course Mth 272 xzUv_ϲУassignment #005
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23:33:03 5.1.12 integrate 3 t^4 dt and check by differentiation
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RESPONSE --> The basic reverse power rule is used for this one y^x+1/x+1 since the deriv is 3t^4 then the anti deriv is 3t^(4+1)/5 then its 3/5 = 3/5t^5 + c confidence assessment: 3
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23:33:23 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 --> got it self critique assessment: 3
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23:38:58 5.1.20 (was 5.1.18) integrate v^-.5 dv and check by differentiation
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RESPONSE --> v ^ -.5 = v^ -.5 +1 / -.5 +1 simple power rule is used = v^1/2 / 1/2 +c this is the same as = 2v^1/2 +c confidence assessment: 3
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23:40:50 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 --> Ive sort of been forgetting to show the derivation 2v^ 1/2 is derived as 1/2 multiplies by 2 to get 1 and 1 is subtracted fron 1/2 to get -1/2 so the deric is v^-1/2 self critique assessment: 3
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23:42:24 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> It was pretty simple and straight forward. The basic power rule of integration comes easy to me so i am able to apply it. I hope the next stuff is as easy... ha yea right.
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