Difference between revisions of "009B Sample Final 1, Problem 2"
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||So, we have | ||So, we have | ||
|- | |- | ||
− | |<math>f(x)=\int_{-1}^{x} \sin(t^2)2t~dt=\int_{1}^{x^2} \sin(u)du=\ | + | | |
+ | ::<math>\begin{array}{rcl} | ||
+ | f(x) & = & \displaystyle{\int_{-1}^{x} \sin(t^2)2t~dt}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{\int_{1}^{x^2} \sin(u)~du}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{-\cos(u)\bigg|_{1}^{x^2}}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{-\cos(x^2)+\cos(1)}\\ | ||
+ | \end{array}</math> | ||
|} | |} | ||
Revision as of 12:34, 10 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) |