Fundamental Theorem of Calculus – Sam Artigliere's Blog

Part I

Let $f(t)$ be a continuous real-valued function on a closed interval $\left[a,b\right]$ (closed interval meaning the numbers $a$ and $b$ at the endpoints of the interval are included).

Recalling that, by definition,

The area under $f(t)$ from $a$ to $b$ equals $\displaystyle\int\limits_a^b f(t)dt=\lim_{n\to\infty}\sum_{i=1}^{n}f(t_i)\Delta t_i$ where $\Delta t = \frac{b-a}{n}$

Let

$F(x)=\displaystyle\int\limits_a^x f(t)dt$

where $\displaystyle\int\limits_a^x f(t)dt$ is the area under $f(t)$ from $a$ to $x$ , $x$ being some number between $a$ and $b$ .

Likewise, the area under $f(t)$ between $a$ and $x + \Delta x$ is:

$F(x+\Delta x)=\displaystyle\int\limits_a^{x +\Delta x}f(t)dt$

It follows, then, that the area under $f(t)$ from $a$ to $x + \Delta x$ minus the area under $f(t)$ from $a$ to $x$ is equal to the area under $f(t)$ from $x$ to $x + \Delta x$ . In mathematical terms:

$\underbrace{\displaystyle\int\limits_a^{x +\Delta x}f(t)dt}_{F(x+\Delta x)}-\underbrace{\displaystyle\int\limits_a^x f(t)dt}_{F(x)}=\underbrace{\displaystyle\int\limits_x^{x +\Delta x}f(t)dt}_{F(\Delta x)}$

Consider the minimum and maximum values of $f(t)$ between $x$ and $\Delta x$ . Designate them $f(m)$ and $f(M)$ , respectively. The area of a rectangle is its base times its height. We know that the area under $f(t)$ on the interval between $x$ and $\Delta x$ has to be greater than the area of a rectangle with height $m$ and base $\Delta x$ and less than a rectangle with height $M$ and base $\Delta x$ . That is,

$m \Delta x \leq F(x+\Delta x)-F(x) \leq M \Delta x$

Next divide through by $\Delta x$ . We have:

$m \leq \frac{F(x+\Delta x)-F(x)}{\Delta x} \leq M$

Now take the limit of this expression as $\Delta x$ goes to zero. The following things happen:

$\displaystyle\lim_{\Delta x \to 0}\frac{F(x+\Delta x)-F(x)}{\Delta x}$ is the definition of the derivative of $F(x)$ , $F^\prime (x)$
Because both $m$ and $M$ are in the interval between $x$ and $x+\Delta x$ , as $\Delta x$ goes to zero, both $f(m)$ and $f(M)$ tend toward $f(x)$ . (This is a result of the squeeze theorem. The intuition behind this theorem can be found here. A formal proof of the theorem can be found here.)
Because $F^\prime$ is between $f(m)$ and $f(M)$ , by the squeeze theorem, $F^\prime$ also goes to $f(x)$ .

Thus,

$\displaystyle\lim_{\Delta x \to 0}\frac{F(x+\Delta x)-F(x)}{\Delta x}=f(x)$

But

$\displaystyle\lim_{\Delta x \to 0}\frac{F(x+\Delta x)-F(x)}{\Delta x}=F^\prime(x)$

Therefore,

$\mathbf{F^\prime=f(x)}$

The above equation is the fundamental theorem of calculus, part 1.

In the above equations, $F(x)$ is called the antiderivative of $f(x)$ . Here are some examples:

$F(x)=\frac{1}{3}x^3+C=\int x^2 dx$
$F(x)=\frac{1}{4}x^5+C=\int x^4 dx$
$F(x)=e^x+C=\int e^x dx$
$F(x)=-\cos{\theta}+C=\int \sin{\theta} dx$

The term $C$ on the left-hand side of each of the above equations is an arbitrary constant. In each case, we can check the correctness of the above statements by taking the derivative of $F(x)$ . If we do, we get $f(x)$ , the function just to the right of the integral sign and to the left of the differentials $dx$ and $d\theta$ .

Why is there a constant, $C$ , added onto every term on the left side of the equations above? It’s because when we take the derivative of the expressions on the left-hand side, the derivative of the non-constant terms gives you $f(x)$ and the derivative of the constant term gives you zero. So we could have any constant added to the non-constant term in these expressions and still come out with zero when we take the derivative of that constant. That means that, for any function $f(x)$ that can be integrated, there are an infinite number of antiderivatives which differ only by the constant, $C$ , that’s added to the end of it.

Part 2

Let $G(x)=\int\limits_a^x F^\prime(t)dt$ . We know from the first part of the fundamental theorem of calculus that $F^\prime(x)=f(x)$ . Therefore, $G(x)=\int\limits_a^x f(t)dt$ . That means that $G(x)$ is an antiderivative of $f(x)$ as is $F(x)$ . This, in turn, means that $G(x)$ and $F(x)$ differ only by a constant, $C$ .

We need to find the value of this constant. To do this, let $x=a$ . Then,

$G(a)=\int\limits_a^a F^\prime(t)dt=0$ . Why is this zero? Because the we’re taking the area under the curve $F^\prime(t)$ at one point – $a$ . That area is zero.