Misunderstandings of the fundamental theorem of calculus?

Question

One part of the fundamental theorem of calculus is that $$F(x)=\int_a^x f(t)\;dt\tag1$$ However, $$\int_a^b f(t)\;dt=F(b)-F(a)\tag2$$ So my first question is why doesn’t equation 1 take the form of $\int_a^x f(t)\;dt=F(x)-F(a)$? Where did $F(a)$ disappear to?

Also, whenever you see an integral in the form of $F(x)=\int_x^a f(t)\;dt$, why is it that you must always change it to $F(x)=-\int_a^x f(t)\;dt$? That is, why is it necessary for $x$ to be the upper limit and not the lower? I know that it’s written in the fundamental theorem of calculus as the upper limit, but why?

Answer 1

You haven't actually stated the theorem. This is where being precise is important.

The first part says:

If $f$ is continuous on $[a,b]$ then the function DEFINED by:

$$F(x)\colon = \int_a^x f(t) dt$$

for $x\in [a,b]$

is differentiable on $(a,b)$ and $F'(x)=f(x)$.

The second part says:

If $F(x)$ is differentiable on $(a,b)$ with derivative $f(x)$ and $f(x)$ is continuous on $[a,b]$ then:

$$\int_a^b f(t) dt = F(b)-F(a)$$

Just stating theorems precisely will help clear up misconceptions.

Answer 2

It seems the underlying difficulty is that due to the coincidence of some symbols used in the two parts of the Fundamental Theorem of Calculus in your textbook, you are linking things together that are not really meant to be linked in that way.

The explanation of the FTC on the MathWorld site does not even have two "parts"--it presents two separate theorems. The "first" theorem says:

If $f$ is continuous on the closed interval $[a,b]$ and $F$ is the indefinite integral of $f$ on $[a,b],$ then $$\int_a^bf(x)\,dx=F(b)-F(a).$$

The "second" theorem (according to MathWorld) says (paraphrasing slightly) that

If $f$ is a continuous function on an open interval $I$ and $a$ is any point in $I$, and if $F$ is defined by $$F(x)=\int_a^xf(t)\,dt, $$ then $F'(x)=f(x) $ at each point in I.

You have not quoted your textbook verbatim, but from the sequence of formulas in your question it would appear that MathWorld's "second" theorem is the first "part" of your book's theorem, and vice versa. This is all fine, because the two theorems (or two parts of "the" theorem) are logically independent of each other--neither one assumes the other as part of its statement.

In particular, all the names of functions and variables in either part of "the" theorem are tied only to functions and variables of the same names in the same part of "the" theorem. We could just as well write:

If $g$ is continuous on the closed interval $[c,d]$ and $G$ is the indefinite integral of $g$ on $[c,d],$ then $$\int_c^d g(x)\,dx=G(d)-G(c).$$

while leaving all the letters $f$, $I$, $a$, $F$, $x$, and $t$ unchanged in the other part of "the" theorem. You don't have to assume the two parts are talking about the same functions and numbers at all.

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To put it another way, the two parts of "the" FTC are doing two separate jobs. One part of "the" FTC tells you how to compute definite integrals of a function if you happen to know an indefinite integral (antiderivative) of the same function. The other part constructs an antiderivative of a function from a particular definite integral of the same function.

Answer 3

There is no inconsistency, because $F(a)=0$ for the $F$ defined in equation (1).

In fact (1) implies (2) whether or not $F$ (constructed as an integral) is an antiderivative of $f(x)$. This is the property $\int_a^b f + \int_b^c f = \int_a^c f$ and it holds in cases where $f$ is not everywhere continuous and $\int_a^x f$ is not everywhere differentiable.

Answer 4

To answer your questions in order:

1) We do not write $\int_a^x f(t)\,dt = F(x) - F(a)$ because $F(x)$ already includes it. As other folks have pointed out, here when we write $F$ we mean any antiderivative of $f$. The value $F(a)$ is constant, so if you subtract it off the derivative will remain the same.

2) Mathematically, there is no reason you have to do it that way. You could just as well write the theorem like this:

$$ -F(x) = \int_x^a f(t)\,dt $$

We don't because it introduces that negative sign, and we find it simpler the other way. So you might ask: why does putting $x$ on the bottom create a negative sign?

I would answer you like this: think of the value on the right as the area under the graph of $f(t)$ on the interval $[x, a]$. We're taking the derivative, so we ask: what's the rate of change of the area as $x$ increases?

Answer 5

If I may start from avid19's more precise wording of the theorem:

If $f$ is continuous on $[a,b]$ then the function DEFINED by:

$$F(x)\colon = \int_a^x f(t) dt$$

for $x\in [a,b]$

is differentiable on $(a,b)$ and $F'(x)=f(x)$.

What this is doing is defining the meaning of the integral sign. It does so using a 'f' for the function and 'F' for the integral of that function. However, the last line is the most important part. It states that the derivative of F is equal to f. If I smoosh the functions together, the theorem states: $$ \frac{d}{dx}_{x=y}(F(x)) = f(y)$$ $$ \frac{d}{dx}_{x=y}\left(\int_a^x f(t) dt\right) = f(y)$$

Note that I added an extra variable for clarity. Since the x's had multiple meanings, I changed it out so that 'x' is now just the free variable used in the derivative and the integral, while y is the value you're evaluating the functions at.

Now its easy to show that there are an infinite number of functions which all have the same derivative (just add a constant, and you get a parallel function with the same derivative). This first half of the theorem deals with the special case where you're taking the derivative with respect to the upper bound of the integral. In this case, it doesn't matter which of those curves you use, because they all have the same derivative. However, in the second half of the equation, we are interested in the case where both upper and lower bounds are fixed numbers:

$$\int_a^b f(t) dt = F(b)-F(a)$$

In this case, by defining the integral in this way, it doesn't actually matter which of the infinite number of paralell functions you choose for F, the subtraction will cancel out whatever constants you might have added.

You need both halves to complete the definition. If you only have the first half, it means you can only operate on integrals in the case where you get to take the derivative of it. If you only have the last half, you show how to operate on integrals in the general case, but you don't have enough information to really define what their computed value actually is. Put the two halves together, and you get something really powerful!

Answer 6

In your equation (1), what is $F(a)$? Or more carefully, what is $\lim\limits_{x \rightarrow a} \int_a^x f(x) \,\mathrm{d}x$? (Since you're not declaring your $f$ to be differentiable or continuous over certain intervals, I'm using the same level of assumption.) Is there any point in subtracting that number from $F(x)$?

(As others have pointed out, you have put the cart in front of the horse: We wouldn't define $F(x) - F(a) = (\text{stuff})$, we'd just define $F(x)$, as was done. If we were to want to define something like $F(x) - F(a)$, we would instead write $F(x) = F(a) + (\text{stuff})$. However, you should see that given your quoted definition of $F(x)$, there's nothing gained or lost by inserting $F(a)$ anywhere.)

For your second question. Picture some function, positive near $x$ and consider the area between the graph of the function and the $x$-axis between $a$ and $x$ for $a < x$. This means we will be changing the right endpoint of the interval. If we increase $x$ slightly, the area increases. If we decrease $x$ slightly, the area decreases. Changing to the language of differentiation (which is what this half of the theorem is explaining): if your derivative is positive, moving to the right, you go up and moving to the left, you go down. For completeness, we should go back to the top of this paragraph and talk about a function negative near $x$ and conclude the relationship between the area decreases/increases as $x$ moves to the right/left respectively, is equivalent to the function value decreases/increases as $x$ moves to the right/left, respectively, when the derivative is negative.

Now repeat the thought experiment, but with $a > x$, so that we're moving the left endpoint. When the function is positive near $x$ and we move $x$ a little to the right, the area between $x$ and $a$ decreases because we've moved $x$ towards $a$. If a little to the left, the area increases. This is exactly backwards from putting $x$ on the right edge of the interval. And if the function is negative near $x$, again the behaviour is backwards. To get the values we expect, we have to have the negative sign you ask about.