I'm confused about the two parts of the Fundamental Theorem of Calculus, as I feel these two parts are somewhat contradictory?
The first part states that:$$ F(x)=\int_{a}^{x}f(t)dt $$
Whereas the second part states that:$$ F(b)-F(a)=\int_{a}^{b}f(t)dt $$
Well if the second part is true in evaluating the integral, then shouldn't the first part be:$$ F(x)-F(a)=\int_{a}^{x}f(t)dt $$
And this is how it works. Like, if I'm evaluating x=2 then I would calculate the area from a to 2, and use F(2)-F(a), for some value a.
So what exactly is the first part of the fundamental theorem of calculus telling us? Looking at the second part, I can't wrap my head on the purpose of the first part, and even to me it seems wrong, that it should be written as:$$ F(x)-F(a)=\int_{a}^{x}f(t)dt $$
$\endgroup$ 74 Answers
$\begingroup$What you call the “First part” is usually called the second part. And it doesn’t state what you say. What is states is that if $f(t)$ is continuous on $[a,b]$, and we define the function $F(x)$ by the formula$$F(x) = \int_a^x f(t)\,dt$$then $F(x)$ is differentiable, and is in fact an antiderivative for $f$; that is, that$$\frac{d}{dx} F(x) = \frac{d}{dx}\int_a^x f(t)\,dt = f(x).$$That’s what that part states.
What you call the second part, which is usually called the first part, also doesn’t quite say what you are writing. What it says is that if $f(x)$ is continuous and $G(x)$ is any antiderivative for $f(x)$ on $[a,b]$, then$$\int_a^b f(t)\,dt = G(b)-G(a).$$
Now, you could try to use the funtion $F(x)$ in your “first part”, but it doesn’t actually give you anything useful: it just says$$\begin{align*} \int_a^b f(t),dt &= F(b)-F(a)\\ &= \int_a^b f(t)\,dt - \int_a^a f(t)\,dt\\ &= \int_a^b f(t)\,dt \end{align*}$$which doesn’t really give you a way to compute the integral.
The “Second part” (usually called the First Part) says:
If you have an antiderivative of the continuous function $f(x)$, then you can use the antiderivative to calculate the integral $\int_a^b f(t)\,dt$
(though implicit is that you need to be able to compute values of the antiderivative independently of the integral...)
The “First part” (usually called the Second Part) says:
If $f(x)$ is continuous, then it definitely has (at least one) antiderivative.
This tells you that the first part is not an empty promise: if you can find a (useful, independently calculable) antiderivative, then you are set, and antiderivatives exist to be found. However, the antiderivative it shows you is not a useful antiderivative as far as the first part is concerned.
$\endgroup$ 11 $\begingroup$The $F$ or the "first part" is actually not the same $F$ as the $F$ of the second part.
...
The gyst of the FTC is that integration and derivation are inverses of each other.
The subtlety in putting this in plain words is distinguishing between "derivatives" as general functions vs. derivatives as specific values about the function at certain values of $x$, and distinguishing between "integrals" as a general indefinite function, vs. a definite integral evaluated between two specific points.
The first part of the theorem is trying to basically say if $F(x) = \int f(x)dx$ then $F'(x) = f(x)$. ... Except $F(x) = \int f(x)dx$ doesn't actually make any sense.
So if $f$ is an intergratable function, and we define a function as $F(x) = \int_a^x f(t)dt$, that is, $F(x)$ for any specific value of $x$ would be evaluated as the integral evaluated from some fixed point $a$ to $x$ were $x$ treated as a specific value.
Then the theorem states $F'(x) = f(x)$.
In other words: the derivative of the integral is the function itself.
....
The "second" part is the other way around. The integral of the derivative is the function itself.
If we start with a function $\mathscr F$ and we know that $\mathscr F' = f$. Then if you integrate $f$ between two points $a,b$ we get $\int_a^b f(x)dx =\mathscr F(x)|_a^b = \mathscr F(b) -\mathscr F(a)$
....
If it were up to me, I'd define the FTC as this.
If $f$ is integratable and we define $G(x) = \int_a^x f(t)dt$ then $G'(x) = f(x)$. (The derivative of the integral is the function.)
ANd if $f$ is differentiable, then $\int_a^b f'(x)dx = f(x)|_a^b = f(b) - f(a)$. (The integral of the derivative is the function.)
$\endgroup$ 1 $\begingroup$One of the fundamental theorems of calculus states that the function $F$ defined by $$ \tag{1} F(x) = \int_a^x f(t) \, dt $$ is an antiderivative of $f$ (assuming that $f$ is continuous).
Since $F$ is an antiderivative of $f$, you are correct to note that the other fundamental theorem of calculus implies that $$ \tag{2} \int_a^x f(t) \, dt = F(x)- F(a). $$ But this does not contradict (1), of course, because $F(a) = 0$.
$\endgroup$ 6 $\begingroup$You have the order of the theorems backwards. Also, they are not referring to the exact same functions.
The first fundamental theorem of calculus states that, if $f$ is continuous on the closed interval $[a..b]$ and $F$ is the indefinite integral of $f$ on $[a..b]$, then $$\int_a^b f(x)~\mathrm d x=F(b)-F(a)\tag 1$$
This refers to an $F$ which is an indefinite integral of a given function $f$.
The second fundamental theorem of calculus holds for $f$, a continuous function on an open interval $I$, and $a$ any point in $I$, and states that if $F$ is defined by $$F(x):=\int_a^x f(t)\mathrm d t\tag 2$$
then $$F'(x)=f(x)\tag 3$$
at each point $x$ in $I$.
This refers to an $F$ which is a definite integral of function $f$ over $[a..x]$.
I usually recall it as simply:
$$f(x)=\dfrac{\mathrm d ~~}{\mathrm d x}\int_a^x f(t)~\mathrm d t$$
$\endgroup$ 3
