Difference between revisions of "The Fundamental Theorem of Calculus"

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* [https://youtu.be/jyRdHbHeUuU The Second Fundamental Theorem of Calculus] Video Lecture by James Sousa.
+
The fundamental theorem of calculus is a critical portion of calculus because it links the concept of a derivative to that of an integral. As a result, we can use our knowledge of derivatives to find the area under the curve, which is often quicker and simpler than using the definition of the integral.
  
* [https://youtu.be/YE9jpfxEFYk Ex 1: The Second Fundamental Theorem of Calculus] Video Lecture by James Sousa.
+
==Mean Value Theorem for Integration==
 +
We will need the following theorem in the discussion of the Fundamental Theorem of Calculus.
  
* [https://youtu.be/gJtkTRjqM6I Ex 2: The Second Fundamental Theorem of Calculus (Reverse Order)] Video Lecture by James Sousa.
+
Suppose <math>f(x)</math> is continuous on <math>[a,b]</math> . Then <math>\frac{1}{b-a}\int\limits_a^b f(x)dx=f(c)</math> for some <math>c\in[a,b]</math>.
  
* [https://youtu.be/X5ke8bUiiQM Ex 3: The Second Fundamental Theorem of Calculus] Video Lecture by James Sousa.
+
===Proof of the Mean Value Theorem for Integration===
 +
<math>f(x)</math> satisfies the requirements of the Extreme Value Theorem, so it has a minimum <math>m</math> and a maximum <math>M</math> in <math>[a,b]</math>. Since
  
* [https://youtu.be/ep52C-MUzDY Ex 4: The Second Fundamental Theorem of Calculus with Chain Rule] Video Lecture by James Sousa.
+
:<math>\int\limits_a^b f(x)dx=\lim_{n\to\infty}\sum_{k=1}^n f(x_k^*)\cdot\frac{b-a}{n}=\lim_{n\to\infty}\frac{b-a}{n}\cdot\sum_{k=1}^n f(x_k^*)</math>
  
* [https://youtu.be/8IrK5JcjMvM Ex 5: The Second Fundamental Theorem of Calculus with Chain Rule] Video Lecture by James Sousa.
+
and since
  
* [https://youtu.be/6p5NQwAeZRc Ex 6: Second Fundamental Theorem of Calculus with Chain Rule] Video Lecture by James Sousa.
+
:<math>m\le f(x_k^*)\le M</math> for all <math>x_k^*\in[a,b]</math>
  
* [https://youtu.be/18qhfQ2IeLU Ex 7: Second Fundamental Theorem of Calculus with Chain Rule] Video Lecture by James Sousa.
+
we have
  
* [https://youtu.be/PGmVvIglZx8 Fundamental Theorem of Calculus Part 1] video by patrickJMT
+
:<math>\begin{align}
 +
&\lim_{n\to\infty}\frac{b-a}{n}\cdot\sum_{k=1}^nm\le\lim_{n\to\infty}\frac{b-a}{n}\cdot\sum_{k=1}^n f(x_k^*)\le\lim_{n\to\infty}\frac{b-a}{n}\cdot\sum_{k=1}^n M\\
  
* [https://youtu.be/yrQ6PnE6D8w PART 1 OF THE DREADED FUNDAMENTAL THEOREM OF CALCULUS!] video by Krista King
+
&\lim_{n\to\infty}mn\cdot\frac{b-a}{n}\le\int\limits_a^b f(x)dx\le\lim_{n\to\infty}Mn\cdot\frac{b-a}{n}\\
  
* [https://youtu.be/aeB5BWY0RlE Fundamental Theorem of Calculus Part 1] video by The Organic Chemistry Tutor
+
&\lim_{n\to\infty}m(b-a)\le\int\limits_a^b f(x)dx\le\lim_{n\to\infty}M(b-a)\\
 +
 
 +
&m(b-a)\le\int\limits_a^b f(x)dx\le M(b-a)\\
 +
 
 +
&m\le\frac{1}{b-a}\int\limits_a^b f(x)dx\le M\end{align}</math>
 +
 
 +
Since <math>f</math> is continuous, by the Intermediate Value Theorem there is some <math>f(c)</math> with <math>c\in[a,b]</math> such that
 +
 
 +
:<math>\frac{1}{b-a}\int\limits_a^b f(x)dx=f(c)</math>
 +
 
 +
==Fundamental Theorem of Calculus==
 +
===Statement of the Fundamental Theorem===
 +
Suppose that <math>f</math> is continuous on <math>[a,b]</math> . We can define a function <math>F</math> by
 +
 
 +
:<math>F(x)=\int\limits_a^x f(t)dt\quad\text{for }x\in[a,b]</math>
 +
 
 +
Suppose <math>f</math> is continuous on <math>[a,b]</math> and <math>F</math> is defined by
 +
:<math>F(x)=\int\limits_a^x f(t)dt</math>
 +
Then <math>F</math> is differentiable on <math>(a,b)</math> and for all <math>x\in(a,b)</math>,
 +
:<math>F'(x)=f(x)</math>
 +
 
 +
When we have such functions <math>F</math> and <math>f</math> where <math>F'(x)=f(x)</math> for every <math>x</math> in some interval <math>I</math> we say that <math>F</math> is the '''antiderivative''' of <math>f</math> on <math>I</math>.
 +
 
 +
Suppose that <math>f</math> is continuous on <math>[a,b]</math> and that <math>F</math> is any antiderivative of <math>f</math> . 
 +
Then
 +
:<math>\int\limits_a^b f(x)dx=F(b)-F(a)</math>
 +
 
 +
[[Image:Fundamental-theorem-1.png|thumb|350px|right|Figure 1]]
 +
 
 +
Note: a minority of mathematicians refer to part one as two and part two as one. All mathematicians refer to what is stated here as part 2 as The Fundamental Theorem of Calculus.
 +
 
 +
===Proofs===
 +
====Proof of Fundamental Theorem of Calculus Part I====
 +
Suppose <math>x\in(a,b)</math> . Pick <math>\Delta x</math> so that <math>x+\Delta x\in(a,b)</math> . Then
 +
:<math>F(x)=\int\limits_a^x f(t)dt</math>
 +
and
 +
:<math>F(x+\Delta x)=\int\limits_a^{x+\Delta x}f(t)dt</math>
 +
 
 +
Subtracting the two equations gives
 +
:<math>F(x+\Delta x)-F(x)=\int\limits_a^{x+\Delta x}f(t)dt-\int\limits_a^x f(t)dt</math>
 +
 
 +
Now
 +
:<math>\int\limits_a^{x+\Delta x}f(t)dt=\int\limits_a^x f(t)dt+\int\limits_x^{x+\Delta x}f(t)dt</math>
 +
so rearranging this we have
 +
:<math>F(x+\Delta x)-F(x)=\int\limits_x^{x+\Delta x}f(t)dt</math>
 +
 
 +
According to the Mean Value Theorem for Integration, there exists a <math>c\in[x,x+\Delta x]</math> such that
 +
:<math>\int\limits_x^{x+\Delta x}f(t)dt=f(c)\cdot\Delta x</math>
 +
Notice that <math>c</math> depends on <math>\Delta x</math> . Anyway what we have shown is that
 +
:<math>F(x+\Delta x)-F(x)=f(c)\cdot\Delta x</math>
 +
and dividing both sides by <math>\Delta x</math> gives
 +
:<math>\frac{F(x+\Delta x)-F(x)}{\Delta x}=f(c)</math>
 +
 
 +
Take the limit as <math>\Delta x\to0</math> we get the definition of the derivative of <math>F</math> at <math>x</math> so we have
 +
:<math>F'(x)=\lim_{\Delta x\to0}\frac{F(x+\Delta x)-F(x)}{\Delta x}=\lim_{\Delta x\to0}f(c)</math>
 +
 
 +
To find the other limit, we will use the '''squeeze theorem'''. <math>c\in[x,x+\Delta x]</math> , so <math>x\le c\le x+\Delta x</math> . Hence,
 +
:<math>\lim_{\Delta x\to0}\Big[x+\Delta x\Big]=x\quad\Rightarrow\quad\lim_{\Delta x\to0}c=x</math>
 +
 
 +
As <math>f</math> is continuous we have
 +
:<math>F'(x)=\lim_{\Delta x\to0}f(c)=f\left(\lim_{\Delta x\to0}c\right)=f(x)</math>
 +
which completes the proof.
 +
 
 +
<math>\blacksquare</math>
 +
 
 +
====Proof of Fundamental Theorem of Calculus Part II====
 +
Define <math>P(x)=\int\limits_a^x f(t)dt</math> . Then by the Fundamental Theorem of Calculus part I we know that <math>P</math> is differentiable on <math>(a,b)</math> and for all <math>x\in(a,b)</math>
 +
:<math>P'(x)=f(x)</math>
 +
So <math>P</math> is an antiderivative of <math>f</math> . Since we were assuming that <math>F</math> was also an antiderivative for all <math>x\in(a,b)</math> ,
 +
:<math>\begin{align}&P'(x)=F'(x)\\&P'(x)-F'(x)=0\\&\Big(P(x)-F(x)\Big)'=0\end{align}</math>
 +
 
 +
Let <math>g(x)=P(x)-F(x)</math> . The Mean Value Theorem applied to <math>g(x)</math> on <math>[a,\xi]</math> with <math>a<\xi<b</math> says that
 +
:<math>\frac{g(\xi)-g(a)}{\xi-a}=g'(c)</math>
 +
for some <math>c</math> in <math>(a,\xi)</math> . But since <math>g'(x)=0</math> for all <math>x</math> in <math>[a,b]</math> , <math>g(\xi)</math> must equal <math>g(a)</math> for all <math>\xi</math> in <math>(a,b)</math> , i.e. g(x) is constant on <math>(a,b)</math> .
 +
 
 +
This implies there is a constant <math>C=g(a)=P(a)-F(a)=-F(a)</math> such that for all <math>x\in(a,b)</math> ,
 +
:<math>P(x)=F(x)+C</math>
 +
and as <math>g</math> is continuous we see this holds when <math>x=a</math> and <math>x=b</math> as well. And putting <math>x=b</math> gives
 +
:<math>\int\limits_a^b f(t)dx=P(b)=F(b)+C=F(b)-F(a)</math>
 +
 
 +
<math>\blacksquare</math>
 +
 
 +
==Notation for Evaluating Definite Integrals==
 +
The second part of the Fundamental Theorem of Calculus gives us a way to calculate definite integrals. Just find an antiderivative of the integrand, and subtract the value of the antiderivative at the lower bound from the value of the antiderivative at the upper bound. That is
 +
:<math>\int\limits_a^b f(x)dx=F(b)-F(a)</math>
 +
where <math>F'(x)=f(x)</math> . As a convenience, we use the notation
 +
:<math>F(x)\bigg|_a^b</math>
 +
to represent <math>F(b)-F(a)</math>
 +
 
 +
==Integration of Polynomials==
 +
Using the power rule for differentiation we can find a formula for the integral of a power using the Fundamental Theorem of Calculus. Let <math>f(x)=x^n</math> . We want to find an antiderivative for <math>f</math> . Since the differentiation rule for powers lowers the power by 1 we have that
 +
:<math>\frac{d}{dx}x^{n+1}=(n+1)x^n</math>
 +
As long as <math>n+1\ne0</math> we can divide by <math>n+1</math> to get
 +
:<math>\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n=f(x)</math>
 +
So the function <math>F(x)=\frac{x^{n+1}}{n+1}</math> is an antiderivative of <math>f</math> . If <math>0\notin[a,b]</math> then <math>F</math> is continuous on <math>[a,b]</math> and, by applying the Fundamental Theorem of Calculus, we can calculate the integral of <math>f</math> to get the following rule.
 +
 
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
:'''Power Rule of Integration I'''
 +
:<math>\int\limits_a^b x^ndx=\frac{x^{n+1}}{n+1}\Bigg|_a^b=\frac{b^{n+1}-a^{n+1}}{n+1}</math> as long as <math>n\ne-1</math> and <math>0\notin[a,b]</math> .
 +
</blockquote>
 +
 
 +
Notice that we allow all values of <math>n</math> , even negative or fractional. If <math>n>0</math> then this works even if <math>[a,b]</math> includes <math>0</math> .
 +
 
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
:'''Power Rule of Integration II'''
 +
:<math>\int\limits_a^b x^ndx=\frac{x^{n+1}}{n+1}\Bigg|_a^b=\frac{b^{n+1}-a^{n+1}}{n+1}</math> as long as <math>n>0</math> .
 +
</blockquote>
 +
 
 +
;Examples
 +
 
 +
*To find <math>\int\limits_1^2x^3dx</math> we raise the power by 1 and have to divide by 4. So
 +
:<math>\int\limits_1^2x^3dx=\frac{x^4}{4}\Bigg|_1^2=\frac{2^4}{4}-\frac{1^4}{4}=\frac{15}{4}</math>
 +
 
 +
*The power rule also works for negative powers. For instance
 +
:<math>\int\limits_1^3\frac{dx}{x^3}=\int\limits_1^3x^{-3}dx=\frac{x^{-2}}{-2}\Bigg|_1^3=\frac{1}{-2}\left(3^{-2}-1^{-2}\right)=-\frac12\left(\frac{1}{3^2}-1\right)=-\frac12\left(\frac19-1\right)=\frac12\cdot\frac89=\frac49</math>
 +
 
 +
*We can also use the power rule for fractional powers. For instance
 +
:<math>\int\limits_0^5\sqrt{x}dx=\int\limits_0^5x^\frac12dx=\frac{x\sqrt{x}}{\frac32}\Bigg|_0^5=\frac23\left(5^\frac32-0^\frac32\right)=\frac{10\sqrt5}{3}</math>
 +
 
 +
*Using linearity the power rule can also be thought of as applying to constants. For example,
 +
:<math>=\int\limits_3^{11}7dx=\int\limits_3^{11}7x^0dx=7\int\limits_3^{11}x^0dx=7x\Bigg|_3^{11}=7(11-3)=56</math>
 +
 
 +
*Using the linearity rule we can now integrate any polynomial. For example
 +
:<math>\int\limits_0^3(3x^2+4x+2)dx=(x^3+2x^2+2x)\Bigg|_0^3=3^3+2\cdot 3^2+2\cdot3-0=27+18+6=51</math>
 +
 
 +
==Exercises==
 +
===Questions===
 +
# Evaluate <math>\int\limits_0^1x^6dx</math>
 +
# Evaluate <math>\int\limits_1^2x^6dx</math>
 +
# Evaluate <math>\int\limits_0^2x^6dx</math>
 +
 
 +
===Solutions===
 +
# <math>\frac17=0.\overline{142857}</math>
 +
# <math>\frac{127}{7}=18.\overline{142857}</math>
 +
# <math>\frac{128}{7}=18.\overline{285714}</math>
 +
 
 +
==Resources==
 +
<strong>The Fundamental Theorem of Calculus, Part 1</strong>
 +
 
 +
* [https://mathresearch.utsa.edu/wikiFiles/MAT1193/The%20Fundamental%20Theorem%20of%20Calculus/Presentation12_DefiniteIntegral%20&%20Antiderivatives.pptx Definite Integral & Antiderivatives] (Slides 6&7). PowerPoint file created by Professor Cynthia Roberts, UTSA.
 +
* [https://mathresearch.utsa.edu/wikiFiles/MAT1193/The%20Fundamental%20Theorem%20of%20Calculus/Presentation13_FTC%20Part%20I.pptx Fundamental Theorem of Calculus Part 1]. PowerPoint file created by Professor Cynthia Roberts, UTSA.
 +
* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Fundamental%20Theorem%20of%20Calculus_/MAT1214-TheFundamentalTheoremPwPt.pptx The Fundamental Theorem of Calculus] PowerPoint file created by Dr. Sara Shirinkam, UTSA.
 +
 
 +
* [https://youtu.be/PGmVvIglZx8 Fundamental Theorem of Calculus Part 1] by patrickJMT
 +
* [https://youtu.be/yrQ6PnE6D8w PART 1 OF THE DREADED FUNDAMENTAL THEOREM OF CALCULUS!] by Krista King
 +
* [https://youtu.be/aeB5BWY0RlE Fundamental Theorem of Calculus Part 1] by The Organic Chemistry Tutor
 +
* [https://youtu.be/jyRdHbHeUuU The Second Fundamental Theorem of Calculus] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/YE9jpfxEFYk Ex 1: The Second Fundamental Theorem of Calculus] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/gJtkTRjqM6I Ex 2: The Second Fundamental Theorem of Calculus (Reverse Order)] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/X5ke8bUiiQM Ex 3: The Second Fundamental Theorem of Calculus] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/ep52C-MUzDY Ex 4: The Second Fundamental Theorem of Calculus with Chain Rule] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/8IrK5JcjMvM Ex 5: The Second Fundamental Theorem of Calculus with Chain Rule] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/6p5NQwAeZRc Ex 6: Second Fundamental Theorem of Calculus with Chain Rule] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/18qhfQ2IeLU Ex 7: Second Fundamental Theorem of Calculus with Chain Rule] by James Sousa, Math is Power 4U
 +
 
 +
 
 +
<strong>The Fundamental Theorem of Calculus, Part 2</strong>
 +
 
 +
* [https://youtu.be/7AteFyUCyZo The Fundamental Theorem of Calculus] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/nHnZVFeQvNQ The Fundamental Theorem of Calculus Part 2] by patrickJMT
 +
* [https://youtu.be/T-J7SkiE39Y PART 2 OF THE FUNDAMENTAL THEOREM OF CALCULUS!] by Krista King
 +
* [https://youtu.be/ns8N1UuXl4w The Fundamental Theorem of Calculus Part 2] by The Organic Chemistry Tutor
 +
 
 +
==Licensing==
 +
Content obtained and/or adapted from:
 +
* [https://en.wikibooks.org/wiki/Calculus/Fundamental_Theorem_of_Calculus Fundamental Theorem of Calculus, Wikibooks: Calculus] under a CC BY-SA license

Latest revision as of 15:20, 28 October 2021

The fundamental theorem of calculus is a critical portion of calculus because it links the concept of a derivative to that of an integral. As a result, we can use our knowledge of derivatives to find the area under the curve, which is often quicker and simpler than using the definition of the integral.

Mean Value Theorem for Integration

We will need the following theorem in the discussion of the Fundamental Theorem of Calculus.

Suppose is continuous on . Then for some .

Proof of the Mean Value Theorem for Integration

satisfies the requirements of the Extreme Value Theorem, so it has a minimum and a maximum in . Since

and since

for all

we have

Since is continuous, by the Intermediate Value Theorem there is some with such that

Fundamental Theorem of Calculus

Statement of the Fundamental Theorem

Suppose that is continuous on . We can define a function by

Suppose is continuous on and is defined by

Then is differentiable on and for all ,

When we have such functions and where for every in some interval we say that is the antiderivative of on .

Suppose that is continuous on and that is any antiderivative of . Then

Figure 1

Note: a minority of mathematicians refer to part one as two and part two as one. All mathematicians refer to what is stated here as part 2 as The Fundamental Theorem of Calculus.

Proofs

Proof of Fundamental Theorem of Calculus Part I

Suppose . Pick so that . Then

and

Subtracting the two equations gives

Now

so rearranging this we have

According to the Mean Value Theorem for Integration, there exists a such that

Notice that depends on . Anyway what we have shown is that

and dividing both sides by gives

Take the limit as we get the definition of the derivative of at so we have

To find the other limit, we will use the squeeze theorem. , so . Hence,

As is continuous we have

which completes the proof.

Proof of Fundamental Theorem of Calculus Part II

Define . Then by the Fundamental Theorem of Calculus part I we know that is differentiable on and for all

So is an antiderivative of . Since we were assuming that was also an antiderivative for all ,

Let . The Mean Value Theorem applied to on with says that

for some in . But since for all in , must equal for all in , i.e. g(x) is constant on .

This implies there is a constant such that for all ,

and as is continuous we see this holds when and as well. And putting gives

Notation for Evaluating Definite Integrals

The second part of the Fundamental Theorem of Calculus gives us a way to calculate definite integrals. Just find an antiderivative of the integrand, and subtract the value of the antiderivative at the lower bound from the value of the antiderivative at the upper bound. That is

where . As a convenience, we use the notation

to represent

Integration of Polynomials

Using the power rule for differentiation we can find a formula for the integral of a power using the Fundamental Theorem of Calculus. Let . We want to find an antiderivative for . Since the differentiation rule for powers lowers the power by 1 we have that

As long as we can divide by to get

So the function is an antiderivative of . If then is continuous on and, by applying the Fundamental Theorem of Calculus, we can calculate the integral of to get the following rule.

Power Rule of Integration I
as long as and .

Notice that we allow all values of , even negative or fractional. If then this works even if includes .

Power Rule of Integration II
as long as .
Examples
  • To find we raise the power by 1 and have to divide by 4. So
  • The power rule also works for negative powers. For instance
  • We can also use the power rule for fractional powers. For instance
  • Using linearity the power rule can also be thought of as applying to constants. For example,
  • Using the linearity rule we can now integrate any polynomial. For example

Exercises

Questions

  1. Evaluate
  2. Evaluate
  3. Evaluate

Solutions

Resources

The Fundamental Theorem of Calculus, Part 1


The Fundamental Theorem of Calculus, Part 2

Licensing

Content obtained and/or adapted from: