I'm reading this text:

enter image description here

I'm a bit lost about why the sentence is in bold. Why is the word 'despite' being used? I can see that the derivative of $\frac{-1}{x} = \frac{1}{x^2}$, but why is this explained with the word 'despite'? What is the point?

Even more so, what is the point of learning about indefinite integrals?

  • 3
    $\begingroup$ The Fundamental Theorem of the Calculus, which is quite a bit about indefinite integrals, is responsible for the modern technological age. $\endgroup$ Apr 11, 2018 at 17:18
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    $\begingroup$ Your title has basically nothing to do with your question. $\endgroup$
    – zhw.
    Apr 11, 2018 at 17:19
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    $\begingroup$ The clarification is due to the fact that the collection of functions $-\frac{1}{x}+C$, if taken defined on all of $\{x\neq 0\}$, doesn't contain all antiderivatives (indefinite integrals) of $1/x^2$. The full collection is the $F$ they defined. $\endgroup$
    – user550675
    Apr 11, 2018 at 17:31
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    $\begingroup$ @zhw. I've edited the title to match the query in the post's body. $\endgroup$
    – ryang
    Oct 3, 2023 at 6:55

2 Answers 2


As pompelle mentioned in the comments, the issue is that the set of functions

$$ -\frac1x+C $$

don't contain all the functions whose derivative is $1/x^2$. It is true that the exemplar is simply $-1/x$, and of course a constant can always be added, but because $-1/x$ is in two disconnected pieces (each of which is continuous), different constants can be applied to each piece.

What is being emphasized is that by convention, we don't have to write

$$ \begin{cases} -\frac1x+C_1 & x < 0 \\ -\frac1x+C_2 & x > 0 \end{cases} $$

Instead, for the sake of brevity, we write just

$$ -\frac1x+C $$

with the provision that each separate piece of the function in question can have a separate constant added to it.

ETA: Here are some pictures to depict what I mean. If you just write

$$ -\frac1x+C $$

then strictly speaking, you only get $-1/x$, along with all the functions that are $-1/x$ shifted up or down by a constant amount across the entire $x$-axis, as shown below:

enter image description here

If, on the other hand, you write

$$ \begin{cases} -\frac1x+C_1 & x < 0 \\ -\frac1x+C_2 & x > 0 \end{cases} $$

you get all of the above, plus in addition, you get those functions where the two halves of $-1/x$ are shifted up or down by different amounts, as shown below:

enter image description here

What the book is saying is that they're engaging in a slight abuse of notation whereby the first expression is understood to mean the same thing as the second, larger (and more rigorously correct) expression.

  • $\begingroup$ If we added the same constant to each piece, isn't that valid? I guess I still don't see how the disconnect makes it so that $\frac{-1}{x}$ isn't the antiderivative of f(x) or the function in the integral. $\endgroup$
    – Jwan622
    Apr 11, 2018 at 18:11
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    $\begingroup$ @Jwan622: It's not that it isn't valid; it's that there are plenty of other functions that are also valid, but aren't strictly speaking captured by the expression $-1/x+C$, because that expression imposes the same constant over both branches of $-1/x$. For instance, a valid function would be $f(x) = -1/x$ for $x < 0$, but $1-1/x$ for $x > 0$. That would correspond to $C_1 = 0, C_2 = 1$ for the complete, branched expression, but there's no single value of $C$ in the expression $-1/x+C$ that covers that example. $\endgroup$
    – Brian Tung
    Apr 11, 2018 at 19:45
  • $\begingroup$ @Jwan622: I've added diagrams that will hopefully make it easier to see what I mean. $\endgroup$
    – Brian Tung
    Apr 11, 2018 at 19:57

"[...] when a formula for an indefinite integral is given, it is valid only on an interval."

The textbook you are using is warning you that the formulas they provide for the specific antiderivative $F(x)=\int f(x)\, dx$ of a specific function $f(x)$ is only defined (and thus is only true) on the interval $I$ for which $f(x)$ and $F(x)$ are defined.

For the square root function: $$f(x)=\sqrt{x}$$ Defined on: $$I=[0,+\infty)\Longleftrightarrow x\geq0$$ Then the book might provide you with the formula:

$$F(x)=\int \sqrt{x}\,dx=\frac{2}{3}\sqrt{x^3}+C$$ Which is also only defined on $I$, meaning only for $x\geq 0$.

[...] this is true despite the fact that the general antiderivative [...]

In their example of $f(x)=\frac{1}{x^2}$, this function is defined on two separate intervals $I_0=(-\infty,0)$ and $I_1=(0,+\infty)$.

And thus, $f(x)=\frac{1}{x^2}$ can be integrated on each of these intervals, resulting in two antidervatives $F_0(x)=-\frac{1}{x}+C_0$ and $F_1(x)=-\frac{1}{x}+C_1$.

They add, despite this, and despite the fact that we cannot say that $C_0=C_1=C$, because these constants can have different values, we still, for the sake of simplicity and quick application, loosely accept the form $F(x)=-\frac{1}{x}+C$.

They are simply making these two points for mathematical rigor. Therefore, they adopt a convention for simplicity and warn the reader.

Even more so, what is the point of learning about indefinite integrals?

The antiderivative's importance is too profound to summarize.

Obviously, it allows you to "reverse", loosely speaking, a derivative. The antiderivative $F$ of $f$ is the function whose derivative is $f$. It allows you to analyze functions as well, just like the derivative does. This is basically what you do in Calculus and Mathematical Analysis.

This has further importance in Differential Equations; solving equations like $f(x)+f'(x)=1$ that describe how $f$ and its rate of change $f'$ behave. Here, you will use the antidervative to try and find a function $f$ which verifies this equation. Differential equations are used in any field describing quantities changing. (mathematics, sciences, engineering, economics, ...)

The antiderivative allows you, in Geometry, to calculate areas relative to curves (under, between, ...); to find surfaces and volumes of shapes even beyond 3-dimensions. Using the integral you can understand where most of the formulas for geometrical shapes (circle, sphere, cube, ...) came from.

It is also used in Probability and Statistics, where you can find the probability of a random variable by integrating it under a probability density function. You can also calculate the mean/average of a function by integrating it in a special way.

And so forth.

As you can see, it helps you solve equations, find expressions, create new functions (non-elementary functions which you cannot write without the integral sign).

Keep in mind, these are all still symbolic solutions. You can extend the power of the antidervative by evaluating it over a range of values, giving you numerical solutions. For any symbolic problem we cannot solve (yet or ever), we can usually get a numerical approximation for it, which opens a lot of possibilities. All thanks to our friend the integral operator.

You can do a few online searches to see the myriad of applications of the integral.

All in all, it elevates your mathematical maturity beyond basic algebra.

Hope this helped.

  • $\begingroup$ Why are the constants not equal? For F0 and F1 why did we label the constants differently? What's the signifcant of this? I thought the constants were just to represent the range of all constant values. Isn't this set of values the same regardless of whether x is > or < 0? $\endgroup$
    – Jwan622
    Apr 11, 2018 at 18:14
  • $\begingroup$ They are each arbitrary constants. Whenever you integrate you add an arbitrary constant, which is any constant. But it does not mean they are the same, it means each of them can be any value within the domain. The set of values is the same, but not these two instances of the set. They can be two different values regardless. $\endgroup$
    – ex.nihil
    Apr 11, 2018 at 18:17
  • $\begingroup$ To simply disprove what you are thinking, if what you say is true, then all constants are the same in any integration, which is clearly not the case. $\endgroup$
    – ex.nihil
    Apr 11, 2018 at 18:18

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