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Can, $$\int_{}^{l}f(t)dt$$ Where $l$ is a function of $x$, be considered as another representation of an indefinite integral?

Let us have the integral function, $y(l)=\int_{a(x) }^{l}f(t)dt$, where $l$ is the argument that takes functions of $x$, to be continuous and differentiable in an interval $I$.

$$y'(l) =l'.f(l)-a'(x).f(a(x))$$

The antiderivative of its derivative is, say, $$g(l)=y(l)+C$$

If the lower limit were specified as $a(x)$, the second part of the Fundamental Theorem of Calculus would give us, $$y(l)=g(l)−g(a(x))$$

$g$ being the antiderivative of $f$. Here in my representation, we know not what $a(x)$ is. Might as well write $y(l)=g(l)−C=g(l)+K$, where $C$ can be any constant.

We know that, an indefinite integral is expressed as a function of $x$ plus a constant. Here, $y(l)=\int_{}^{l}f(t)dt$, seems like an appropriate representation of the indefinite integral. I find this approach comforting because it helps me accept the integral notation used in the definition of the indefinite integral. Could I be right about this?

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    $\begingroup$ You can do what you like, as long as you justify its use and define it correctly. Don't expect others to conform though ;) $\endgroup$ – Edward Evans Jun 22 '17 at 2:56
  • $\begingroup$ I fear I could be living with a wrong understanding. $\endgroup$ – R004 Jun 22 '17 at 3:01
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When an integral has bounds, it means that we are integrating over a region. In Calc 3 with multiple integrals we have regions that are purely functions. So when you have just one bound like your notation suggests it doesn't make too much sense with the integral notation itself. Consider the following $\ln x=\int_{1}^{x}\frac{1}{t}\,\mathrm{d}t$. As a representation for $\ln x$, we can express functions like this, however notice we have both bounds. Of course, I am guilty of making up notation for convenience like $\ln^2 x$ instead of $(\ln x)^2$ it is purely up to the worker. However, it is customary to see one bound, the lower bound that describes the region such as $\displaystyle\oint\limits_{\mathbb{R}}\frac{1}{1+x^2}\,\mathrm{d}x$, then again, that is a definite integral. So what this amounts to is essentially no, due to the second fundamental theorem of calculus $\int_{a}^{b}f'(x)dx=f(b)-f(a)$. 2 bounds or a subscript that describes the region that will be integrated over. Hope this helps. (PS also we don't need to know what a(x) other than that it is a function of x)

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  • $\begingroup$ So, why do they have the integral notation for the collection of antiderivatives called the indefinite integral? $\endgroup$ – R004 Jun 22 '17 at 3:14
  • $\begingroup$ For definite integrals, for the beauty of solving it all, for the art it sometimes takes to work out a long integral. $\endgroup$ – Teh Rod Jun 22 '17 at 3:22
  • $\begingroup$ In one of the paper sets put forward by UC Davis Mathematics, they stated that $\int_{a}^{x}f(t)dt$ is an indefinite integral. This made me think. This does not take me away from my question posted above. I really want to know. $\endgroup$ – R004 Jun 22 '17 at 3:23
  • $\begingroup$ @R004 I have not seen that, I don't think I'm wrong. Bounds means definite. Indefinite means no bounds and a constant at the end. Do you have a link to this? $\endgroup$ – Teh Rod Jun 22 '17 at 3:24
  • $\begingroup$ here, in the opening lines of the paper. $\endgroup$ – R004 Jun 22 '17 at 3:43

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