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My friend is tutoring high school mathematics, and one of the techniques taught is to let an integral be $I$ then get $I = abc - I$ so that $I = abc/2.$ 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 and $x=-u.$ Then $$ I := \int g(x) dx = \int g(-u)d(-u) = \int g(u) (-du) = -\int g(u)du = -\int g(x) dx = -I. $$ So, $I = 0.$ Putting $g(x) \equiv 1$ gives $x + \text{constant} =0,$ which is ridiculous. But I cannot spot any mistake in this. 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 the main issue is in not getting $I$ again but in applying the Dummy Variable Rule in indefinite integrals. Surely we have u-substitutions, but I think it is not true in general that $\int f(u)du = \int f(x)dx$ ?

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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 yesterday
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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 yesterday
  • $\begingroup$ @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 yesterday

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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 (to correct that final step, you'd simply perform integration-by-substitution in the reverse direction).

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: $$\int g(\color{blue}u)\,\mathrm d\color{blue} u = G(\color{blue} u)+C_1,\\ \int g(\color{brown} x)\, \mathrm d\color{brown} x= G(\color{brown} x)+C_2.$$

Tangentially related.

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  • $\begingroup$ Indeed, the argument above correctly shows that $G(x) = -G(u) + C$. Since $u = -x$, we have $G(x) = -G(-x) + C$, i.e. the antiderivative is an odd function plus a constant, which is true. $\endgroup$ Commented 22 hours ago
  • $\begingroup$ Indeed, every antiderivative of an even function is a vertical translation of a common odd function. $\endgroup$ Commented 22 hours ago
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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 yesterday
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Any apparent inconsistency in your argumentation is one of notation, not of concept. We can consider two equivalent interpretations of the integral, namely the antiderivative and the area, to resolve the issue.

First, the meaning of $\int f(x)\,dx$ as antiderivative: the notation $\int f(x)\,dx$ denotes a function whose derivative is $f(x)$. In this notation, the fundamental theorem of calculus is $$ F'(x)=\frac{d}{dx}\int f(x)\,dx=f(x). $$ Now if $g$ is an even function, then the change of variables $u=-x$ gives $$ G(x)=\int g(x)\,dx=\int g(-u)\,d(-u)=-\int g(u)\,du=-G(u)=-G(-x). $$ All this says is that the antiderivative of an even function is an odd function, not that it is zero. So, there is no inconsistency.

A second perspective is provided by the area interpretation of the integral. Thus, the area under the graph of an even function $g$, say, from the origin to an arbitrary point $x$ is $$ A=\int^{x}_{0}g(s)\,ds=\int^{-x}_{0}g(-u)\,d(-u)=-\int^{-x}_{0}g(u)\,du=\int^{0}_{-x}g(u)\,du, $$ Here, $s$ is a genuine dummy variable and can be substituted by any other symbol, like $u=-s$ in the above. Now this equation says that the area is symmetric with respect to the vertical axis. Thus, the integral (area) to the left of the vertical axis equals to the area on the right, and not its negative. Further peace of mind comes from applying the fundamental theorem of calculus in its more precise form $$ \frac{d}{dx} \int _a ^x f(s)\, ds = f(x) $$ to obtain $$ g(x)=\frac{d}{dx}\int^{x}_{0}g(s)\,ds=\frac{d}{dx}\int^{-x}_{0}g(-u)\,d(-u)=-\frac{d}{dx}\int^{-x}_{0}g(u)\,du=g(-x), $$ which indeed confirms that $g$ is an even function.

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