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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:

enter image description here

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!

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    What does the integration part in the second one means? By checking the units of both side in second one: $(kg) = (kg)(sec)$. Is that formula correct? – 2017-02-01
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    Oh, I see what you mean. So this is one of the reasons why the second formulation is wrong. But what if it is $\int \frac{d\theta}{dt}dt$. Good point! Then it would be like the first one right? Is there any way to use an integral in such a case? – 2017-02-01
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    The question is tagged "definite-integral," but since $m_\text{vapor}$ is a function of time, one can immediately eliminate a definite integral as a possibility. In general, one of your tools for figuring out whether to use an integral is to consider what's a function and what's just a number. – 2017-02-01
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    @Ben well I'd consider this a definite integral with a variable limit. That not the same as an indefinite integral (with its arbitrary constant), right? – 2017-02-01
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    I recommend you take a look: [Riemann Sum](https://en.wikipedia.org/wiki/Riemann_sum), then you will know what I mean "as though they're linear" in my answer. And you *should* be aware the unit issue I've pointed out. – 2017-02-02

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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.

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    Finally! An answer from someone who carefully read the OP instead of just assuming that $\theta(t)$ was something other than what was said! – 2017-02-01
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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.

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    Thank you. What about the comment from N1ng? Do the units make the integral formulation invalid? – 2017-02-01
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    Sure. Natural examples need knowing _dimensions_ and this helps for better formulation. – 2017-02-01
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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.

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    Thank you. It makes more sense now, but what about the comment from N1ng? Do the units make the integral formulation invalid? – 2017-02-01
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    Not at all. If you think about it, the area is the sum of all the masses within the boundary, so the units will remain unchanged. – 2017-02-01
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    +1 I would add a generalization: the integral does more than find areas. You use it to _add things up_. In this case you want the _total_ mass of vapour, given the _rate_ at which it's formed. The units tell you that the integral is right since $m \cdot \theta(t)$ is measured in grams per second and $dt$ in seconds. The integral sign is a stretched out "S" for "sum". – 2017-02-01
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    @EthanBolker, Linus Choy - The OP states that $\theta(t)$ is unitless, and is the ratio of the mass of vapor in the air to the original mass of water. It is not a rate. In this case, the integral formula gives a physically meaningless value with units (kg)(sec), and it is $m\theta(t)$ that tells you what is the total mass of water vapor that has entered the air since the experiment started. – 2017-02-01
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    @PaulSinclair I think you're right. I misread $\theta$ as the percent of $m$ that was evaporating at time $t$. In the correct interpretation there is indeed no integral - just read the value of $\theta$ at the time $t$ and multiply by the mass $m$. The real problem is getting the physics right in order to figure out what math applies. – 2017-02-01
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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.