course Mth 272 7/28 11:45 am 021.
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Given Solution: `a The integral as stated here diverges. You need to take the limit as t -> infinity of INT(x^2 e^(-x^3), x from -t to t ). Using the obvious substitution we see that the result is the same as the limiting value as t -> infinity of INT( 1/3 e^(-u), u from -t to t ). Using -1/3 e^(-u) as antiderivative we get -1/3 e^(-t)) - (-1/3 e^(-(-t))); the second term is 1/3 e^t, which approaches infinity as t -> infinity. The first term approaches zero, but that doesn't help. The integral approaches infinity. 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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 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: To find the present value of the farm, we first need to set up the Present value equation. Since the farm is earning 75,000 per year, c(t) is 75,000. So our present value = Int(75000 e^-.08t) dt. Using our integration rules, we find that this integral = -75000/.08 e^-.08t. To find the present value at 20 years, we can solve the integral for [0,20]. We find that -75000/.08 e^(-.08*20) – -75000/.08e^(.08*0) = 748222.02. To find the value for forever, we can solve for [0,2]. We can do this easily by picking a very large number for infinity and plugging that into the integral. For instance, we can use 1000. So -75000/.08 e^(-.08*1000) – 75000/.08 e^(-.08*0) = 937500. So we can assume that for forever, the present value of the farm will remain about 937500. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: To find the present value at 20 years, we can solve the integral for [0,20]. We find that -75000/.08 e^(-.08*20) – -75000/.08e^(.08*0) = 748222.02. To find the value for forever, we can solve for [0,2]. We can do this easily by picking a very large number for infinity and plugging that into the integral. For instance, we can use 1000. So -75000/.08 e^(-.08*1000) – 75000/.08 e^(-.08*0) = 937500. So we can assume that for forever, the present value of the farm will remain about 937500. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. "