Difference between revisions of "The Fundamental Theorem of Calculus"

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:<math>\begin{align}&P'(x)=F'(x)\\&P'(x)-F'(x)=0\\&\Big(P(x)-F(x)\Big)'=0\end{align}</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 [[Calculus/Some_Important_Theorems#Mean_Value_Theorem|Mean Value Theorem]] applied to <math>g(x)</math> on <math>[a,\xi]</math> with <math>a<\xi<b</math> says that
+
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>
 
:<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> .
 
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> .
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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.
 
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.
  
{{Calculus/Def|text='''Power Rule of Integration I'''
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<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> .}}
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:'''Power Rule of Integration I'''
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:<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> .
 
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> .
  
{{Calculus/Def|text='''Power Rule of Integration II'''
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<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> .}}
+
:'''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
 
;Examples
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==Exercises==
 
==Exercises==
{{question-answer|question=1. Evaluate <math>\int\limits_0^1x^6dx</math> . Compare your answer to the answer you got for exercise 1 in section {{Calculus/map page|Definite integral}}.|answer={{noprint|<math>\frac17=0.\overline{142857}</math>}}}}
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===Questions===
 
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# Evaluate <math>\int\limits_0^1x^6dx</math>
{{question-answer|question=2. Evaluate <math>\int\limits_1^2x^6dx</math> . Compare your answer to the answer you got for exercise 2 in section {{Calculus/map page|Definite integral}}.|answer={{noprint|<math>\frac{127}{7}=18.\overline{142857}</math>}}}}
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# Evaluate <math>\int\limits_1^2x^6dx</math>
 
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# Evaluate <math>\int\limits_0^2x^6dx</math>
{{question-answer|question=3. Evaluate <math>\int\limits_0^2x^6dx</math> . Compare your answer to the answer you got for exercise 4 in section {{Calculus/map page|Definite integral}}.|answer={{noprint|<math>\frac{128}{7}=18.\overline{285714}</math>}}}}
 
{{noprint|[[Calculus/Fundamental_Theorem_of_Calculus/Solutions|Solutions]]}}
 
 
 
  
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===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>
 
<strong>The Fundamental Theorem of Calculus, Part 1</strong>
  
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* [https://youtu.be/T-J7SkiE39Y PART 2 OF THE FUNDAMENTAL THEOREM OF CALCULUS!] by Krista King
 
* [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
 
* [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: