# Antiderivatives and the indefinite integral

Thomas Calculus defines the indefinite integral to be the collection of all antiderivatives.

The indefinite integral is mathematically defined as, $$y=\int_{a}^{x} f(t)dt$$

The second part of the Fundamental Theorem of Calculus states that, $$y=G(x)-G(a)$$ where $G$ is the antiderivative of $f$ and $y$ is the indefinite integral defined above.

Here, the indefinite integral $y$ is given in terms of $G(x)+C$, where $C=-G(a)$. We see that $G(a)$ is a single value because $a$ is a fixed constant. This tells me that $y=G(x)-G(a)$ is just ONE curve and not a family of curves( collection of antiderivatives ). My understanding here seems to contradict the contents of the book.

Where could I have gone wrong with my understanding?

Edit: The definition of the indefinite integral comes from UC Davis Mathematics

• Keep in mind that different authors may use different definitions. This may be a case of one author's preference versus another's. – Matthew Leingang Jun 18 '17 at 12:36
• Yes, it seems that way. – Andre Jun 18 '17 at 12:48
• Different definitions of the indefinite integral lead us into a confusion, do they not? – R004 Jun 18 '17 at 12:59
• True, but each author will have their own reasons for their definitions. By the time mathematics gets into textbooks it's pretty well settled as to what the best definitions are, but there will always be edge cases. That's one of the reasons I recommend students read from a single text at a time, or at least exercise care when googling for resources. – Matthew Leingang Jun 19 '17 at 0:24

The definite integral $\int_a^b f(x) dx$ can be expressed, via antiderivatives and the fundamental theorem as $$\int_a^b f(x) dx = G(b) - G(a).$$ Here you have specified from where to where you want to integrate, making the integral definite. On the other hand, the indefinite integral is the collection of all these integrals, for all possible limit points. So once you make a choice about the integration limits, everything is perfectly well-defined and single-valued, but if you do not do that, you can only determine the (indefinite) integral up to a constant: $$\int f(x) dx = G + C$$ Since no matter which number $C$ you plug into this equation, the derivative of the right-hand side will always be just $f(x)$. In this way, the indefinite integral is really a collection of functions: $$\int f(x)dx = \{ G + C ~|~ C \in \mathbb{R}\}.$$ But for the matter of convenience, we leave away the brackets etc and write the integral as if it was a function with an additive, unspecified constant. So I think your misunderstanding lies within the distinction of indefinite (no limits) and definite (with limits) integrals.

Think of it like this:

1. A definite integral has the form $$\int_a^b f(x) dx = G(b) - G(a),$$ where $a, b$ are fixed real numbers. The result will be a real number.

2. Now you allow one of them, say $b$, to be variable. This then gives you an integral FUNCTION of the form $$I_a (x) = \int_a^x f(t) dt = G(x) - G(a).$$ Two words on notation: I changed the integration variable from $x$ to $t$ to avoid confusion as $x$ now denotes the upper limit of the integral. Furthermore, since the integral function depends on the lower limit $a$ (different $a$'s will give different functions), I stressed this dependence by writing $I_a$ for the function.

3. Now if you let $a$ remain unspecified, you will get a set of functions which is the indefinite integral. So it is a set as written above, and once we specify the lower bound $a$ and hence the constant $C = -G(a)$ we get one particular function $I_a$. If we then, finally, ask for the value of this function in a point $b$, we get the definite integral from $a$ to $b$ which is a number.

So: a definite integral is a number, an integral function is a function and an indefinite integral is a collection (or set) of functions.

So in the two sources that you provide, the authors each use a different definition for "indefinite integral". The first one uses the definition of step 3 above whereas the second one chose to stop at step 2.

Let me give you an example: Consider the function $$f(x) = x^2 + 1.$$ An antiderivative is given by $$G(x) = \frac{1}{3} x^{3} + x$$ but also for any $C \in \mathbb{R}$ by $$G_{C}(x) = \frac{1}{3} x^{3} + x + C = G(x) + C.$$ Now to evaluate the integral with known bounds is relatively straightforward, you just have to plug the values of one of the antiderivatives into the formula (it doesn't matter which one since you subtract! So the extra constant $C$ will cancel in the evaluation). You could, for example, calculate $$\int_{3}^{6} f(t) dt = G(6) - G(3) = G(6) - 12 = 75 - 12 = 63.$$ Now let us fix the lower bound, i.e. we let $a = 3$, and vary the upper bound, i.e. $b = x$. Then we will get a function $$I_{3}(x) := \int_{3}^{x} f(t) dt = G(x) - G(3) = G(x) - 12 = \frac{1}{3} x^{3} + x - 12.$$ As you can see when looking at the right-hand side of the equation, this is a perfectly normal function and you can perform all kinds of calculations with it. This function can be thought of as a special case of the function $G_{C}$ defined above where $C = - 12$. If we now choose to let our integrals start at a different lower bound, say $a = 1$, we will get a related, but different function: $$I_{1}(x) := \int_{1}^{x} f(t) dt = G(x) - G(1) = G(x) - (\frac{1}{3} + 1) = \frac{1}{3} x^{3} + x - \frac{4}{3}.$$ This corresponds to $G_{\frac{4}{3}}$ in the above notation. So you see that in this way you really get a collection of functions, via the process described in steps 1-3 above, but you can stop this process at any of the steps and it will be clear how to get to the previous/next step.

Hope that helps.

Andre

• UC Davis Mathematics defines the indefinite integral in a way I have put forward. The definition is reasonable because $x$ in the limits of integration is a variable, thereby suggesting that the area need not be definite. – R004 Jun 18 '17 at 11:55
• You can access the paper here at google.com/url?sa=t&source=web&rct=j&url=https://… – R004 Jun 18 '17 at 11:56
• While the definite integral gives you a number, the indefinite integral gives you a function of $x$. – R004 Jun 18 '17 at 11:58
• What you have defined in your post, i.e. the integral with fixed lower limit but variable upper limit is what is sometimes called the "integral function" (at least in German, I actually don't know the name of this concept in English) of $f$. It has to be distinguished from the indefinite integral, which is an integral where NONE of its limits is fixed. – Andre Jun 18 '17 at 12:14
• I rewrote and expanded my answer to provide much greater clarity. I hope that it is understandable now :-) – Andre Jun 18 '17 at 12:30