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I am having trouble understanding the validity of integrating both sides of an equation. I understand that an operation/manipulation can be performed to both sides of an equation, preserving the equality, eg. if two sides of an equation are equal, their derivatives are equal and hence it is valid to differentiate both sides.

However, when it comes to integrating both sides of an equation I feel a bit uneasy. I will try to explain my concern through an example.

Take the equation $x=x$

integrating both sides,

$\int x$ = $\int x$

$x^2/2 + c$ = $x^2/2 + d$

where c and d are constants.

Surely this only makes sense if c=d, but I see no reason why this should be the case as c and d can assume any constant following integration. Thus I don't see why integrating both sides of an equation is a legal thing to do (although I know it is). I'm hoping that someone can resolve this for me. Thank you.

Jacob G.
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  • An indefinite integral produces a family of functions, not a single specific function. – Clayton Sep 26 '22 at 23:15
  • An indefinite integral is not a specific function, it is a class of functions which differ only by a constant amount. So to say that $\int x,dx=x^2/2+c$ means that $c$ can be any constant. To say that $\int x,dx=x^2/2+d$ means that $d$ can be any constant. So the only diference is the name that you are giving the general constant term. – John Wayland Bales Sep 26 '22 at 23:16
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    Integrating a function gives you a family of functions, each member of which is defined by its unique constant in $\mathbb{C}$ (or $\mathbb{R}$ if you prefer). It represents infinitely many possible cases, the specific form of which depends on more information (sometimes called initial conditions), e.g. $\int x dx=\left{\frac{x^2}{2}+c\mid c\in\mathbb{C}\right}$. – pshmath0 Sep 26 '22 at 23:17
  • https://math.stackexchange.com/q/1473685/96384, https://math.stackexchange.com/q/2475233/96384, https://math.stackexchange.com/q/234537/96384, https://math.stackexchange.com/q/182344/96384, ... – Torsten Schoeneberg Sep 26 '22 at 23:19
  • Thank you for the comments, I am reading everything and thinking about it. I understand everything that is said here and yes I am familiar with the use of integration for solving DEs. Everything that has been said here including the answers all make complete sense to me, so I'm trying to piece it together – Jacob G. Sep 26 '22 at 23:24

3 Answers3

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If you feel unsure about integrating both sides of $$f'(x)=g'(x),$$ you can just move everything to the LHS and integrate the LHS: $$ f'(x)-g'(x)=0\qquad\Longleftrightarrow\qquad f(x)-g(x)=C, $$ using the fact that a function has zero derivative iff it is constant. Notice you would reach the same conclusion if you hadn't moved everything to the LHS.

grand_chat
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More generally, if you have two derivatives of functions equal, say $f'(x)=g'(x)$. It does not automatically follow that $f(x)=g(x)$.

Although, assuming that $f'(x)=g'(x)$ on some interval, say $[a,b]$, then we will have $\int_a^b f'(x)dx=\int_a^b g'(x)dx$.

jdods
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  • Referring to the first sentence, which I understand. If $f'(x) = g'(x)$ then $f(x) = g(x) + c$. I.e. the functions differ by the addition of a constant term.

    If I were to apply this to the example which I gave though, I would surely have that $x^2/2 = x^2/2 + c$ which only makes sense if c=0.

     – Jacob G. Sep 26 '22 at 23:31
  • @RichbertFeynstein Yes, in that case, c=0. It's not they differ by every constant, they differ by A constant – Alan Sep 27 '22 at 03:53
  • @Alan Thank you, that has cleared this up for me – Jacob G. Sep 27 '22 at 08:39
  • @RichbertFeynstein If you start with $x=x$ and you want to integrate, look at it this way: You have two functions $f$ and $g$ such that $f'(x)=g'(x)=x$. Then you know that $f(x)$ and $g(x)$ are both identical to $x$ plus some constant. We don't know what the constants are, they could be the same or different. There is no way to know what the constants are without, say, having the value of the functions at some points, e.g. maybe we know $f(0)=5$ and $g(-3)=8$, in which case we then can solve for $f(x)$ and $g(x)$ explicitly. This is then a differential equation initial value problem. – jdods Sep 27 '22 at 16:15
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The issue here is that indefinite integral is not a 1-1 operation, but 1 to an infinite family. So integrating both sides doesn't get you equality, it gets you equality up to a constant.

The same issue applies when we do any non 1-1 operation to both sides of an equation, for instance when we take a step of squaring both sides of an equation we take the risk of putting in fake solutions because $(-x)^2=x^2$ but $-x\neq x$.

Now, the trick is, we can combine the constants of integration, so we can say one is equal to the other plus a constant.

Here's a concrete example: There are 3 easy ways to integrate

$$\int \sin x \cos x dx$$ If you let $u=\sin x$, then you get to $\frac 1 2 \sin^2 x$. If you let $u=\cos x$ you get $\frac 1 2 \cos^2 x$. Or you could do the substitution from $\sin (2x)=2\sin x\cos x$ and get $\frac 1 4 \sin(2x)$. You'll note if you graph these that the functions aren't identical. They are, however, all the same UP TO the addition of a constant, in other words, their difference is a constant.

For the first 2, it's obvious, $\frac 1 2 \sin^2(x)-\frac 1 2 \cos^2 x=\frac 1 2$, you can play around with the others.

Alan
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