Query 5

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course Mth 272

If your solution to a stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

004. `query 4

005. `query 5

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Question: `q 5.1.5 (previously 5.1.12) integrate 3 t^4 dt and check by differentiation

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Your solution:

3t^5/5+C

confidence rating #$&*: 2.5

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Given Solution:

`a An antiderivative of the power function t^4 is one power higher so it will be a multiple of t^5. Since the derivative of t^5 is 5 t^4 an antiderivative of t^4 is be t^5 / 5. By the constant rule the antiderivative of 3 t^4 is therefore 3 * t^5 / 5. Adding the arbitrary integration constant we end up with general antiderivative3 t^5 / 5 + c.

The derivative of 3/5 t^5 is 3/5 * 5 t^4 = 3 t^4), verifying our antiderivative. **

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Self-critique (if necessary): I am sooooo happy! I finally understood this and got the problem correct! I had to look up videos to review chain rule and how to integrate/do antiderivatives in general form and it worked, right down to adding the constant at the end!

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Self-critique Rating: 3

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I'm happy that you've got it.

I also like it that you're happy.

*@

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Question: `q 5.1.7 (previously 5.1.20 (was 5.1.18)) integrate v^-.5 dv and check by differentiation

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Your solution:

V^.5/.5+C or sqrt V/.5+C

confidence rating #$&*: 2.5

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Given Solution:

`a An antiderivative of this power function is a constant multiple of the power function which is one power higher. The power of the present function is -.5 or -1/2; one power higher is +.5 or 1/2. So you will have a multiple of v^.5. Since the derivative of v^.5 is .5 v^-.5 an antiderivative will be v^.5 / .5 = v^(1/2) / (1/2) = 2 v^(1/2). Adding the arbitrary integration constant we end up with general antiderivative 2 v^(1/2) + c.

The derivative of 2 v^(1/2) is 2 * (1/2) v^(-1/2) = v^(-1/2), verifying our antiderivative. **

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Self-critique (if necessary): I had the answer correct until I tried to simplify it by bringing in the square root into the function. I think I will just leave it in its general form without reducing it. It will definitely save me some time and give me a better shot at getting it correct.

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Self-critique Rating: 3

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Question: `q Add comments on any surprises or insights you experienced as a result of this assignment.

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Self-critique (if necessary): I was very intimidated by antiderivatives, integration, and differentiation, but I looked at many videos and examples online and found it to actually be quite simple. I am extremely satisfied with how I did on this assignment.

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Self-critique Rating: 3 

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Very good. Check my note(s).

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