$$\lim_{x\to 0} \int_x^{x+1} \sqrt {\arctan {t}}\space dt$$
I think it does not exist because we can't talk about limit from $x\to 0^-$, but what if we just look for $\lim_{x\to 0^+}$? I see there less and less area of the graph so it should be $0$, but how we show it?
Let $F(x)=\int_{0}^x \sqrt{\arctan t}dt$ Then, $$\lim_{x\to 0+}\int_{x}^{x+1}\sqrt{\arctan t}dt=\lim_{x\to 0+}(F(x+1)-F(x))\\=\lim_{x\to 0+}(F(x+1)-F(1))-\lim_{x\to 0+}(F(x)-F(0))+F(1)-F(0)\\=\int_{0}^1\sqrt{\arctan t}dt\\=\int_{0}^{\pi/4}\sqrt{\theta}\sec^2\theta d\theta$$ I am not sure if this can be further simplified.