Query 5 

#$&*

course Mth 272

6/14 6 pm

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

 

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Your solution:

 

3 t^ (4+1)

 (3 t^5) / 5

 3/5 t^5 + C

 

 Differentiation:

 5 * (3/5) t^5

 3 t^(5-1)

 3 t^4

 

 

confidence rating #$&*: 3

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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):

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Self-critique Rating: Ok

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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 + 1)

 v^ .5

 (v^.5)/ .5

 2v^.5 + c

 

 Check:

 (.5)(2) v^(.5-1)

 v^-.5

 

 

  

 

 

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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):

 Self-critique Rating: Ok

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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):

This was an easy query.

 Self-critique Rating: Ok "

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

 v^ (-.5 + 1)

 v^ .5

 (v^.5)/ .5

 2v^.5 + c

 

 Check:

 (.5)(2) v^(.5-1)

 v^-.5

 

 

  

 

 

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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):

 Self-critique Rating: Ok

*********************************************

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):

This was an easy query.

 Self-critique Rating: Ok "

Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Your work looks very good. Let me know if you have any questions. &#