Difference between revisions of "009B Sample Final 1, Problem 2"
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!Step 1: | !Step 1: | ||
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− | |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>. |
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!Step 1: | !Step 1: | ||
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− | |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: | !Step 2: | ||
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− | |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>. |
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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: |
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Review -substitution |
Solution:
(a)
Step 1: |
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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: |
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So, we have |
|
(b)
Step 1: |
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From part (a), we have . |
Step 2: |
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If we take the derivative, we get . |
(c)
Step 1: |
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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: |
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The Fundamental Theorem of Calculus, Part 2 |
Let be continuous on and let be any antiderivative of . |
Then, |
(d)
Step 1: |
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By the Fundamental Theorem of Calculus, Part 1, |
Final Answer: |
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(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) |