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My friend is tutoring high school mathematics, one of the skills is to let the integral as $I$ and then get $I = abc - I$ and then $I = abc/2$. For example, $$ I = \int e^x\cos{x} dx = eˣ \sin{x} - \int e^x \sin{x} dx = e^x\sin{x} + e^x\cos{x} - \int e^x\cos{x} = e^x(\sin{x} + \cos{x}) - I, $$ so $I = \frac{1}{2}e^x(\sin{x} + \cos{x}) + \text{constant}$.

However, this technique does not seem to be true in general, for example, let $g(x)$ be an even function, let $x=-u$, $$ I = \int g(x) dx = \int g(-u)d(-u) = \int g(u) (-du) = -\int g(u)du = -I, $$ so $2I = 0$. If we put $g(x) \equiv 1$, then we have $x = \int 1 dx = \text{constant}$, which is ridiculous.

I cannot spot the mistake in it. In my own thoughts, if we add the bounds this is clearly false because letting $x = -u$ changes the bounds so we do not have this issue. I think getting $I$ again is not the main issue, the issue is applying the Dummy Variable Rule in indefinite integrals.

Surely we have u-substitutions, but I think it is not true that $\int f(u)du = \int f(x)dx$ in general, is it the issue here?

I have not touched calculations of integrals for a long time, all we have are definite integrals, never seen indefinite integrals for a long time.

Any help would be appreciated.

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    $\begingroup$ It's true that $\int g(x)\ dx=-\int g(u)\ du$, but you implicitly assume that $\int g(x)\ dx=\int g(u)\ du$, which is not the case. For example, if $g(x)=x^2$, then $\int g(x)\ dx=\frac{1}{3}x^3+C$ and $\int g(u)\ du=\frac{1}{3}u^3+C=-\frac{1}{3}x^3+C.$ $\endgroup$ Commented 9 hours ago
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    $\begingroup$ I think you're onto something pointing out that the assumption that $\int g(x)\ dx=\int g(u)\ du$ might be wrong, but I'm having the devil of a time saying why. A lot of calculus notation is a bit iffy, and when pushed to the limit even basic function notation breaks, (like is $f(x)$ the function or the evaluation of the function at some point)? $\endgroup$ Commented 9 hours ago
  • $\begingroup$ @JonathanZ Maybe we shall ask when $\int g(x)\ dx$ is equal to $\int g(u)\ du$. Set $u=f(x)$. If $\int g(x)\ dx=\int g(u)\ du$, then $$\int g(x)\ dx=\int g(f(x)) f'(x)\ dx.$$ So we should have $g(x)=g(f(x))\cdot f'(x)$. $\endgroup$ Commented 9 hours ago

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for example, let $x=-u.$ Then $$ \int g(x)\, \mathrm dx = -\int g(u)\,\mathrm du \color\red= - \int g(x)\, \mathrm dx.$$

Your mistake is in asserting the second equality above (coloured red), which is false: unlike with definite integrals, where$\displaystyle\int_3^7 e^u \,\mathrm du$ indeed equals $\displaystyle\int_3^7 e^x \,\mathrm dx,$ in your example $u$ and $x$ are not dummy variables.

Rather, the variable of integration in an indefinite integral, which is a family of antiderivatives, is not bound within the integral, so remains present in the resulting antiderivatives: $$\int g(\color{cyan}u)\,\mathrm d\color{cyan} u = G(\color{cyan} u)+C_1,\\ \int g(\color{brown} x)\, \mathrm d\color{brown} x= G(\color{brown} x)+C_2.$$

Tangentially related.

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First off, if you're doing an indefinite integral, $I$ is a function, so let's write $I(x)$. Second, as pointed out, once we make a substitution in your second example, we no longer have the same function $I()$.

So in your second example, under the overall assumption that $u=-x$ and being stricter about functions, you'd be making the argument

$$I(x) = .... = J(u) = J(-x)$$

where $J()$ is the name introduced for the new function we have after substitution.

Now in some cases we do have $J(-x) = I(x)$ and we can do the "bring over to the other side" trick and get a final answer for $I$. But in the cases where the trick goes bad we can see that $J(-x)$ is not equal to $I(x)$ , and the trick wasn't permitted in the first place.

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Well as Bowei Tang said in the comments your reason is right until you "assume" that $$\int f(x)\, dx=\int f(u)\, du$$ and the reasoning here is the fact that for $\textbf{indefinite}$ integrals the variables are NOT dummy variables, since the integral does not give you a number but a function (or more specifically a family of functions). The problem here is that $$\int_{a}^{b} f(x)\, dx\text{ and }\int f(x)\ dx$$ are two completely different things and it happens, thanks to the Fundamental Theorem of Calculus, that these two things agree in the form

$$\int_{a}^{b} f(x)=F(b)-F(a),$$ for all primitives $F(x)$ given by the indefinite integral.

So the problem is that variables are dummy in the case of definite integral thanks to the Fundamental Theorem, because even if you change variable you will change integration extremes and the result has to be the same. In the case of indefinite Integrals there is no concept of "Dummy Variable", because for a variable to be Dummy, the result cannot depend on the variable itself.

And it is important also to know that the indefinite integral is the "inverse problem" of the derivate and definite integral is "just" used to calculate areas under the curves. It happens that those two things are linked, when conditions meet, under the Fundamental Theorem of Calculus. So at the end the problem is more linked with how the symbols behind an integral are much more deep than a simple calculation.

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  • $\begingroup$ @Dennis Yes, thanks it is a typo!!!! $\endgroup$ Commented 8 hours ago

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