For, $n\ge 1$ let, $g_n(x)=\sin^2(x+1/n)$ , $x\in [0,\infty)$ and $f_n(x)=\int_0^xg_n(t)\,dt.$
Then, which is(/are) correct ?
(A) $\{f_n(x)\} $ converges pointwise to a function $f$ on $[0,\infty)$ but does not converge uniformly on $[0,\infty)$.
(B) $\{f_n(x)\} $ does not converge pointwise to any function on $[0,\infty)$.
(C) $\{f_n(x)\} $ converges uniformly on $[0,1]$.
(D) $\{f_n(x)\} $ converges uniformly on $[0,\infty)$.
I found that, $$f_n(x)=\frac{1}{2}\left[x-\frac{1}{2}\sin(x+1/n)+\frac{1}{2}\sin(1/n)\right].$$
So, $f(x)=\lim_{n}f_n(x)=\frac{x}{2}-\frac{1}{4}\sin x$.
Now, $|f_n(x)-f(x)|\le 3/4$. So, by M-test, $f_n(x)$ is not uniformly convergent, but point-wise convergent. So option (A) is correct..Am I right??