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I would like to know the whole purpose of adding a constant termed **constant of integration** everytime we integrate an indefinite integral $\int f(x)dx$. I am aware that this constant “goes away” when evaluating definite integral $\int_{a}^{b}f(x)dx $. What has that constant have to do with anything? Why is it termed as the **constant of integration**? Where does it come from?

The motivation for asking this question actually comes from solving a differential equation $$x \frac{dy}{dx} = 5x^3 + 4$$ By separation of $dy$ and $dx$ and integrating both sides, $$\int dy = \int\left(5x^2 + \frac{4}{x}\right)dx$$ yields $$y = \frac{5x^3}{3} + 4 \ln(x) + C .$$

I’ve understood that $\int dy$ represents adding infinitesimal quantity of $dy$’s yielding $y$ but I’am doubtful about the arbitrary constant $C$.

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Sometimes you need to know all antiderivatives of a function, rather than just one antiderivative. For example, suppose you’re solving the differential equation

$$

y'=-4y

$$

and there’s an initial condition $y(0)=5$. You get

$$

\frac{dy}{dx}=-4y

$$

$$

\frac{dy}{y} = -4\;dx

$$

$$

\int\frac{dy}{y} = \int -4\;dx

$$

$$

\log_e y = -4x + C.

$$

Here you’ve added a constant. Then

$$

y = e^{-4x+C} =e^{-4x}e^C

$$

$$

5=y(0)= e^{-4\cdot0}e^C = e^C

$$

So

$$

y=5e^{-4x}.

$$

(And sometimes you need only one antiderivative, not all of them. For example, in integration by parts, you may have $dv=\cos x\;dx$, and conclude that $v=\sin x$.)

The key concept to note here is that when you differentiate a constant you get 0, this is due to the fact that the slope of the tangent line of a constant function, say $f(x) = 4$, will simply be a horizontal line spanning the x-axis with a slope of zero everywhere.

When you differentiate a particular function that has a constant at the end, say $f(x) = x^2 +2x +4$ to get $f'(x) = 2x +2$, you have no way, given only the derivative $f'(x)$, to recover the “constant information” about the original function.

This is precisely why you have to have a slope field representation of the anti-derivative of a function. When you integrate a particular function, you must add that $+C$ because it says that, the anti-derivative of the function could be one of any of the slope field lines in $\mathbb{R}^2$. The particular value for $C$ collapses it to exactly one of these slope field lines. Here is a graphical representation of a slope field and three particular values of $C$ that draw out the purple functions. The slope field anti-derivative is given by:

$$ \int \cos(x) dx = \sin(x) +C$$

The three purple functions drawn out are:

$$f(x) = \sin(x) -2$$

$$f(x) = \sin(x) + 0$$

$$f(x) = \sin(x) +2 $$

It may also make more sense when you take a differential equations course, but this should be a sufficient explanation.

The indefinite integral $\int f(x)dx$ is the function $F(x)$ such that $\frac{d}{dx}F(x) = f(x)$. Problem is, there are multiple (in fact infinitely many) such functions $F(x)$. We cannot allow the expression $\int f(x)dx$ to refer to multiple functions, so to get around this we introduce the *constant of integration* $C$. This fixes things, because though there are infinitely many $F(x)$ such that $\frac{d}{dx}F(x) = f(x)$, if we pick any single such function $F(x)$ then all solutions to $\frac{d}{dx}G(x) = f(x)$ are of the form $G(x) = F(x) + C$ for some value of $C$, while for any value of $C$ the function $F(x)+C$ satisfies $\frac{d}{dx}(F(x)+C) = f(x)$. Hence we say $\int f(x)dx = F(x) + C$.

The indefinite integral $\int f(x) dx$ denotes, in this particular context, the *set* of all primitives of $f(x)$. If you find one primitive, say $F(x)$, then you know all other primitives have the form $F(x) + C$, where $C$ is any constant. This is what we mean when we write $\int f(x) dx = F(x) + C$: the set of all primitives of $f(x)$ is the set of all the functions of the form $F(x) + C$.

The indefinite integral $\int f(x)\,dx$ is defined to be the general class of *functions* whose derivatives are $f(x)$.

Now, given any function $F$ with $F'=f$, it follows that $F+C$ is also an antiderivative of $f$:

$$

(F+C)'=F'+C'=F'+0=F'=f.

$$

Conversely if $G$ is an antiderivative of $f$, then $G$ has the form $F+C$ for some constant $C$.

That’s why we write, for example:

$$

\tag{1}

\int x^2\,dx={x^3\over 3}+C.

$$

The derivative of ${x^3\over 3}+C$ is $x^2$. Moreover if $F'(x)=x^2$, then $F$ must have the form $F(x)={x^3\over3}+C$.

So, the general antiderivative of $f(x)=x^2$ has the form $F(x)={x^3\over3}+C$, and equation (1) is just stating this fact.

The differential equation you are considering has more than one solution. The *general solution* is as you have, with the arbitrary parameter $C$. It does belong there.

Any function of that form will be a solution to the equation.

If you were given an initial value problem where, say, you needed $y(1)=2$, then that constant $C$ could be determined.

The “integration constant” $C$ does have the “purpose” to make a seemingly true equation at least halfway true. In fact it is what many people call a “dangling variable”, similar to the $i$ and $k$ when we talk about a “matrix $\bigl[a_{ik}\bigr]\ $”. As noted in other answers, an *indefinite integral* is not a function, but an infinite set of functions, any two of them differing by a constant. Writing $\langle F(x)\rangle$ (or similar) instead of $F(x)+C$ for the set of all functions differing from the term $F(x)$ by a constant, one could write, e.g.,

$$\int (5x -6x^4)\ dx =\langle {5\over2} x^2-{6\over5} x^5\rangle\ ,$$

and the mysterious constant has disappeared.

It is simply a special case of the ubiquitous *linear* principle that the *general* solution of a *nonhomogeneous* linear equation is given by adding any *particular* solution to the general solution of the associated *homogeneous* equation. Namely, if $\rm\:D\:$ is a linear map then one easily proves

**Lemma** $\ \ $ If $\rm\ D\:f_1\ =\ g\ $ then $\rm\ D\:f_2\ =\ g\ \iff\ 0\ =\ D\:f_1 – D\:f_2\ =\ D\:(f_1-f_2)$

Therefore $\rm\ D^{-1}(g)\ =\ f_1\ +\ ker\ D\ =\:\: $ particular + homogeneous solution, as in linear algebra.

In particular $\rm\quad\:\: \int g\ =\ f_1\: +\ c,\ $ for $\rm\ c\in ker\ \dfrac{d}{d\:x}\: =\:\: $ constants w.r.t. the derivation $\rm\ D\: =\: \dfrac{d}{d\:x}\:.$

Compare this to $\rm \ x\ =\ 3\ +\ 5\ \mathbb Z\:,\:$ the solution of $\rm\ 2\: x\ \equiv\ 6\pmod{10}\:,\: $ with particular solution $\rm x \equiv 6/2 \equiv 3\:,\: $ and homogeneous: $\rm\ 2\: x\:\equiv 0\pmod{10}\iff 10\:|\:2\:x\iff 5\:|\:x\iff x\in 5\ \mathbb Z\:.$

There are many great answers here, but I just wanted to chime in with my favorite example of how things can go awry if one forgets about the constant of integration.

Consider

$$\int \sin(2x) dx.$$

We will find antiderivatives in two ways. First, a substitution $u=2x$ yields:

$$\int \frac{\sin(u)}{2}du = -\frac{\cos(u)}{2} = -\frac{\cos(2x)}{2}.$$

Second, we use the identity $\sin(2x)=2\sin(x)\cos(x)$ and a substitution $v=\sin(x)$:

$$\int \sin(2x)dx = \int 2\sin(x)\cos(x)dx = \int 2vdv =v^2 = \sin^2(x).$$

Thus, we have found two antiderivatives of $\sin(2x)$ that are completely different! Namely

$$F(x)=\sin^2(x) \quad \text{and} \quad G(x)=-\frac{\cos(2x)}{2}.$$

Notice that $F(x)=\sin^2(x)\neq -\cos(2x)/2=G(x)$. For instance, $F(0)=\sin^2(0) = 0$ but $G(0)=-\cos(2\cdot 0)/2=-1/2$. So, what happened?

We forgot about the constant of integration, that’s what happened. The theory of integration tells us that all antiderivatives differ by a constant. So, if $F(x)$ is an antiderivative, then any other antiderivative $G(x)$ can be expressed as $G(x)=F(x)+C$ for some constant $C$. In particular, our antiderivatives above must differ by a constant. Indeed, the constant $C$ in this case is exactly $C=-\frac{1}{2}$:

$$F(x)+C = F(x) – \frac{1}{2} = \sin^2(x)-\frac{1}{2} = \frac{(1-\cos(2x))}{2}-\frac{1}{2} = -\frac{\cos(2x)}{2} = G(x),$$

where we have used the trigonometric identity $\sin^2(x) = (1-\cos(2x))/2.$

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