I do not see how the fundamental theorem of calculus establishes that integration and differentiation are inverses of each other

Question

I am trying to properly understand the first fundamental theorem of calculus but there are several little details that are giving me a lot of trouble. So I figured here I would describe my understanding of it and explain what gives me trouble.

Suppose I let the function $A(x)=\int_a^xf(t)dt$ define the area bound by a curve $f(t)$ and the lines $t=a$ and $t=x$, such that $a\leq x$, and the horizontal axis. $A(x)$ is a function of $x$. The fundamental theorem of calculus states, that $\frac{d}{dx}A(x)=\frac{d}{dx}\int_a^xf(t)dt=f(x)$.

1. We say that $A(x)$ is the antiderivative of f(x), does this mean that it is the indefinite integral of f(x)? If so, is the theorem using a definite integral of f(t) to get to un indefinite integral of $f(x)$?

2. Is $f(t)$ the same function as $f(x)$? What I mean by this, is that if $f(t)=t\times cos(2t^2-3t)$, would $f(x)=x\times cos(2x^2-3x)$? If so, is this an assumption we need to make to define the fundamental theorem of calculus, or is it a consequence of it?

3. Does the fact that both functions are given the letter $f$ meant that they are both the same? If this is the case, could we maybe not use $t$ is a variable, and use $x$ instead? I know it might get a little confusing since $x$ is also one of the boundaries of the integral, but could it be done in principle?

4. This is I think the most crucial question I have. The first part of the fundamental theorem of calculus is used to prove that integration and differentiation are inverses of each other. But my problem is that the area that we are finding is bound by the function $f(t)$, not the function $f(x)$. Essentially, what I think the theorem is saying, is that the derivative of the function that gives us the area under the curve $f(t)$ on an interval between $t=a$ and $t=x$ is $f(x)$. If we really wanted to prove that integration is the inverse of differentiation, wouldn't we have to prove that the area function gives us the area under $f(x)$?

I guess that the crux of my issue is that I struggle to figure out how the two variables $t$ and $x$ relate to each other. Because if both $f(t)$ and $f(x)$ can be used to find area $A(x)$, then they must be related to each other in some way right?

Answer 1

This is correct - you do grasp the essence of this form of the fundamental theorem of calculus.

Essentially, what I think the theorem is saying, is that the derivative of the function that gives us the area under the curve $f(t)$ on an interval between $t=a$ and $t=x$ is $f(x)$=.

I think your problem is with variable names. There really is no difference between $f(x)$ and $f(t)$. You have a function $f$ that maps numbers to numbers. $f(x)$ isn't the function, it's the value of the function $f$ at the number $x$. The expression $$ \int_a^b f(t)dt $$ can and often is written simply as $$ \int_a^b f \ . $$ It can be interpreted as the area under the graph of $f$ between the values $a$ and $b$ on the first coordinate axis, which is traditionally called the $x$-axis.

When stating the fundamental theorem of calculus you want to consider how that area changes as $b$ changes. Since that upper limit $b$ is to be thought of as changing, you call it $x$. Then you differentiate the way that area changes as a function of $x$. Now if you want to use some variable and its differential in the integrand you need another name for it, so write $f(t)dt$.

Related: Why can't the second fundamental theorem of calculus be proved in just two lines?

Answer 2

Q1
"Indefinite integral" is simply a convenience term that signifies that the indefinite integral $\int f(t) dt = A(t) + C,$ where $C$ is the constant of integration.

In practice, when $A(t) + C$ is evaluated at $t=a$ and $t=x$, the constant of integration "cancels out". This is why you never see it in definite integrals, but you do see it in indefinite integrals.

Has this answered your question here?

Q2
This is an area of confusion.

Given a function $f$, the use of $t$ in the expression $f(t)$ is a placeholder for any value.

In the $A(x) = \int_a^x f(t)dt$ arena, you must not employ the variable $x$ with the function $f$, because the variable $x$ is already being used as one of the endpoints of the integral.

So you don't want to over-load the $x$ variable.

Outside of that scenario, if you have the function $f(t)$, and the variables $x$ or (for that matter $y$) are not otherwise being used, the alternate expressions $f(x)$ or $f(y)$ would make sense.

Q3
In the $A(x) = \int_a^x f(t)dt$ arena,
specifying that $f(x) = A(x)$ is both sloppy and dead wrong.

In this arena, the function that relates to the integral is $A(x),$
and $f(t)$ is the function being integrated.

Q4
When $A(x) = \int_a^x f(t)dt,$ then $\frac{d}{dx}A(x)$ is $f(x).$

That is, regarding $f$ as the function being integrated, and $A$ as the corresponding function that represents the area under the curve,
at a specific point $x_0, A'(x_0)$ is $f(x_0)$.
What this signifies, is that at $x_0$, the rate of change of $A(x)$ is
equal to the height under the curve at $x=x_0$, which corresponds to $f(x_0).$

Regarded that way, the association between the Area function $A(x)$, and the height function $f(x)$ makes sense.

"If we really wanted to prove that integration is the inverse of differentiation, wouldn't we have to prove that the area function gives us the area under $f(x)$".

No, you would instead have to prove, as Calculus books generally do prove, that the derivative of the Area function at $x=x_0$ is equal to the height function $f(x_0).$

So, the relationship is $A'(x_0) = f(x_0),$ and the
antiderivative (in general) of $f(t)$ is $A(t) + C.$