does Integrating both sides of an equation in dx will Invalidates the equality?

Question

I'm struggling to grasp the justification behind integrating both sides of an equation. While I understand that operations can be applied to both sides, maintaining equality, it appears that this principle doesn't apply here.

given the equality $x=y$ integrating both sides by dx would give $\frac{x^2}{2} = xy$

but this seems not to be valid since if I start from x=y=5 , I would get $\frac{25}{2}=25 $ that's not true.

given for instance $log(a)=log(b)$ if I integrate both sides by $da$ I get: $alog(a)-a=log(b)a$

if $a=b=2$ then

$alog(a)-a=-0.61..$ and

$log(b)a=1.38$

that are different, what I'm doing wrong here?

Answer 1

It is not the case that $\int ydx = xy + C$ here. Such an equality happens only when $y$ does not depend on $x$ (in which case, it can be treated as a constant when integrating).

Here, we have the relation $y = x$, so that $y$ is actually a function of $x$. That is,

$$\int y(x)dx = \int xdx = \frac{x^2}{2} + C = \frac{y^2}{2} + C$$

So in fact, integrating your equality yields the equation

$$\frac{x^2}{2} = \frac{y^2}{2}$$ or just $x^2=y^2$.

Answer 2

If you anti-differentiate two equal functions the answers differ only by a constant. This is because the derivative of a constant function is zero.

You do need to be careful with integrating expressions involving $x$ and $y$ if one of them depends on the other. Integrating $y$ with respect to $x$ and getting $xy+C$ assumes that $y$ remains fixed while you change $x$, but if $y=x$, then this is not the case.

For comparison, in general the derivative of $xy$ with respect to $x$ is not $y$, but $$\frac{d(xy)}{dx}=y+x\frac{dy}{dx} $$ If $y$ does not depend on $x$ the derivative in the RHS is zero and we get just $y$.