The Fundamental Theorem of Calculus - Revision history

Khanh: /* Integration of Polynomials */

2021-10-28T21:20:26Z

Integration of Polynomials

2021-10-28T21:19:39Z

Proof of Fundamental Theorem of Calculus Part II

2021-10-28T21:18:56Z

Integration of Polynomials

2021-10-28T21:17:59Z

2021-10-28T21:15:25Z

Questions

2021-10-28T21:12:17Z

Questions

2021-10-28T21:10:44Z

Exercises

2021-10-28T19:14:43Z

Exercises

2021-09-28T16:39:45Z

Fundamental Theorem of Calculus

2021-09-28T16:25:15Z

Mean Value Theorem for Integration

@@ Line 118: / Line 118: @@
 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.
 :'''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> .

@@ Line 93: / Line 93: @@
 :<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>
 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> .

@@ Line 127: / Line 127: @@
 :'''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> .
 ;Examples

@@ Line 118: / Line 118: @@
 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'''
+:'''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> .}}
+:<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> .
 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'''
+<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'''
 ;Examples

@@ Line 145: / Line 145: @@
 ==Exercises==
 ===Questions===
-# Evaluate <math>\int\limits_0^1x^6dx</math>.
+# Evaluate <math>\int\limits_0^1x^6dx</math>
 # Evaluate <math>\int\limits_1^2x^6dx</math>
-# Evaluate <math>\int\limits_1^2x^6dx</math>.
+# Evaluate <math>\int\limits_0^2x^6dx</math>
 ===Solutions===

@@ Line 144: / Line 144: @@
 ==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>}}}}
+===Questions===
-{{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>}}}}
+# Evaluate <math>\int\limits_1^2x^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>}}}}
+# Evaluate <math>\int\limits_0^2x^6dx</math>.
 ==Resources==

@@ Line 152: / Line 152: @@
+==Resources==
 <strong>The Fundamental Theorem of Calculus, Part 1</strong>
@@ Line 178: / Line 178: @@
 * [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

@@ Line 33: / Line 33: @@
 ==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
@@ Line 39: / Line 38: @@
 :<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> ,
+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>}}
+:<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>}}
+:<math>\int\limits_a^b f(x)dx=F(b)-F(a)</math>
 [[Image:Fundamental-theorem-1.png|thumb|350px|right|Figure 1]]

@@ Line 28: / Line 28: @@
 &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 [[Calculus/Continuity#Intermediate_Value_Theorem|Intermediate Value Theorem]] there is some <math>f(c)</math> with <math>c\in[a,b]</math> such that
+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>