course MTH 272 03/03/10,2020 assignment #005005. `query 5
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18:54:57 5.1.12 integrate 3 t^4 dt and check by differentiation
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RESPONSE --> 3 t^4 dt 3(t^5/5) + C (3/5)t^5 + C (3/5)(5)t^(5-1) 3t^4 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:55:03 An antiderivative of the power function t^4 is one power higher so it will be a multiple of t^5. Since the derivative of t^5 is 5 t^4 an antiderivative of t^4 is be t^5 / 5. By the constant rule the antiderivative of 3 t^4 is therefore 3 * t^5 / 5. Adding the arbitrary integration constant we end up with general antiderivative3 t^5 / 5 + c. The derivative of 3/5 t^5 is 3/5 * 5 t^4 = 3 t^4), verifying our antiderivative. **
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RESPONSE --> ------------------------------------------------ Self-critique rating #$&*:ent: 3
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18:57:22 5.1.20 (was 5.1.18) integrate v^-.5 dv and check by differentiation
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RESPONSE --> v^-.5 dv (v^.5)/.5 + C 2 v^.5 + C 2 sqrt v + C 2 v^.5 + C 2(1/2)v^(.5-1) v^(-1/2) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:57:30 An antiderivative of this power function is a constant multiple of the power function which is one power higher. The power of the present function is -.5 or -1/2; one power higher is +.5 or 1/2. So you will have a multiple of v^.5. Since the derivative of v^.5 is .5 v^-.5 an antiderivative will be v^.5 / .5 = v^(1/2) / (1/2) = 2 v^(1/2). Adding the arbitrary integration constant we end up with general antiderivative 2 v^(1/2) + c. The derivative of 2 v^(1/2) is 2 * (1/2) v^(-1/2) = v^(-1/2), verifying our antiderivative. **
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RESPONSE --> ------------------------------------------------ Self-critique rating #$&*:ent: 3
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18:57:37 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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