For any $s\in[0,T)$, the variance in question is $$\int_s^T f(t)^2\,dt.$$ This is shown just as in the case $s=0$.

Alternatively, the case $s\in(0,T)$ can be reduced to the case $s=0$ by noting that the random function $[0,T-s]\ni u\mapsto W^s_u:=W_{s+u}-W_s$ is a standard Wiener process on the interval $[0,T-s]$ and $\int_s^T f(t)\,dW_t=\int_0^{T-s} f(s+u)\,dW^s_u$ and the variance of $\int_0^{T-s} f(s+u)\,dW^s_u$ is $\int_0^{T-s} f(s+u)^2\,du=\int_s^T f(t)^2\,dt$.

For any $s\in[0,T)$, the variance in question is $$\int_s^T f(t)^2\,dt.$$ This is shown just as in the case $s=0$.

Alternatively, the case $s\in(0,T)$ can be reduced to the case $s=0$ by noting that

• the random function $[0,T-s]\ni u\mapsto W^s_u:=W_{s+u}-W_s$ is a standard Wiener process on the interval $[0,T-s]$;

• $\int_s^T f(t)\,dW_t=\int_0^{T-s} f(s+u)\,dW^s_u$;

• the variance of $\int_0^{T-s} f(s+u)\,dW^s_u$ is $\int_0^{T-s} f(s+u)^2\,du=\int_s^T f(t)^2\,dt$.

For any $s\in[0,T)$, the variance in question is $$\int_s^T f(t)^2\,dt.$$ This is shown just as in the case $s=0$.

For any $s\in[0,T)$, the variance in question is $$\int_s^T f(t)^2\,dt.$$ This is shown just as in the case $s=0$.

Alternatively, the case $s\in(0,T)$ can be reduced to the case $s=0$ by noting that the random function $[0,T-s]\ni u\mapsto W^s_u:=W_{s+u}-W_s$ is a standard Wiener process on the interval $[0,T-s]$ and $\int_s^T f(t)\,dW_t=\int_0^{T-s} f(s+u)\,dW^s_u$ and the variance of $\int_0^{T-s} f(s+u)\,dW^s_u$ is $\int_0^{T-s} f(s+u)^2\,du=\int_s^T f(t)^2\,dt$.