When to use integral? I have a question concerning when to use integral and what is the difference between two formulations, one with integral and another one without. I have formulated a simple example:
Let's assume we have m = 1 kg of water that is heated. And because of this, there is some vapor forming. Now let's say that the fraction of vapor ($\theta$) takes values from 0 to 1 and has the profile shown below:

Now it comes. For calculating the mass of the vapor ($m_{vapor}$), there are two formulations in my mind:


*

*$m_{vapor} = m \cdot \theta(t)$

*$m_{vapor} = m \cdot \int_t \theta (t) dt$


But the problem is that I don't know why would I use one over another and here is where I need help. Can anyone please help me to understand why would one use the integral formulation and why not? What makes them different? 
I have serious problems understanding the function of the integral, other than the fact that it represents the area under a curve. But when to use it? I would highly appreciate if anyone can explain it to me in detail.
Thank you in advance!
 A: Briefly we use integrals to obtain anti-derivatives to use them various place, in mathematics or science. Two integrals we generally use

indefinite integral:

An integral expressed without limits.

definite integral:

An integral expressed as the difference between the values of the integral at specified upper and lower limits of the independent variable.
\begin{cases}
\text{areas}\\
\text{volumes}\\
\text{approximations}\\
\cdots
\end{cases}
Mainly, in science, integrals obtained by varieties. In your example you want to release a formula shows vapor volume forming in times. We don't matter here is true or false, but at first glance vapor amount change in time with parameter $k$ which depend on many things (temperature of oven and environment, dish volume, liquid type even water has various density-, and any others), that are important. We write
$$\Delta m=k(\Delta t)(\Delta \theta)$$
These varieties make differentials like 
$$d m=k(dt)(d\theta)$$
These differentials with initial and boundary values make integrals to give us final solutions.

That was for when we use integrals and at last, the difference between two formulations come from our assumptions. Powerful and precise assumptions make the formulation better in sense of nature behaviors, and a tiny differences in formulation might make wrong results.

A: To put it qutie simply, you use the integral to find the total area bounded underneath a curve. In this example, I can use the interal $m_{vapor} = m \cdot \int_t \theta (t) dt$ to find the total mass of the vapour from the start to a value $\theta (t)$. Whereas if you simply used the formula $m_{vapor} = m \cdot \theta(t)$ you would only find the mass of the vapour at that particular of  value $\theta (t)$ instead of a range.
A: Personally, I don't use actually use either.  Instead, I try to remember what the graph means.  As my teacher says, there are two things you need to remember:


*

*Slope

*Area
As per units and what each means, slope is division and area is multiplication.
For this particular graph, the slope has units $\theta/s$ and the area has units $\theta s$, where $s$ is seconds for time.
From there, I decide which I need, and in this case, I'll want the area.
The last step I do is decide if I can tackle the area with geometric shapes, and since its a triangle, I can easily use $\frac12bh$ for the area.  Integrals in basic physics courses should generally be avoided when possible.
A: You can't tell whether to use an integral just by looking at
the graph of $\theta(t)$.
Whether or not to use an integeral depends on two things:


*

*What does the formula mean?

*What is the answer you need?


In this case, you were told $\theta(t)$ is the fraction of vapor as a mass
$m$ of water is heated.
I take this to be the following definition of $\theta(t)$:
$$
\theta(t) = \frac{m_\mathrm{vapor}(t)}{m}.
$$
From this, simple algebra tells us immediately that
$m_\mathrm{vapor}(t) = m \cdot \theta(t).$
Therefore no integral is necessary.
If instead of the fraction of vapor, you had some measurement of the rate at which the water was being turned into vapor, then you could use an integral
to determine how much water was turned to vapor between two times.
The two times could be "before any water was vaporized"
and "now", if that's appropriate to what is being asked.
