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

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|'''The Fundamental Theorem of Calculus, Part 1'''
 
|'''The Fundamental Theorem of Calculus, Part 1'''
 
|-
 
|-
|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>.
+
|
 +
: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>
 
|-
 
|-
|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>.
+
|
 +
: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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|'''The Fundamental Theorem of Calculus, Part 2'''
 
|'''The Fundamental Theorem of Calculus, Part 2'''
 
|-
 
|-
|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>.
+
|
 +
: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>.
+
|
 +
:Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 
|}
 
|}
 
'''(b)'''
 
'''(b)'''
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|Let <math style="vertical-align: -15px">F(x)=\int_0^{\cos (x)}\sin (t)~dt</math>. The problem is asking us to find <math style="vertical-align: -5px">F'(x)</math>.
+
|Let <math style="vertical-align: -15px">F(x)=\int_0^{\cos (x)}\sin (t)~dt.</math> The problem is asking us to find <math style="vertical-align: -5px">F'(x).</math>
 
|-
 
|-
|Let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -14px">G(x)=\int_0^x \sin(t)~dt</math>.
+
|Let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -14px">G(x)=\int_0^x \sin(t)~dt.</math>
 
|-
 
|-
|Then, <math style="vertical-align: -5px">F(x)=G(g(x))</math>.
+
|Then, <math style="vertical-align: -5px">F(x)=G(g(x)).</math>
 
|}
 
|}
  
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|Now, <math style="vertical-align: -5px">g'(x)=-\sin(x)</math> and <math style="vertical-align: -5px">G'(x)=\sin(x)</math> by the '''Fundamental Theorem of Calculus, Part 1'''.
 
|Now, <math style="vertical-align: -5px">g'(x)=-\sin(x)</math> and <math style="vertical-align: -5px">G'(x)=\sin(x)</math> by the '''Fundamental Theorem of Calculus, Part 1'''.
 
|-
 
|-
|Since <math style="vertical-align: -6px">G'(g(x))=\sin(g(x))=\sin(\cos(x))</math>, we have <math style="vertical-align: -5px">F'(x)=G'(g(x))\cdot g'(x)=\sin(\cos(x))\cdot(-\sin(x))</math>.
+
|Since <math style="vertical-align: -6px">G'(g(x))=\sin(g(x))=\sin(\cos(x)),</math> we have  
 +
|-
 +
|
 +
::<math style="vertical-align: -5px">F'(x)=G'(g(x))\cdot g'(x)=\sin(\cos(x))\cdot(-\sin(x)).</math>
 
|}
 
|}
  
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| Using the '''Fundamental Theorem of Calculus, Part 2''', we have
 
| Using the '''Fundamental Theorem of Calculus, Part 2''', we have
 
|-
 
|-
| &nbsp;&nbsp; <math>\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan(x)\biggr|_{0}^{\pi/4}</math>
+
|
 +
::<math>\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan(x)\biggr|_{0}^{\pi/4}.</math>
 
|}
 
|}
  
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|-
 
|-
 
|So, we get  
 
|So, we get  
|-
 
| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan \bigg(\frac{\pi}{4}\bigg)-\tan (0)=1</math>.
 
 
|-
 
|-
 
|
 
|
 +
::<math style="vertical-align: -16px">\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan \bigg(\frac{\pi}{4}\bigg)-\tan (0)=1.</math>
 
|}
 
|}
  
Line 100: Line 107:
 
|'''The Fundamental Theorem of Calculus, Part 1'''
 
|'''The Fundamental Theorem of Calculus, Part 1'''
 
|-
 
|-
|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>.
+
|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>
 
|-
 
|-
|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>.
+
|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'''
 
|'''The Fundamental Theorem of Calculus, Part 2'''
 
|-
 
|-
|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>.
+
|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>.
+
|Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 
|-
 
|-
|'''(b)''' &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x))</math>.
+
|'''(b)''' &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x))</math>
 
|-
 
|-
|'''(c)''' <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx\,=\,1</math>.
+
|'''(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 14:42, 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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