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

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(Created page with "<span class="exam"> This problem has three parts: ::<span class="exam">a) State the Fundamental Theorem of Calculus. ::<span class="exam">b) Compute   <math style="ve...")
 
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|Review the Fundamental Theorem of Calculus.
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|'''1.''' What does Part 1 of the Fundamental Theorem of Calculus say about&nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt?</math>
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::Part 1 of the Fundamental Theorem of Calculus says that&nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt=\sin(x).</math>
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|'''2.''' What does Part 2 of the Fundamental Theorem of Calculus say about <math style="vertical-align: -15px">\int_a^b\sec^2x~dx,</math> where <math style="vertical-align: -5px">a,b</math> are constants?
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::Part 2 of the Fundamental Theorem of Calculus says that <math style="vertical-align: -15px">\int_a^b\sec^2x~dx=F(b)-F(a),</math> where <math style="vertical-align: 0px">F</math> is any antiderivative of <math style="vertical-align: 0px">\sec^2x.</math>
 
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Revision as of 15:14, 8 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) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{\pi/4}\sec^2 x~dx\,=\,1} .

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