The function $f$ gives the instantaneous rate of change in the amount of rainwater in a given Vessel as a function of time $t$ over the course of $60$ minutes.
The graph of $f$ is :
At time $t = 0$ there are $R_0$ $cm^3$ rainwater in this vessel.
(a) Create a term to calculate the amount of rainwater (in $cm^3$) in this vessel for the time $t = 60$.
(b) Someone claims that at the time $t = 0$ there is at most $10 \ cm^3$ rainwater in the container. Use the graph to explain why this claim must be false.
$$$$
For (a) I have done the following :
Let $R(t)$ is the function that calculates the rainwater.
It holds that $f(t)=R'(t)$ and for $t_0$ there is $R_0$ in the vessel.
Then we have that \begin{align*}\int_0^{60}f(t)\, dt=\int_0^{60}R'(t)\, dt &\Rightarrow \int_0^{60}f(t)\, dt=R(60)-R(0) \\ & \Rightarrow \int_0^{60}f(t)\, dt=R(60)-R_0\\ & \Rightarrow R(60)=\int_0^{60}f(t)\, dt+R_0 \end{align*} Is that correct?
As for (b) : The area of the graph of $f$ is the amount of rainwater, or not? But which part of the are do we have to use here?
Your answer to question a) is correct.
On the other hand, the area of the graph is $0$, since the graph is $1$-dimensional. But the amount of rainwater at time $t=20$ is$$R_0+\int_0^{20}f(t)\,\mathrm dt=R_0-\int_0^{20}-f(t)\,\mathrm dt.$$And $\int_0^{20}-f(t)\,\mathrm dt$ is greater than the area of the triangle whose vertices are $(0,0)$, $(20,0)$ and $(10,2)$. But the area of this triangle is $20~\textrm{cm}^3$. So, if the original amount was $\leqslant10~\textrm{cm}^3$, the amount when $t=20$ would be negative, which makes no sense.