Difference between revisions of "Fundamental Theorem of Calculus"
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+ | The '''Fundamental Theorem of Calculus''' establishes a link between the two central operations of [[calculus]]: [[derivative|differentiation]] and [[integral|integration]]. | ||
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== Introductory Problems == | == Introductory Problems == | ||
− | This section is for people who know what | + | This section is for people who know what [[integral]]s 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: <math>\int_2^5 x^3 dx</math> and <math>\int_{.2}^{.4} \cos(x) dx</math>. (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 <math>{t}</math> is <math>e^{t^2}</math> meters/second. (Yes, probably no object really moves this way, but just pretend.) ''Approximately'' how far does the object move between times <math>t=1 </math> second and <math>t=2</math> seconds? (I picked <math>e^{t^2}</math> because I wanted a function that doesn't have a nice anti-derivative.) Interpret the distance that the object travels between times <math>t=1</math> and <math>t=2</math> geometrically, as an area under a curve. | ||
+ | # An object is moving along a straight line, and its velocity at time <math>{t}</math> is <math>t^3</math> m/s. Exactly how far does the object go between times <math>t=2</math> sec and <math>t=5</math> sec? Interpret this distance geometrically, as an area under a curve. | ||
+ | # Same as question 3, but this time the object's velocity at time <math>{t}</math> is <math>\cos{t}</math>, and you want to find out exactly how far the object moved between times <math>t=.2</math> and <math>t=.4</math>. 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 <math>f(x)= \frac{x^4}{4}</math> change over the interval from <math>x=2</math> to <math>x=5</math>? Obviously, the answer is <math>\frac{5^4}{4} - \frac{2^4}{4}</math> . But, there's another way to look at it: ''the total change is the sum of all the little changes''. Break the interval from <math>2</math> to <math>5</math> up into <math>30</math> little subintervals. Using the derivative, tell me: approximately how much does <math>{f}</math> 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 <math>{f}</math> over the whole interval. This Riemann sum also approximates a certain integral. What is that integral? | |
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The remaining question deals with the "second" Fundamental Theorem of Calculus. | The remaining question deals with the "second" Fundamental Theorem of Calculus. | ||
− | + | # Let <math>f: \mathbb{R} \to \mathbb{R}</math> be [[continuous function | continuous]]. Define <math>F(x) = \int_0^x f(s) ds</math> for all <math>x \ge 0</math>. Draw a picture to explain this definition of <math>{F}</math>. <math>F'(x)= \lim_{\Delta x \to 0} \frac{F(x+\Delta x) - F(x)}{\Delta x}</math>. Assuming <math>\Delta x</math> is a small number, represent <math>F(x+\Delta x) - F(x)</math> geometrically in your picture. Using your picture, approximately how large is <math>\frac{F(x+\Delta x) - F(x)}{\Delta x}</math>? Now the final question: What is <math>F'(x)</math>? | |
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== Statement== | == Statement== | ||
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Second Fundamental Theorem of Calculus: | Second Fundamental Theorem of Calculus: | ||
− | Let <math>{a}</math>, <math>{b} \in \mathbb{R}</math>, <math>{a}<{b}</math>. Suppose <math>f:[a,b] \to \mathbb{R}</math> is continuous on the whole interval <math>[a,b]</math>. Let <math>F(x) = \int_a^x f(s) ds</math> for all <math>x \in [a,b]</math>. Then <math>F</math> is differentiable on the whole interval <math>[a,b]</math> (using limits from the right and from the left for the derivatives at a and b, respectively), and <math>F'(x) = f(x)</math> for all <math>x \in [a,b]</math>. | + | Let <math>{a}</math>, <math>{b} \in \mathbb{R}</math>, <math>{a}<{b}</math>. Suppose <math>f:[a,b] \to \mathbb{R}</math> is [[continuous function | continuous]] on the whole interval <math>[a,b]</math>. Let <math>F(x) = \int_a^x f(s) ds</math> for all <math>x \in [a,b]</math>. Then <math>F</math> is differentiable on the whole interval <math>[a,b]</math> (using limits from the right and from the left for the derivatives at a and b, respectively), and <math>F'(x) = f(x)</math> for all <math>x \in [a,b]</math>. |
− | == Intuitive | + | == Intuitive explanation == |
The first Fundamental Theorem of Calculus basically says that "the total change is the sum of all the little changes." | The first Fundamental Theorem of Calculus basically says that "the total change is the sum of all the little changes." | ||
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How much does <math>F</math> change over the tiny interval <math>[x_i,x_{i+1}]</math>? Of course, the answer is <math>F(x_{i+1})-F(x_i)</math>. But this is approximately <math>F'(x_i)\Delta x</math>. If <math>\Delta F_i</math> is the amount that <math>F</math> changes on the interval <math>[x_i, x_{i+1}]</math>, then <math>\Delta F_i \approx F'(x_i)\Delta x</math>. | How much does <math>F</math> change over the tiny interval <math>[x_i,x_{i+1}]</math>? Of course, the answer is <math>F(x_{i+1})-F(x_i)</math>. But this is approximately <math>F'(x_i)\Delta x</math>. If <math>\Delta F_i</math> is the amount that <math>F</math> changes on the interval <math>[x_i, x_{i+1}]</math>, then <math>\Delta F_i \approx F'(x_i)\Delta x</math>. | ||
− | The total change of <math>F</math> over the interval <math>[a,b]</math> , <math>F(b)-F(a)</math>, is exactly equal to <math>F_0 + F_1 +\cdots + F_{n-1}</math> . Thus, | + | The total change of <math>F</math> over the interval <math>[a,b]</math> , <math>F(b)-F(a)</math>, is exactly equal to <math>\Delta F_0 + \Delta F_1 +\cdots + \Delta F_{n-1}</math> . Thus, |
<math>F(b)-F(a) \approx F'(x_0)\Delta x + F'(x_1) \Delta x + \cdots + F'(x_{n-1}) \Delta x</math>. | <math>F(b)-F(a) \approx F'(x_0)\Delta x + F'(x_1) \Delta x + \cdots + F'(x_{n-1}) \Delta x</math>. | ||
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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 <math>F(b)-F(a) = \int_a^b F'(x) dx</math>. | 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 <math>F(b)-F(a) = \int_a^b F'(x) dx</math>. | ||
− | This can be made into a rigorous proof, if you use the | + | 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.) | 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.) | ||
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== Proof == | == Proof == | ||
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Here's a proof of the first Fundamental Theorem of Calculus. | Here's a proof of the first Fundamental Theorem of Calculus. | ||
− | Let <math>{a}</math>, <math>{b} \in \mathbb{R}</math> , <math>a<b</math>. Suppose <math>F:[a,b] \to \mathbb{R}</math> is differentiable on the whole interval <math>[a,b]</math> (using limits from the right and left for the derivatives at <math>{a}</math> and <math>{b}</math>, respectively), and suppose that <math>F'</math> is Riemann integrable on <math>[a,b]</math>. | + | Let <math>{a}</math>, <math>{b} \in \mathbb{R}</math> , <math>a<b</math>. Suppose <math>F:[a,b] \to \mathbb{R}</math> is differentiable on the whole interval <math>[a,b]</math> (using limits from the right and left for the derivatives at <math>{a}</math> and <math>{b}</math>, respectively), and suppose that <math>F'</math> is [i]Riemann integrable[/i] on <math>[a,b]</math>. |
− | Let <math>\epsilon > 0</math>. There exists <math>\delta > 0</math> such that if <math>a=x_0, x_1, x_2, \ldots, x_n=b</math> is a partition of <math>[a,b]</math> , and <math>\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}</math> , and <math> | + | Let <math>\epsilon > 0</math>. There exists <math>\delta > 0</math> such that if <math>a=x_0, x_1, x_2, \ldots, x_n=b</math> is a partition of <math>[a,b]</math> , and <math>\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}</math> , and <math>\Delta x_i < \delta</math> for all <math>i</math>, <math>0 \le i < n</math>, and <math>\xi_i</math> is in <math>[x_i,x_{i+1}]</math> for all <math>i</math>, <math>0 \le i < n</math>, Then the [i]Riemann sum [/i] |
<math>F'(\xi_0) \Delta x_0 + F'(\xi_1) \Delta x_1 + \cdots + F'(\xi_{n-1}) \Delta x_{n-1}</math> | <math>F'(\xi_0) \Delta x_0 + F'(\xi_1) \Delta x_1 + \cdots + F'(\xi_{n-1}) \Delta x_{n-1}</math> | ||
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is within <math>\epsilon</math> of <math>\int_a^b F'(x) dx</math>. | is within <math>\epsilon</math> of <math>\int_a^b F'(x) dx</math>. | ||
− | (In fact, this is sometimes taken as the definition of the statement "<math>F'</math> is Riemann integrable on the interval <math>[a,b]</math>.") | + | (In fact, this is sometimes taken as the definition of the statement "<math>F'</math> is [i]Riemann integrable[/i] on the interval <math>[a,b]</math>.") |
So, let <math>a=x_0, x_1,\ldots, x_n=b</math> be a partition of <math>[a,b]</math> such that <math>\Delta x_i < \delta</math> for all <math>i</math>, <math>0 \le i < n</math>. | So, let <math>a=x_0, x_1,\ldots, x_n=b</math> be a partition of <math>[a,b]</math> such that <math>\Delta x_i < \delta</math> for all <math>i</math>, <math>0 \le i < n</math>. | ||
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Thus, | Thus, | ||
− | <math>F(b)-F(a) = \Delta F_0 + \Delta F_1 + \cdots + \Delta F_{n- | + | <math>F(b)-F(a) = \Delta F_0 + \Delta F_1 + \cdots + \Delta F_{n-1}</math> |
<math>= F'(\xi_0) \Delta x_0 + F'(\xi_1)\Delta x_1 + \cdots + F'(\xi_{n-1}) \Delta x_{n-1}</math>. | <math>= F'(\xi_0) \Delta x_0 + F'(\xi_1)\Delta x_1 + \cdots + F'(\xi_{n-1}) \Delta x_{n-1}</math>. | ||
− | This last expression, on the right, is a Riemann sum, and it is within \epsilon of \int_a^b F'(x) dx. | + | This last expression, on the right, is a Riemann sum, and it is within <math>\epsilon</math> of <math>\int_a^b F'(x) dx</math>. |
− | + | Therefore, we have found that <math>F(b) - F(a)</math> is within <math>\epsilon</math> of <math>\int_a^b F'(x) dx</math>. | |
But <math>\epsilon > 0</math> was arbitrary. The only way that <math>F(b) - F(a)</math> could be within <math>\epsilon</math> of <math>\int_a^b F'(x) dx</math> for any <math>\epsilon > 0</math> is if <math>F(b)-F(a)</math> is actually equal to <math>\int_a^b F'(x) dx</math>. | But <math>\epsilon > 0</math> was arbitrary. The only way that <math>F(b) - F(a)</math> could be within <math>\epsilon</math> of <math>\int_a^b F'(x) dx</math> for any <math>\epsilon > 0</math> is if <math>F(b)-F(a)</math> is actually equal to <math>\int_a^b F'(x) dx</math>. | ||
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So that means <math>\int_a^b F'(x) dx = F(b)-F(a)</math>. | So that means <math>\int_a^b F'(x) dx = F(b)-F(a)</math>. | ||
− | + | 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.) | 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.) | ||
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+ | As recommended by the original poster, the following proof is taken from <math>\textit{Spivak's}</math> Calculus 4th edition. | ||
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+ | Proof of the Fundamental Theorem of Calculus Part <math>2:</math> | ||
+ | |||
+ | Let <math>P = \{t_0,...,t_n\}</math> be any partition of <math>[a, b].</math> By the Mean Value Theorem there is a point <math>x_i</math> in <math>[t_{i-1}, t_i]</math> such that | ||
+ | |||
+ | <cmath>g(t_i) - g(t_{i-1}) = g'(x_i)(t_i - t_{i-1})</cmath> | ||
+ | <cmath>= f(x_i)(t_i - t_{i-1}).</cmath> | ||
+ | |||
+ | If | ||
+ | <cmath>m_i = \text{inf}\{f(x) : t_{i-1} \le x \le t_i\},</cmath> | ||
+ | <cmath>M_i = \text{sup}\{f(x) : t_{i-1} \le x \le t_i\},</cmath> | ||
+ | then clearly | ||
+ | <cmath>m_i(t_i - t_{i-1}) \le f(x_i)(t_i - t_{i-1}) \le M_i(t_i - t_{i-1}),</cmath> | ||
+ | that is, | ||
+ | <cmath>m_i(t_i - t_{i-1}) \le g(t_i) - g{(t_{i-1})} \le M_i(t_i - t_{i-1}).</cmath> | ||
+ | |||
+ | Adding these equations for <math>i = 1,..., n</math> we obtain | ||
+ | |||
+ | <cmath> \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})</cmath> | ||
+ | so that | ||
+ | <cmath>L(f, P) \le g(b) - g(a) \le U (f, P)</cmath> | ||
+ | <math>\textit{for every partition P}.</math> But this means that | ||
+ | <cmath>g(b) - g(a) = \int_{a}^{b} f.</cmath> | ||
+ | |||
+ | ==Generalizations== | ||
+ | |||
+ | ==See also== | ||
+ | |||
+ | [[Category:Calculus]] |
Latest revision as of 15:23, 2 January 2022
The Fundamental Theorem of Calculus establishes a link between the two central operations of calculus: differentiation and integration.
Contents
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:
and
. (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
is
meters/second. (Yes, probably no object really moves this way, but just pretend.) Approximately how far does the object move between times
second and
seconds? (I picked
because I wanted a function that doesn't have a nice anti-derivative.) Interpret the distance that the object travels between times
and
geometrically, as an area under a curve.
- An object is moving along a straight line, and its velocity at time
is
m/s. Exactly how far does the object go between times
sec and
sec? Interpret this distance geometrically, as an area under a curve.
- Same as question 3, but this time the object's velocity at time
is
, and you want to find out exactly how far the object moved between times
and
. 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
change over the interval from
to
? Obviously, the answer is
. But, there's another way to look at it: the total change is the sum of all the little changes. Break the interval from
to
up into
little subintervals. Using the derivative, tell me: approximately how much does
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
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
be continuous. Define
for all
. Draw a picture to explain this definition of
.
. Assuming
is a small number, represent
geometrically in your picture. Using your picture, approximately how large is
? Now the final question: What is
?
Statement
First Fundamental Theorem of Calculus:
Let ,
,
. Suppose
is differentiable on the whole interval
(using limits from the right and left for the derivatives at
and
, respectively), and suppose that
is Riemann integrable on
. Then
.
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 ,
,
. Suppose
is continuous on the whole interval
. Let
for all
. Then
is differentiable on the whole interval
(using limits from the right and from the left for the derivatives at a and b, respectively), and
for all
.
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 change over an interval
? Obviously, the answer is
. But there's another way to look at it. Break the interval
up into a whole bunch of tiny subintervals, each having a tiny width
. Let's say
has been broken up into
subintervals, so
. Let
.
How much does change over the tiny interval
? Of course, the answer is
. But this is approximately
. If
is the amount that
changes on the interval
, then
.
The total change of over the interval
,
, is exactly equal to
. Thus,
.
What we have here on the right is a Riemann sum. It approximates a certain integral. What integral does it approximate? Well, it approximates .
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 .
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 ,
,
. Suppose
is differentiable on the whole interval
(using limits from the right and left for the derivatives at
and
, respectively), and suppose that
is [i]Riemann integrable[/i] on
.
Let . There exists
such that if
is a partition of
, and
, and
for all
,
, and
is in
for all
,
, Then the [i]Riemann sum [/i]
is within of
.
(In fact, this is sometimes taken as the definition of the statement " is [i]Riemann integrable[/i] on the interval
.")
So, let be a partition of
such that
for all
,
.
For each ,
, let
be the amount that
changes over the interval
. Then
.
According to the Mean Value Theorem, for each ,
, there exists
such that
.
(This is similar to the part of the intuitive argument where we said that . However, this is better. The Mean Value Theorem, fortunately, gives us exact equality, rather than just an approximation.)
Thus,
.
This last expression, on the right, is a Riemann sum, and it is within of
.
Therefore, we have found that is within
of
.
But was arbitrary. The only way that
could be within
of
for any
is if
is actually equal to
.
So that means .
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 Calculus 4th edition.
Proof of the Fundamental Theorem of Calculus Part
Let be any partition of
By the Mean Value Theorem there is a point
in
such that
If
then clearly
that is,
Adding these equations for we obtain
so that
But this means that