I am learning Calculus. I am trying to understand the fundamental theorem of calculus. I am following this wikipedia article: https://en.wikipedia.org/wiki/Integral.
I am having a hard time understanding what they refer to as the the Fundamental theorem of Calculus. Could someone kindly explain to me what it is in plain english. The wikipedia article is quite gibberish.
The Fundamental Theorem of Calculus (there are two parts, but it seems you're focusing on the second part) essentially says that we can compute an integral using anti-derivatives (As J.W. Tanner says in the comments). Here is the exact text of the Wikipedia article:
The integrals discussed in this article are those termed definite integrals. It is the fundamental theorem of calculus that connects differentiation with the definite integral: if $f$ is a continuous real-valued function defined on a closed interval $[a, b]$, then, once an antiderivative $F$ of $f$ is known, the definite integral of $f$ over that interval is given by $$\int_a^b f(x) \text{ d}x = F(b)-F(a)$$
A definite integral is your classic "area under the curve integral." When calculus was first being (discovered/invented?), the definite and indefinite integral were thought of as completely separate. The indefinite integral finds the antiderivative of a function Essentially, this reverses differentiation. Whereas the derivative of $f(x)=x^2$ is $f'(x)=2x$, the antiderivative of $f'(x)=2x$ is $f(x)=x^2$. This is represented symbolically as $\int2x \text{ d}x = x^2$.
A definite integral, however, comes from the Riemann Sum. It allows you to calculate the area under a curve, essentially. It is defined over a closed interval, which is represented by $a$ and $b$ in the above integral. Now, what the Fundamental Theorem of Calculus (FTC) shows us is a method of calculating a definite integral. Although Wikipedia says that the FTC connects integration and differentiation (which it does), the more important idea is the connection between indefinite and definite integration. Let's do an example to demonstrate this.
Calculate the area under the curve $f(x)=2x$ over the interval [1,2]
Now the first thing we need to do is represent this problem symbolically,
$$\int_1^2 2x \text{ d}x$$
Here's where the FTC comes in. The above integral is a definite integral, but we need to know the antiderivative of $2x$ (remember, the antiderivative is the opposite of a derivative. The antiderivative of $2x$ is the function whose derivative is $2x$)
We can represent the antiderivative symbolically,
$$\int 2x \text{ d}x$$
Notice the lack of bounds on the above integral. This is because it is an indefinite integral. We can solve using the power rule
$$\int 2x \text{ d}x = x^2$$
Now, we can check this by differentiating $x^2$ using the power rule (for derivatives). Remember, the antiderivative of $2x$ is the function whose derivative is $2x$, so the derivative of $x^2$ should be $2x$. You'll find that the derivative of $x^2$ is, in fact, $2x$. Thus, $F(x) = x^2$
Now we can apply the FTC
$$\int_1^2 2x \text{ d}x = F(2) - F(1)$$ $$\int_1^2 2x \text{ d}x = 2^2 - 1^2$$ $$\int_1^2 2x \text{ d}x = 4 - 1$$ $$\int_1^2 2x \text{ d}x = 3$$
At the most basic level, from a purely conceptual point of view, and omitting all required conditions.
Let the area under the curve of a function $f$ be limited by the fixed point $(a,0) $ and the moving point $(x,0)$,
$FTC$ :
the ( instantaneous) rate of growth of this area is nothing else than $f(x)$( the value of $f$ at $x$).
Since the area function $A$ is the indefinite integral of $f$ ( namely, $A(x)=\int_{a}^{x} f(t)dt$) and since the (instantaneous) rate of change of the area is ( by defnition) the derivative of this indefinite integral, we have :
$FTC :$
$A'(x)=f(x)$.
Now, maybe explaining what you can do with this theorem will allow you to understand better what it means .
In plain english FTC can be stated as follows
(1) you can find indirectly the derivative of a function by finding the function whereof it is an integral ( i. e. if $f_1$ is the integral of $f_2$, then the derivative of $f_1$ is simply $f_2$)
(2) you can find indirectly a primitive of a function by finding the integral of this function ( if $f_1$ is an indefnite integral of $f_2$, then $f_1$ is a primitive of $f_2$, and therefore $f_1$ is identical to any primitive $F$ of $f_1$, but for a constant).
(3) you can find indirectly the definite integral of a function $f$ from $a$ to $b$ (namely the number $\int_{a}^{b}f(x)dx$) by just computing the difference $F(b)-F(a)$ , $F$ being any primitive of $f$.
EDIT :
(1) added this point : function F is identical to function A but for a constant ( this is always the case for 2 primitives of the same function).
(2) also added a 3rd case that is the most common statement of the FTC in College Calculus books.
The FTC just says that
If $ f $ is a function differentiable at the intervalle $ [a,b ]$ and if its derivative $ f ' $ is integrable at $ [a,b] $ THEN we have
$$\int_a^bf '(x)dx = \Bigl[ f(x) \Bigr]_a^b=$$ $$ f(b)-f(a)$$
This theorem allows to compute usual integrals and in particular, to use by parts integration.
This is a way to compute integrals by just subtracting a value from the other.
It says the total change of a function (integral of the function's differential over an interval) is equal to the difference in the values of the function at the endpoints of the interval.
That is, given the integral $$\int_a^bf'(x)\mathrm dx,$$ since $f'(x)\mathrm dx$ is the differential of $f(x),$ then the integral may be rewritten as $$\int_a^b \mathrm d(f(x)),$$ and this may be computed by taking the difference $f(b)-f(a).$ That's the fundamental theorem of calculus.
Given an inerval $[a,b]$ and a function $f: \>[a,b]\to{\mathbb R}$ there is something like the "total impact of $f$ on $[a,b]$". This "total impact" is called the integral of $f$ over $[a,b]$, and is denoted by $$\int_a^b f(x)\>dx\ .$$ When $f(x)>0$ on $[a,b]$ this "total impact" is intuitionally represented by the area between $y=0$ and $y=f(x)$ over the interval $[a,b]$.
This setup indicates that we want $\int_a^b f(t)\>dt\geq0$ when $f(t)\geq0$, then $$\int_a^b \bigl(\lambda f(t)+\mu g(t)\bigr)\>dt=\lambda \int_a^b f(t)\>dt+\mu\int_a^b g(t)\>dt$$ as well as $$\int_a^b f(t)\>dt=\int_a^c f(t)\>dt+\int_c^b f(t)\>dt\qquad(a<c<b)\ .$$ Thinking about the whole situation one arrives at the Riemann integral $$\int_a^b f(t)\>dt=\lim_\ldots\sum_{k=1}^N f(\xi_k)(x_k-x_{k-1})\tag{1}\ ,$$ a complicated limit. Of course we want to compute this integral in many cases. When $f$ is given only numerically as a data set then we can use $(1)$ for a numerical approximation of the integral.
But often the function $f$ is given as an analytical expression, and we hope that the value of the integral can then also be expressed "analytically". That's where the FTC comes in. This theorem says that the above integrals are connected with the so-called primitives of $f$. Such a primitive is a function $F$ tied to $f$ by the condition $F'=f$. When $f$ is given by an analytical expression in the variable $x$ then it is often possible to find another analytical expression $F(x)$ satisfying $F'(x)\equiv f(x)$, e.g., $\sin'(x)\equiv\cos x$.
The FTC then says the following: If $F$ is a primitive of $f$ valid over the interval $[a,b]$ then $$\int_a^b f(t)\>dt=F(b)-F(a)\ .$$ This theorem is not a "rewording of definitions". It is a miracle. It allows the computation of the interesting limit $(1)$ by the evaluation of $F$-values. But we need to know the "analytical expression" of $F$ when $f$ is given as such an expression.
The FTC says that integration and differentiation are inverse operations. If you differentiate the right kind of integral, then you get the integrand back. If you integrate a derivative, you get the original function back.
D(I(f)) = f
I(D(f)) = f.
