Difference between revisions of "009B Sample Final 1, Problem 2"

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!Step 1:    
 
!Step 1:    
 
|-
 
|-
|We proceed using <math>u</math>-substitution. Let <math>u=t^2</math>. Then, <math>du=2tdt</math>.  
+
|We proceed using <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: 0px">u=t^2</math>. Then, <math style="vertical-align: 0px">du=2tdt</math>.  
 
|-
 
|-
 
|Since this is a definite integral, we need to change the bounds of integration.  
 
|Since this is a definite integral, we need to change the bounds of integration.  
 
|-
 
|-
|Plugging in our values into the equation <math>u=t^2</math>, we get <math>u_1=(-1)^2=1</math> and <math>u_2=x^2</math>.
+
|Plugging in our values into the equation <math style="vertical-align: 0px">u=t^2</math>, we get <math style="vertical-align: -5px">u_1=(-1)^2=1</math> and <math style="vertical-align: -3px">u_2=x^2</math>.
 
|}
 
|}
  
Line 52: Line 52:
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|From part (a), we have <math>f(x)=-\cos(x^2)+\cos(1)</math>.
+
|From part (a), we have <math style="vertical-align: -5px">f(x)=-\cos(x^2)+\cos(1)</math>.
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|If we take the derivative, we get <math>f'(x)=\sin(x^2)2x</math>.
+
|If we take the derivative, we get <math style="vertical-align: -5px">f'(x)=\sin(x^2)2x</math>.
 
|}
 
|}
  

Revision as of 12:03, 22 February 2016

We would like to evaluate

.

a) Compute .

b) Find .

c) State the fundamental theorem of calculus.

d) Use the fundamental theorem of calculus to compute without first computing the integral.

Foundations:  
Review -substitution

Solution:

(a)

Step 1:  
We proceed using -substitution. Let . Then, .
Since this is a definite integral, we need to change the bounds of integration.
Plugging in our values into the equation , we get and .
Step 2:  
So, we have

(b)

Step 1:  
From part (a), we have .
Step 2:  
If we take the derivative, we get .

(c)

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,

(d)

Step 1:  
By the Fundamental Theorem of Calculus, Part 1,
Final Answer:  
(a)
(b)
(c) 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, .
(d)

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