Is representing indefinite integration as definite integration with the independent variable accurate?

Question

Say we want to indefinitely integrate the function $f(x) = x^2$.

The usual way I would do it would be indefinite integration:

$ \int{dxf(x)} =\int{dx^2} =\frac{1}{3}x^3 + C $

However, I've seen a few people do it a different (interesting) way, using definite integration, which is:

$ \int_{0}^{x}{dtf(t)} =\int_{0}^{x}{dtt^2} =[\frac{1}{3}t^3 + C]^{x}_{0} =(\frac{1}{3}(x)^3 + C) - (\frac{1}{3}(0)^3 + C) =\frac{1}{3}x^3 $

Abstractly, the second method sort of makes sense to work. The problem I have here is that the integration constant has been lost in the second method. Where have I made an incorrect assumption with the second method?

Answer 1

Say we want to indefinitely integrate the function $f(x) = x^2$. I've seen a few people do it using definite integration: \begin{align} \int_{0}^{x}{f(t)}\,\mathrm dt =\int_{0}^{x}{t^2}\,\mathrm dt\\ =[\frac{1}{3}t^3 + C]^{x}_{0}\tag1\\ =(\frac{1}{3}(x)^3 + C) - (\frac{1}{3}(0)^3 + C) =\frac{1}{3}x^3 \end{align} The problem I have here is that the integration constant has been lost

1. It's not that the parameter $C$ was lost but that it should never have been there in step $(1),$ which is invoking the Fundamental Theorem of Calculus, which technically asks for a particular antiderivative (that is, an indefinite integral whose $C$ has been instantiated with a value—typically $0$ but any value is fine) to be plugged in.

2. Since an indefinite integral is the general specification of its integrand's infinitely many antiderivatives, instantiating an indefinite integral simply returns one of the antiderivatives. Instantiating an indefinite integral does not create a definite integral: the latter denotes a signed area that varies with the inputs $a$ and $b,$ whereas the former, and every antiderivative, varies with the input $x.$

3. Why does $\displaystyle∫^x_0f(t)\,\mathrm dt$ not contain the parameter $C?$

   As this object is literally a definite integral (albeit now varying with only $x$ rather than $a$ and $b$), it contains no arbitrary constant $C.$

   Simultaneously, if its integrand $f$ is continuous on $[0,x],$ then, by the Fundamental Theorem of Calculus, this object is an (one particular) antiderivative of $f,$ which makes it not a family of functions, that is, not an indefinite integral. To be clear: an antiderivative of a function $f$ is literally a function whose derivative is $f,$ and the FTC says that $\displaystyle\frac{\mathrm d}{\mathrm dx}\int_0^xf(t)\,\mathrm dt=f(x).$

4. This point will be initially confusing, but ultimately illuminating: a definite integral isn't technically a particular instance of an indefinite integral: for example, for \begin{align}f(x)&= \begin{cases} 0, &x\in[-1,0)\cup(0,1];\\ 1, &x=0,\end{cases}\end{align}$\displaystyle\int_{-1}^1 f(x)\,\mathrm dx=0$ even while $f$ has no antiderivative and so no indefinite integral.

5. Say we want to indefinitely integrate the function $f(x) = x^2$. I've seen a few people do it using definite integration

   To “indefinitely integrate” $f$ (that is, to antidifferentiate $f$) by (definite-)integrating $f$ is actually being circular: you would be antidifferentiating $f$ by (definite-)integrating $f$ by applying the FTC by antidifferentiating $f.$

Answer 2

$\int_a^x f(t) \, dt$ gives you an antiderivative for any constant $a$, and different choices of $a$ correspond to different values of $C$.