Difference between revisions of "009B Sample Midterm 2, Problem 2"

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!Final Answer:    
 
!Final Answer:    
 
|-
 
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|'''(a)'''
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|   '''(a)'''
 
|-
 
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|'''The Fundamental Theorem of Calculus, Part 1'''
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|   '''The Fundamental Theorem of Calculus, Part 1'''
 
|-
 
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|Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -14px">F(x)=\int_a^x f(t)~dt.</math>
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|&nbsp;&nbsp; Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -14px">F(x)=\int_a^x f(t)~dt.</math>
 
|-
 
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|Then, <math style="vertical-align: 0px">F</math> is a differentiable function on <math style="vertical-align: -5px">(a,b),</math> and <math style="vertical-align: -5px">F'(x)=f(x).</math>
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|&nbsp;&nbsp; Then, <math style="vertical-align: 0px">F</math> is a differentiable function on <math style="vertical-align: -5px">(a,b),</math> and <math style="vertical-align: -5px">F'(x)=f(x).</math>
 
|-
 
|-
|'''The Fundamental Theorem of Calculus, Part 2'''
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|&nbsp;&nbsp; '''The Fundamental Theorem of Calculus, Part 2'''
 
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|Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f.</math>
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|&nbsp;&nbsp; Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f.</math>
 
|-
 
|-
|Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
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|&nbsp;&nbsp; Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 
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|'''(b)''' &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x))</math>
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|&nbsp;&nbsp; '''(b)''' &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x))</math>
 
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|'''(c)''' <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx\,=\,1</math>
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|&nbsp;&nbsp; '''(c)''' <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx\,=\,1</math>
 
|}
 
|}
 
[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 15:05, 18 April 2016

This problem has three parts:

a) State the Fundamental Theorem of Calculus.
b) Compute   .
c) Evaluate .


Foundations:  
1. What does Part 1 of the Fundamental Theorem of Calculus say about 
Part 1 of the Fundamental Theorem of Calculus says that 
2. What does Part 2 of the Fundamental Theorem of Calculus say about where are constants?
Part 2 of the Fundamental Theorem of Calculus says that where is any antiderivative of

Solution:

(a)

Step 1:  
The Fundamental Theorem of Calculus has two parts.
The Fundamental Theorem of Calculus, Part 1
Let be continuous on and let
Then, is a differentiable function on and
Step 2:  
The Fundamental Theorem of Calculus, Part 2
Let be continuous on and let be any antiderivative of
Then,

(b)

Step 1:  
Let The problem is asking us to find
Let and
Then,
Step 2:  
If we take the derivative of both sides of the last equation, we get by the Chain Rule.
Step 3:  
Now, and by the Fundamental Theorem of Calculus, Part 1.
Since we have

(c)

Step 1:  
Using the Fundamental Theorem of Calculus, Part 2, we have
Step 2:  
So, we get
Final Answer:  
   (a)
   The Fundamental Theorem of Calculus, Part 1
   Let be continuous on and let
   Then, is a differentiable function on and
   The Fundamental Theorem of Calculus, Part 2
   Let be continuous on and let be any antiderivative of
   Then,
   (b)  
   (c)

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