Let $f(x)$ be an odd continuous function which is periodic with period 2. If $g(x)=\int\limits_0^x f(t) d t$, then

$g(2 n)=0$ for all $n \in N$

It is given that $f(x)$ is an odd function. Therefore, $g(x)=\int\limits_0^x f(t) d t$ is an even function.

$\Rightarrow g(x+2)=\int\limits_0^{x+2} f(t) d t$

$\Rightarrow g(x+2)=\int\limits_0^2 f(t) d t+\int\limits_2^{x+2} f(t) d t$

$\Rightarrow g(x+2)=g(2)+\int\limits_0^x f(t) d t$       [∵ f(x) is periodic with period 2]

$\Rightarrow g(x+2)=g(2)+g(x)$ for all $x$            ......(i)

Now,

$g(x+2)=g(2)+g(x)$

$\Rightarrow g(1)=g(2)+g(-1)$           [Replacing x by -1]

$\Rightarrow g(2)=g(1)-g(-1)$

$\Rightarrow g(2)=\int\limits_0^1 f(t) d t-\int\limits_0^{-1} f(t) d t$

$\Rightarrow g(2)=\int\limits_0^1 f(t) d t+\int\limits_{-1}^0 f(t) d t$

$\Rightarrow g(2)=\int\limits_{-1}^1 f(t) d t$

$\Rightarrow g(2)=0$               [∵ f is an odd function]

Putting g(2) = 0 in (i), we get

$g(x+2)=g(x)$ for all x

$\Rightarrow g(x)$ is periodic with period 2 units

Also,

$g(x+2)=g(x)$ for all x

$\Rightarrow g(2)=g(4)=g(6)=...=g(2 n)$

$\Rightarrow g(2 n)=0$ for all $n \in N$