# Let $f: \Bbb{R} \to [0; \infty)$ .Prove that $\forall n \in \Bbb{N}$ $\forall y \in \Bbb{R}$ $\exists t=t(n;y)$ such that $\int_{y}^{t}f(x)dx=n$

The problem goes like this:

Let $f: \Bbb{R} \to [0; \infty)$ be a continuous function such that $\lim_{x \to \infty}f(x)=\infty$. Prove that $\forall n \in \Bbb{N^{*}}$ and $\forall y \in \Bbb{R}$ there exists $t=t(n;y)$ such that $\int_{y}^{t}f(x)dx=n$.

As far as I tried, it seems obvious from the continuity that if we fix $n \in \Bbb{N^{*}}$, then for every real number $y$, there exists $t \in \Bbb{R}$ which fulfills the conditions, although I don't see how I can turn it into a function of 2 variables.

Hint: For a fixed $y$, you can show that the function $$F(t) = \int_y^t f(x)dx$$ is a continuous function which has a value $F(y)=0$ and a limit of $\displaystyle\lim_{t\to\infty} F(t) = \infty$. What you are trying to show then follows from the IVT.
Hint: For $y\in \mathbb{R}$, define $$G(t) := \int_y^t f(x)dx$$ Then, $G$ is continuous, $G(y) = 0$ and prove that $$\lim_{t\to \infty} G(t) = +\infty$$ Now apply Intermediate Value theorem.