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Scenario 1:

$$f(x):=\int{\cos(x)dx}\tag{1}$$

Scenario 2:

$$g(x):=\int_{a}^{x}{\cos(t)dt}\tag{2}\\ [\text{edited after seeing @hardmath's comment}]\\ [\text{a is a constant}]$$

Comments:

Both $(1)$ and $(2)$ are functions. An antiderivative/indefinite integral is a function, while a definite integral is a number. So,

$$f(x)=g(x)=\sin(x)+c\tag{3}\\ [\text{c is a constant}]$$

Is line $(3)$ correct? Are both $(1)$ & $(2)$ antiderivatives/indefinite integrals?

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    $\begingroup$ (2) should be $\int_a^x\cos(t)dt$. $\endgroup$ Commented Jan 10, 2022 at 17:16
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    $\begingroup$ Your second integral is an "abuse of notation" because the variable $x$ that appears inside the integral gets reused as a limit of integration (and as the argument to $g(x)$). It is better to replace the variable of integration with a "dummy variable" such as: $$ g(x) := \int_a^x \cos(t) \mathrm{d}t $$ $\endgroup$ Commented Jan 10, 2022 at 17:17
  • $\begingroup$ @hardmath yes you're right sir; my bad, should I edit to correct it? $\endgroup$ Commented Jan 10, 2022 at 19:02
  • $\begingroup$ No, I just wanted to comment (or as some might say, "nitpick"). I think the Question makes sense, and it got a good Answer from Greg. I hope you saw the (closely) Related link (shown to the right on the desktop site): Textbooks that use notation ... for "indefinite integrals." $\endgroup$ Commented Jan 10, 2022 at 19:23
  • $\begingroup$ @hardmath Sir, I'm confused about this question still, so I posted another one. If you want, you can check it out. $\endgroup$ Commented Jan 14, 2022 at 6:12

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Not quite. (1) is definitely (no pun intended) an indefinite integral. Note that evaluating an indefinite integral does not actually lead to a function, but rather to a family of functions differing by a constant. In the case of (1), we have $f(x) = \sin x + C$ (a family of functions).

(2) is simply a function for any given $a$: by the Fundamental Theorem of Calculus, $g(x) = \sin x - \sin a$. Note that this really is one function, not a family of functions. (If one takes a family of parameters $a$, then that would result in a family of functions $g(x) = g_a(x)$; but it still wouldn't result in every possible function in the family from (1), since $\sin a$ is bounded.)

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  • $\begingroup$ I see. I had this concept from an earlier question. Please see EricTowers' 2nd comment kind sir. It seems like you are disagreeing with him. I'm confused. $\endgroup$ Commented Jan 10, 2022 at 6:54
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    $\begingroup$ A definite integral can have constants as endpoints, in which case it results in a numerical answer, or it can have variables as endpoints, in which case it results in an answer that is a function of those variables. $\endgroup$ Commented Jan 10, 2022 at 8:13
  • $\begingroup$ Thanks for the clarification kind sir. I had another question: $\int{\cos xdx}=\sin x+c$. $\sin x+c$ is an anti-derivative/indefinite integral. Now, c is a constant, and it could have any of the following values $c=\pi, \frac{1}{3}, 4, 5,3,-3,...$. Now, if I specify the value of $c$ ($c=3$ for example), will $\sin x+3$ still be called an indefinite integral? $\endgroup$ Commented Jan 10, 2022 at 10:47
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    $\begingroup$ "Antiderivative" and "indefinite integral" are not synonyms. An indefinite integral is literally an integral, and evaluating an indefinite integral leads to the set of all antiderivatives of the integrand (the set of all functions whose derivative equals the integrand). So: $\int \cos x\,dx$ is an indefinite integral; $\sin x + C$ is the set of all antiderivatives of $\cos x$; and $\sin x + 3$ is one antiderivative of $\cos x$. $\endgroup$ Commented Jan 10, 2022 at 17:11
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    $\begingroup$ You have now met a calculus instructor who does not treat them as synonymous. Note that I would never fault a student who used those terms interchangeably in practice—but the OP has asked several questions about fine distinctions in the terminology. $\sin x+C$ is not an indefinite integral—it's not an integral of any sort. $\int \cos x\,dx$ is not an antiderivative—it's not a function of any sort. By the way, your claim that "indefinite integral" is not legitimate mathematical terminology and that no mathematican uses it is demonstrably false. $\endgroup$ Commented Jan 11, 2022 at 20:50

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