{2})=0$; $∴F(x+T)=F(x)$