So I was solving a question which asked me to tell whether the function $f (x)=\int_{0}^{x} t\sin (1/t) \,dt $ is continuous. The answer given is that the function is continuous.
I want to know the reason behind it. What I think that since the function $x \sin (1/x) $ is continuous, its area is also going to be a continuous and hence the function $f (x) $ is continuous. Is my reasoning correct? Is there any other way to prove it?
Yes, note that $$\big|f (x)-f(y)\big|=\\\bigg|\int_{0}^{x} t\sin (1/t) \,dt -\int_{0}^{y} t\sin (1/t) \,dt\bigg|\\=\bigg|\int_{x}^{y} t\sin (1/t) \,dt\bigg|\le \int _x^y|t\sin (1/t) |\,dt\\\le \int_{x}^{y} |t|\,dt=\frac {1}{2}|x^2-y^2|$$