course Mth 272 5/1/10 at 14:00 021.
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Given Solution: `a The integral as stated here diverges. We take the limit as t -> infinity of INT(x^2 e^(-x^3), x from -t to t ). Using the obvious substitution we obtain antiderivative -1/3 e^(-x^3). Thus INT(x^2 e^(-x^3), x from -t to t ) = -1/3 ( e^(- (t^3) ) - e^(-(-t)^3) )= -1/3 ( e^(-t^3 ) - e^(t^3 ) ) The second term in parentheses is e^(t^3). As t -> infinity, t^3 approaches infinity so e^(t^3) approaches infinity. The first term e^(-t^3) is equal to 1 / e^(t^3)), and as t approaches infinity this term approaches zero. However that doesn't help. The antiderivative still approaches infinity as t approaches infinity, and adding something that approaches zero doesn't change this. The integral from -infinity to infinity therefore diverges. It wasn't asked, but note that the integral from 0 to infinity converges: We take the limit as t -> infinity of INT(x^2 e^(-x^3), x from 0 to t ), which using the same steps as before gives us the limit as t -> infinity of -1/3 e^(-t) - (-1/3) e^0. The first term approaches zero, the second is just 1/3. So the limiting value is 1/3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??????????Not sure what I am doing here. Having a hard time with the hole section ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qQuery problem 6.5.50 (7th edition 6.6.40) (was 6.6.38) farm profit of $75K per year, 8% continuously compounded, find present value of the farm for 20 years, and forever. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P = 75000, r = - 0.08 INT[75000 / -0.08 * e^(-0.08t)] from 0 to 20 = [75000 / -0.08 * e^(-0.08* 20)] – [75000 / -0.08 * e^(-0.08 * 0)] = [-937500 * 0.2018] – [-937500 * 1] = [-189277.9856 - -937500.000] = $748,220.01 from in 20 years Let b = infinity INT [75000 / -0.08 * e^(-0.08t)] from 0 to b = [75000 / -0.08 * e^(-0.08* b] – [75000 / -0.08 * e^(-0.08 * 0)] = 75000 / -0.08 *e^0.08b – 1 e^0.08b = 0 =(75000 / -0.08) * 0-1 = -937500 * -1 = $937,500 value for ever confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The correct antiderivative is [75,000 /-0.08 e^ -0.08t]. For 20 years you evaluate the change in this antiderivative between t = 0 and t = 20, and I believe you obtain $ 748,222.01 To get the present value forever you integrate from 0 to b and let b -> infinity. The integral from 0 to b is 75,000 / (-.08) e^(-(.08 b)) - 75,000 / (-.08) e^(0.08 * 0) = 75,000 / -.08 * (e^(-.08 b) - e^0). e^0 is 1 and as b -> infinity e^(-.08 b) -> 0. So the integral is 75,000 / -.08 ( 0 - 1) = 75,000 / .08 = 937,500. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qWhat is the present value of the farm for 20 years, and what is its present value forever? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: = $748,220.01 from in 20 years = $937,500 value for ever confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The present value for 20 years is $ 748,222.01 Forever $ 937,500.00 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. ???????? I believe I Udnerstant the problems in section 5 just need to see an example worked like problem number one. I believe I had the right thought process with this problem.