Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus establishes a link between the two central operations of calculus: differentiation and integration.

Introductory Problems

This section is for people who know what integrals are but don't know the Fundamental Theorem of Calculus yet, and would like to try to figure it out. (Actually there are two different but related Fundamental Theorems of Calculus. Questions 0 through 5 correspond to the "first" Fundamental Theorem of Calculus. The last question corresponds to the "second" Fundamental Theorem of Calculus.)

Evaluate: $\int_2^5 x^3 dx$ and $\int_{.2}^{.4} \cos(x) dx$ . (The next few questions are meant as hints for how to do this.)

An object is moving along a straight line, and its velocity at time ${t}$ is $e^{t^2}$ meters/second. (Yes, probably no object really moves this way, but just pretend.) Approximately how far does the object move between times $t=1$ second and $t=2$ seconds? (I picked $e^{t^2}$ because I wanted a function that doesn't have a nice anti-derivative.) Interpret the distance that the object travels between times $t=1$ and $t=2$ geometrically, as an area under a curve.
An object is moving along a straight line, and its velocity at time ${t}$ is $t^3$ m/s. Exactly how far does the object go between times $t=2$ sec and $t=5$ sec? Interpret this distance geometrically, as an area under a curve.
Same as question 3, but this time the object's velocity at time ${t}$ is $\cos{t}$ , and you want to find out exactly how far the object moved between times $t=.2$ and $t=.4$ . Interpret the distance that the object moved geometrically, as an area under a curve.

Can you do the main problem now?

Here's a slightly different way to think about the main problem, that doesn't use physics. How much does the function $f(x)= \frac{x^4}{4}$ change over the interval from $x=2$ to $x=5$ ? Obviously, the answer is $\frac{5^4}{4} - \frac{2^4}{4}$ . But, there's another way to look at it: the total change is the sum of all the little changes. Break the interval from $2$ to $5$ up into $30$ little subintervals. Using the derivative, tell me: approximately how much does ${f}$ change over each of these little subintervals? Adding up all of these little changes gives you a Riemann sum that approximates the total change of ${f}$ over the whole interval. This Riemann sum also approximates a certain integral. What is that integral?

The remaining question deals with the "second" Fundamental Theorem of Calculus.

Let $f: \mathbb{R} \to \mathbb{R}$ be continuous. Define $F(x) = \int_0^x f(s) ds$ for all $x \ge 0$ . Draw a picture to explain this definition of ${F}$ . $F'(x)= \lim_{\Delta x \to 0} \frac{F(x+\Delta x) - F(x)}{\Delta x}$ . Assuming $\Delta x$ is a small number, represent $F(x+\Delta x) - F(x)$ geometrically in your picture. Using your picture, approximately how large is $\frac{F(x+\Delta x) - F(x)}{\Delta x}$ ? Now the final question: What is $F'(x)$ ?

Statement

First Fundamental Theorem of Calculus:

Let ${a}$ , ${b} \in \mathbb{R}$ , $a<b$ . Suppose $F:[a,b] \to \mathbb{R}$ is differentiable on the whole interval $[a,b]$ (using limits from the right and left for the derivatives at ${a}$ and ${b}$ , respectively), and suppose that $F'$ is Riemann integrable on $[a,b]$ . Then $\int_a^b F'(x)dx = F(b) - F(a)$ .

In other words, "the total change (on the right) is the sum of all the little changes (on the left)."

Second Fundamental Theorem of Calculus:

Let ${a}$ , ${b} \in \mathbb{R}$ , ${a}<{b}$ . Suppose $f:[a,b] \to \mathbb{R}$ is continuous on the whole interval $[a,b]$ . Let $F(x) = \int_a^x f(s) ds$ for all $x \in [a,b]$ . Then $F$ is differentiable on the whole interval $[a,b]$ (using limits from the right and from the left for the derivatives at a and b, respectively), and $F'(x) = f(x)$ for all $x \in [a,b]$ .

Intuitive explanation

The first Fundamental Theorem of Calculus basically says that "the total change is the sum of all the little changes."

How much does a function $F$ change over an interval $[a,b]$ ? Obviously, the answer is $F(b)-F(a)$ . But there's another way to look at it. Break the interval $[a,b]$ up into a whole bunch of tiny subintervals, each having a tiny width $\Delta x$ . Let's say $[a,b]$ has been broken up into $n$ subintervals, so $\Delta x = \frac{b-a}{n}$ . Let $x_0 = a, x_1 = a + \Delta x, x_2 = a + 2\Delta x, \ldots, x_i = x_0 + i\Delta x, \ldots, x_n = a + n\Delta x = b$ .

How much does $F$ change over the tiny interval $[x_i,x_{i+1}]$ ? Of course, the answer is $F(x_{i+1})-F(x_i)$ . But this is approximately $F'(x_i)\Delta x$ . If $\Delta F_i$ is the amount that $F$ changes on the interval $[x_i, x_{i+1}]$ , then $\Delta F_i \approx F'(x_i)\Delta x$ .

The total change of $F$ over the interval $[a,b]$ , $F(b)-F(a)$ , is exactly equal to $\Delta F_0 + \Delta F_1 +\cdots + \Delta F_{n-1}$ . Thus,

$F(b)-F(a) \approx F'(x_0)\Delta x + F'(x_1) \Delta x + \cdots + F'(x_{n-1}) \Delta x$ .

What we have here on the right is a Riemann sum. It approximates a certain integral. What integral does it approximate? Well, it approximates $\int_a^b F'(x) dx$ .

If we repeat this process, using more and more subintervals, then our approximations will get better and better, and the Riemann sums will approximate that integral better and better, and in the limit we will find that $F(b)-F(a) = \int_a^b F'(x) dx$ .

This can be made into a rigorous proof, if you use the Lagrange's mean value theorem. The standard proof of the first Fundamental Theorem of Calculus, using the Mean Value Theorem, can be thought of in this way.

In order to get an intuitive understanding of the second Fundamental Theorem of Calculus, I recommend just thinking about problem 6. The idea presented there can also be turned into a rigorous proof. (The standard proof can be thought of in this way.)

Proof

Here's a proof of the first Fundamental Theorem of Calculus.

Let ${a}$ , ${b} \in \mathbb{R}$ , $a<b$ . Suppose $F:[a,b] \to \mathbb{R}$ is differentiable on the whole interval $[a,b]$ (using limits from the right and left for the derivatives at ${a}$ and ${b}$ , respectively), and suppose that $F'$ is [i]Riemann integrable[/i] on $[a,b]$ .

Let $\epsilon > 0$ . There exists $\delta > 0$ such that if $a=x_0, x_1, x_2, \ldots, x_n=b$ is a partition of $[a,b]$ , and $\Delta x_0=x_1-x_0, \Delta x_1 = x_2-x_1, \ldots, \Delta x_{n-1}=x_n - x_{n-1}=b-x_{n-1}$ , and $\Delta x_i < \delta$ for all $i$ , $0 \le i < n$ , and $\xi_i$ is in $[x_i,x_{i+1}]$ for all $i$ , $0 \le i < n$ , Then the [i]Riemann sum [/i]

$F'(\xi_0) \Delta x_0 + F'(\xi_1) \Delta x_1 + \cdots + F'(\xi_{n-1}) \Delta x_{n-1}$

is within $\epsilon$ of $\int_a^b F'(x) dx$ .

(In fact, this is sometimes taken as the definition of the statement " $F'$ is [i]Riemann integrable[/i] on the interval $[a,b]$ .")

So, let $a=x_0, x_1,\ldots, x_n=b$ be a partition of $[a,b]$ such that $\Delta x_i < \delta$ for all $i$ , $0 \le i < n$ .

For each $i$ , $0\le i < n$ , let $\Delta F_i = F(x_{i+1})-F(x_i)$ be the amount that $F$ changes over the interval $[x_i,x_{i+1}]$ . Then

$F(b) - F(a)=\Delta F_0 + \Delta F_1 + \cdots + \Delta F_{n-1}$ .

According to the Mean Value Theorem, for each $i$ , $0 \le i < n$ , there exists $\xi_i \in (x_i, x_{i+1})$ such that $\Delta F_i = F(x_{i+1}) - F(x_i) = F'(\xi_i)\Delta x_i$ .

(This is similar to the part of the intuitive argument where we said that $\Delta F_i \approx F'(x_i) \Delta x_i$ . However, this is better. The Mean Value Theorem, fortunately, gives us exact equality, rather than just an approximation.)

Thus,

$F(b)-F(a) = \Delta F_0 + \Delta F_1 + \cdots + \Delta F_{n-1}$ $= F'(\xi_0) \Delta x_0 + F'(\xi_1)\Delta x_1 + \cdots + F'(\xi_{n-1}) \Delta x_{n-1}$ .

This last expression, on the right, is a Riemann sum, and it is within $\epsilon$ of $\int_a^b F'(x) dx$ .

Therefore, we have found that $F(b) - F(a)$ is within $\epsilon$ of $\int_a^b F'(x) dx$ .

But $\epsilon > 0$ was arbitrary. The only way that $F(b) - F(a)$ could be within $\epsilon$ of $\int_a^b F'(x) dx$ for any $\epsilon > 0$ is if $F(b)-F(a)$ is actually equal to $\int_a^b F'(x) dx$ .

So that means $\int_a^b F'(x) dx = F(b)-F(a)$ .

This concludes the proof of the first Fundamental Theorem of Calculus.

For a proof of the second Fundamental Theorem of Calculus, I recommend looking in the book Calculus by Spivak. (Hopefully I or someone else will post a proof here eventually.)

As recommended by the original poster, the following proof is taken from $\textit{Spivak's}$ Calculus 4th edition.

Proof of the Fundamental Theorem of Calculus Part $2:$

Let $P = \{t_0,...,t_n\}$ be any partition of $[a, b].$ By the Mean Value Theorem there is a point $x_i$ in $[t_{i-1}, t_i]$ such that

$g(t_i) - g(t_{i-1}) = g'(x_i)(t_i - t_{i-1})$ $= f(x_i)(t_i - t_{i-1}).$

If $m_i = \text{inf}\{f(x) : t_{i-1} \le x \le t_i\},$ $M_i = \text{sup}\{f(x) : t_{i-1} \le x \le t_i\},$ then clearly $m_i(t_i - t_{i-1}) \le f(x_i)(t_i - t_{i-1}) \le M_i(t_i - t_{i-1}),$ that is, $m_i(t_i - t_{i-1}) \le g(t_i) - g{(t_{i-1})} \le M_i(t_i - t_{i-1}).$

Adding these equations for $i = 1,..., n$ we obtain

$\sum_{i=1}^{n}m_i(t_i - t_{i-1}) \le g(b) - g(a) \le \sum_{i=1}^{n}M_i(t_i - t_{i-1})$ so that $L(f, P) \le g(b) - g(a) \le U (f, P)$ $\textit{for every partition P}.$ But this means that $g(b) - g(a) = \int_{a}^{b} f.$

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Fundamental Theorem of Calculus

Contents

Introductory Problems

Statement

Intuitive explanation

Proof

Generalizations

See also