Assignment 5

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course Mth 272

September 21 around 11:50

005. `query 5

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Question: `q 5.1.5 integrate 3 t^4 dt and check by differentiation

Your solution:

Antiderivative of t^4 is t^5/5

3t^4=3t^5/5

3t^5/5+c

Derivative of 3/5t^5=(3/5)*5t^4=3t^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): ok

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Self-critique rating #$&* 3

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Question: `q 5.1.7 integrate v^-.5 dv and check by differentiation

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Your solution:

Antiderivative of v^-0.5 is v^0.5

v^-0.5=v^0.5/0.5=2v^0.5

2v^0.5+c

Derivative of 2v^0.5=2(0.5)v^-0.5=v^-0.5

confidence rating #$&* 3

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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): ok

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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 got these by using the book to help me. I know I need to review how to do it to be able to do it without doubt.

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Self-critique rating #$&* 3

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&#Good work. Let me know if you have questions. &#

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