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ÉÄ°\Ï÷Ÿ†öí¼Ã|èçr»z”ïÐassignment #005

005. `query 5

Applied Calculus II

06-29-2008

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07:46:35

5.1.12 integrate 3 t^4 dt and check by differentiation

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RESPONSE -->

you want to use the constant multiple rule here.

=3 integral sign t^4dx

=3(t^5/5) + C

= (3/5)t^5 +C

confidence assessment: 2

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07:48:39

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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RESPONSE -->

ok i forgot to check with differentiation.

the derivative of 3/5 t^5 is 3/5 * 5 t^4 = 3 t^4)

self critique assessment: 2

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07:53:09

5.1.20 (was 5.1.18) integrate v^-.5 dv and check by differentiation

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RESPONSE -->

= integral sign v^.5/(.5) + C

= (1/2)v^.5 + C

confidence assessment: 2

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07:56:56

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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RESPONSE -->

ok i see where I slightly messed up.

it should be 2 instead of (1/2) when you move it up from the denominator.

Again the the derivative of 2 v^(1/2) = 2* (1/2) v^(-1/2) = v^(-1/2) verifying the antiderivative

self critique assessment: 2

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08:00:33

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

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RESPONSE -->

the only thing was that I thought when you have

v^(1/2) / (1/2) + C = 1/2v^(1/2)

but it should be 2v^(1/2)

could you explain why this is so please?

confidence assessment: 3

I'm not completely sure what you are asking, but:

The derivative of v^(1/2) with respect to v is 1/2 v^(-1/2) = 1 / (2 sqrt(v)).

An antiderivative of v^(-1/2) is 2 v^(1/2) + C, because the derivative of 2 v^(1/2) is 2 * (1/2 v^(-1/2) ) = v^((-1/2).

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This looks good. See my notes. Let me know if you have any questions. &#