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As we know, antiderivative or indefinite integral is the function the derivative of which gives the actual function. Let $F(x)$ be the derivative of $f(x)$ ie. the instantaneous rate of change of $f(x)$ with respect to $x$ is $F(x)$ . $$\dfrac{d{f(x)}}{dx} = F(x).$$ Now $$ d{f(x)} = F(x)\,dx,$$ right? Then writing indefinite integral on both side, we get, $$\int d{f(x)} = \int F(x)\,dx.$$ Then abruptly many books do like that $$ f(x) = \int F(x)\,dx + C.$$ Really? What does this $\int$ mean? Summation, right? Then how can summation of $d{f(x)}$ give the function? It is the change and not the function. What about $C$? My physics book tells that it is the initial value i.e. $v_0, a_0 , x_0$ . But it can be anything, right? So, how can this process give the function?

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  • $\begingroup$ You must understand that when you write down the differential equation, the function $f(x)$ is only defined upto an additive constant. This is basically equivalent to saying the solution of an inhomogenous ODE is the sum of the complementary solution and a particular integral, the complementary solution solving the homogenous equation. In your case, this turns out to be a constant. There is nothing surprising about it. $\endgroup$ – surajshankar Nov 20 '14 at 6:25
  • $\begingroup$ @surajshankar: Sir, I am not of your level. I am just on Class-XI . Can you please simply explain??? $\endgroup$ – user36790 Nov 20 '14 at 6:38
  • $\begingroup$ As for the first part of your question, $f = \int df$ for the same reason that $x = \int dx$. $\endgroup$ – Jold Nov 20 '14 at 7:00
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The integral $\int$ is indeed a continuous version of summation. There are two ways of looking at this:

As an indefinite integral, you have $\int df(x)=\int F(x)dx$. As we are dealing with indefinite integral, the right side after evaluation still depends on $x$. So naturally, it's a function of $x$. On the left, the notation with $(x)$ is perhaps confusing to you, but you can just write $df$ instead of $df(x)$, and you just get $f$ on the left (that is then obviously dependent on $x$ that is found in the solution of the right-side integral). Also, you could just expand the differential of the function over $x$: $df(x)=f'(x)dx$ so you have $\int f'(x)dx=\int F(x)dx$ and now it's probably more obvious to you that it equals to $f(x)$ because integration and differentiation are roughly speaking inverse operations (up to a constant). With indefinite integral, you get this free constant $C$ that has to be manually determined by plugging in the initial conditions. And yes, you usually get something like "initial velocity" there.

As a definite integral with a free upper bound. This is the more logical way of dealing with things. With definite integrals, you have two boundary conditions. So, we start by writing $$\int_a^b df=\int_{A}^{B} F(x)dx$$ Now, what does that mean? Imagine following what's happening to both sides when we go from initial to final condition. On the right, we start at initial $x=x_0$ where we know about our initial condition (depending on what you are integrating, it could be initial position, initial time, or something like that -- you very commonly choose your coordinates so that $x_0=0$). At this chosen initial condition, the value of $f$ is also the initial one, $f(x_0)$. Then, you imagine integrating slowly over the region, the right side calculates the small contributions $F(x)dx$ that are added on the left as increments $df$. There are now two possibilities. If you actually want to follow what's happening in the middle -- the entire curve that $f$ traces when you are adding more and more contributions on the right, then you use an "arbitrary $x$" in the upper limit on the right, and of course, $f(x)$ on the left, because that's how far $f$ got by that "time" (or position, or whatever). So, you get

$$\int_{f(x_0)}^{f(x)}df=f|_{f(x_0)}^{f(x)}=f(x)-f(x_0)=\int_{x_0}^x F(t)dt$$ Notice that I renamed the integration variable, because it's inappropriate to use the same symbol as in the limits. $t$ is the thing that goes from the initial $x_0$ to the current $x$ when the integral performs the summation. Each chosen $x$ results in a different integration interval and also different value of $f(x)$. So, we performed a definite integral, but left a variable upper limit so we now still have a functional dependence: $$f(x)=f(x_0)+\int_{x_0}^x F(t)dt$$ Which you read as: function $f$ starts at its initial value $f(x_0)$ and changes when you go to $x$ for the amount given by the integral.

Sometimes, you don't care what happens in the middle, and you actually just want the definite integral over a pre-determined interval (such as "how far did we get in one hour"). In that case, you can write $$f(x_1)=f(x_0)+\int_{x_0}^{x_1} F(t)dt$$ where $x_1$ is just a number. Note how there is absolutely no difference between these two cases and you get the "definite integral over a fixed range" just by putting the number of $x_1$ into the function $f(x)$ that you calculated before.

In many aspects, the definite integral with variable upper limit is the same as the indefinite integral. However, with indefinite integral, you have absolutely no idea what $C$ before you take into account the initial conditions, while with definite integral, you see it's equal to $f(x_0)$ in this case. Also, the notation is better because in definite integrals, the integrating variable has nothing to do with the limits. I could have written it as $\int_{f(x_0)}^{f(x)}d\xi=\int_{x_0}^x F(\chi)d\chi$ and it wouldn't have made a difference. This probably answers your confusion about $df(x)$: it just tells you that you want to capture variation of the upper limit on the left when you vary the upper limit at the right. Also, it's much easier to deal with physical units, because lower and upper limit of the integral have the same units (are the same physical quantity as the differential under the integral sign) while the $C$ is just a placeholder for the thing you don't know yet.

If you differentiate the definite integral over the upper limit it's the same thing as differentiating the result of the indefinite integral. In both cases you get $f'(x)=F(x)$.

To sum up: indefinite integral is the mathematically formal procedure that brings you to the correct result. Definite integral with variable upper limit is the same thing and has the physical interpretation of following the "summation" (following your position when making steps in time with changing velocity, or something like that). Definite integral with fixed numerical limits is just one read-out of the function you get from the integrals discussed above.

I hope this answers your question.

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exactly as you said, if you only know $F(x)$, which is how the function "changes", then you can only recover the original "shape" of $f(x)$, because the change will be the same if you move ("shift") $f(x)$ all up or all down the y axis. So if $f(x)$ is a solution, then $f(x)+C$ must also be a solution. Regarding why C is the initial condition: it is by definition. $F(0)$ is a constant thus $\int F(0).dx=xF(0)$. Thus $f(0)=C$, thus $C$ is by definition the initial condition .

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  • $\begingroup$ +1 . That's great. $\int f(x)dx$ is the function of area while $C$ gives the initial condition to give the actual function, right? But what about $\int_a^x f(x)dx$ ? From Fundamental Theorem of Calculus 1, it is also indefinite integral,right?? But how can it be an indefinite integral as it has lower limit?? Can you please explain??? $\endgroup$ – user36790 Nov 20 '14 at 9:28
  • $\begingroup$ No, it in this case is definite, you just treat the upper limit of the integration interval as a variable, but is no longer an indefinite integral (the choice of $a$ fixes the value of $C=-f(a)$). $\endgroup$ – Wolphram jonny Nov 20 '14 at 14:31
  • $\begingroup$ Sir, now I have understood that indefinite integral is equivalent to definite integral having a variable limit. So, you are right. Thanks:-D $\endgroup$ – user36790 Nov 21 '14 at 5:33
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Let's say for simplicity $F(x) = 3 x^2$, just so we have an example to talk about. As you wrote $$f(x) = \int F(x) \mathrm d x + C = x^3 +C\,.$$ And now the initial conditions come into play (the $x_0$, $v_0$ and so on) or in some cases boundary conditions. Those are always needed to find a specific solution to a differential equation. So if I also know, that $f(0) = f_0$, then we can just plug that in: $$f(0) = 0^3 +C = C = f_0$$ So we see, if $f(x)$ should also fulfill the condition $f(0) = f_0$, $C$ can not have any value, but needs to be $f_0$ in this specific example - in general (for other $f(x)$'s) you always need to calculate the $C$ that you need to fulfill your initial conditions.

Here some more notes:

A lot of times, when you solve a differential equation you try to find all or a whole bunch of solutions (that's the $x^3 +C$ above) and then pick one of them, that fits your initial/boundary conditions.

If you have higher order DGLs (differential equations) like $F = m \ddot x$ (where each dot means a derivative with respect to time), then you need more than one initial condition to solve this for $x(t)$ (the initial position and initial velocity).

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  • $\begingroup$ How does your argument work if F(x)=sin(x) ? $\endgroup$ – Wolphram jonny Nov 20 '14 at 6:52
  • $\begingroup$ the integral is $-cos(x)$, so that $f(0)=-1+C=f_0$ and thus $C=f_0+1$ $\endgroup$ – jens_bo Nov 20 '14 at 6:56
  • $\begingroup$ but you said before that "C can not have any value, but needs to be $f_0$". And now it can be $f_0+1$? $\endgroup$ – Wolphram jonny Nov 20 '14 at 6:59
  • $\begingroup$ $C$ depends on the initial conditions and of course on the function. So you always need to find the right $C$ for your problem. $\endgroup$ – jens_bo Nov 20 '14 at 7:01
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    $\begingroup$ I agree, I just pointed out that you contradicted yourself, perhaps you should change the text a little bit. Or did you mean that "C can not have any value, but needs to be f0" only for that example? $\endgroup$ – Wolphram jonny Nov 20 '14 at 7:03

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