This is Apostol calculus chapter 10.24 exercise 21.
Let $$f:[1,\infty)\to \mathbb R$$ such that $$\exists I_{n}=\int_{1}^{n}f(x)dx \forall n\in \mathbb N$$If $$\lim_{x\to \infty}f(x)=0; \lim_{n\to \infty}I_{n}=A$$ then $$\Rightarrow \int_{1}^{\infty}f(x)dx=A$$
I just want you to tell me if this proposition is true or false, My classmate told me is false but I cant find a counterexample, I would appreciate your help :)
Yes, it is true. The difference between the integral and the closest $I_n$ is bounded by the maximum of $|f(x)|$ in the interval of length $\frac 12$. It would not be true if you didn't have $f(x) \to 0$. You could have $f(x)=\sin (2\pi x)$, where all the $I_n$ are $0$, but the integral oscillates and has no limit.