#$&* course Mth 272 6/14 6 pm Question: `q 5.1.5 (previously 5.1.12) integrate 3 t^4 dt and check by differentiation
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Given Solution: `a An antiderivative of the power function t^4 is one power higher so it will be a multiple of t^5. Since the derivative of t^5 is 5 t^4 an antiderivative of t^4 is be t^5 / 5. By the constant rule the antiderivative of 3 t^4 is therefore 3 * t^5 / 5. Adding the arbitrary integration constant we end up with general antiderivative3 t^5 / 5 + c. The derivative of 3/5 t^5 is 3/5 * 5 t^4 = 3 t^4), verifying our antiderivative. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: Ok ********************************************* Question: `q 5.1.7 (previously 5.1.20 (was 5.1.18))integrate v^-.5 dv and check by differentiation YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v^ (-.5 + 1) v^ .5 (v^.5)/ .5 2v^.5 + c Check: (.5)(2) v^(.5-1) v^-.5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This was an easy query. Self-critique Rating: Ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!