# Surprised over notation in fundamental theorem

So I'm looking into the fundamental theorem of calculus and I'm a bit weirded out by the notation used in part two.

$$\int_{a}^b f(t) dt = G(B)-G(a)$$

Why don't we say $F(b)-F(a)$? We are just talking about the anti-derivative of $a$ and $b$, right? • Part two from... where? – naslundx Mar 20 '14 at 16:19
• Part two of the fundamental theorem of calculus, as stated. – Paze Mar 20 '14 at 16:23
• Yes, but I could read about it in a thousand different places. – naslundx Mar 20 '14 at 16:24
• What is the relationship between $f$, $G$, and $F$? – wckronholm Mar 20 '14 at 16:24
• Calculus - A complete course, 7th edition. Do you want me to take a screenshot maybe? – Paze Mar 20 '14 at 16:25

$G$ is any anti-derivative. The distinction here is important because if a function $f(x)$ has one anti-derivative $F(x)$, then $G(x) = F(x) + C$ is also an anti-derivative for $f$ for any constant $C$. So, there are many, and any one of them suffices to compute the definite integral $\int_a^b f(x)\;dx$.

It is usually written with $F(b) - F(a)$ as you say, for example on Wikipedia and Mathworld.

However, while $F$ usually denotes the antiderivative of $f$, this is not neccesarily true ($f$ and $F$ may be completely unrelated depending on the context). You could specify $G' = f$ and instead use the notation you have written. But if the book hasn't (EDIT: I see now that it did), I understand your confusion.

The author defines a specific function $F(x)$ by $F(x) = \int_a^xf(x)dx$ and shows that it is an antiderivative of $f(x)$.

He does not assume or claim that it is the only antiderivative, so he doesn't say "the" antiderivative.

He then goes on to handle the case for any other antiderivative that might exist, calling it $G(x)$ just to give it a name in order to state the result.

Note that your calling $F$ "the" antiderivative is incorrect in that sense. I understand what you mean (and everybody else does too), and the result stated shows that it doesn't matter which antiderivative you choose as long as you are subtracting two values of it, which causes the arbitrary constant to disappear in the end. The real content is that all of the antiderivatives, in the end, only differ by a constant.

To elaborate on what has already been said, the notation $F(x) = \int_a^x f(t) dt$ means the following: sketch a graph of $f$, and find the area under the part of the graph that starts at an input value of $a$ and ends at an input value of $x$. The area you compute is $F(x)$.

For example, you can figure out that $\int_1^x 2t dt = x^2 - 1$ by just sketching the function and using the formula for the area of trapezoids. So in ths example, $f(x) = 2x$, and $F(x)$ in part 1 of the theorem has to be the function $F(x) = x^2-1$.

The second part of the theorem says that if you want to find $\int_a^b f(x) dx$, that is, the area under the graph of $f$ starting at an input value of $a$ and ending at an input value of $b$, you can do the following: choose any antiderivative $G(x)$, for $f(x)$, and then take $G(b) - G(a)$. The result is the area you seek to find. For example, suppose you want to find $\int_2^3 x^2 dx$. Again, $f(x) = 2x$, but you can take $G(x)$ to be $x^2 - 1$ or $x^2 + 3$ or $x^2$, since all of these functions have $2x$ as their derivative. No matter which one of these you choose, $G(3) - G(2)$ will give you the area under the graph of $f(x) = 2x$ from $x = 2$ to $x = 3$.