course Mth 272 7/28 10:20 pm 016.
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Given Solution: `a The volume over an interval `dx will be approximately equal to pi ( x e^x)^2 `dx, so we will be integrating pi x^2 e^(2x) with respect to x from x = 0 to x = 1. Integrating just x^2 e^(2x) we let u = x^2 and dv = e^(2x) dx so that du = 2x dx and v = 1/2 e^(2x). This gives us u v - int(v du) = 2x (1/2 e^(2x)) - int(1/2 * 2x e^(2x) ). The remaining integral is calculated by a similar method and we get a result that simplifies to e^(2x) ( 2 x^2 - 2x + 1) / 4. Our antiderivative is therefore pi e^(2x) (2 x^2 - 2x + 1) / 4. This antiderivative changes between x = 0 and x = 1 by pi ( e^2 / 4 - 1/4) = 5.02, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qProblem 6.2.58 (was 6.2.56) revenue function 410.5 t^2 e^(-t/30) + 25000 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: In order to find the first quarter revenue, we need to take the integral of the given function. We can do this by identifying u, ud, dv, and v. We find that u = t^2, du = 2t, dv = e^(-t/30), and v = -30e^(-t/30). We can now place these quantities into the formula, which makes the integral = (t^2)(-30e^(-t/30) – Int(-30e^(-t/30) (2t))dt. We can now find the second half of that expression by finding u, du, dv, and v for that integral. We find that u = 2t, du = 2, dv = -30e^(-t/30), and v = 900e^(-t/30). So now, our whole integral equals (t^2)(-30e^(-t/30) – ((2t(900e^(-t/30)) – Int(2(900e^(-t/30))). This simplifies to -30t^2 e^(-t/30) -1800t e^(-t/30) + 1800 (-30e^(-t/30) + C. Now we can use our integral to find the area from 0 to 90, and then divide that total by 90 for the daily average. So -30(90)^2 e^(-90/30) -1800(90) e^(-90/30) + 1800 (-30e^(-90/30) – (-30(0)^2 e^(-0/30) -1800(0) e^(-0/30) + 1800 (-30e^(-0/30)) = -22852.264 + 54000 = 31147.736. 31147.736/90 = 346.086. 346.086*410.5 = 142068.285.
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Given Solution: `a If we integrate the revenue function from t = 0 to t = 90 we will have the first-quarter revenue. If we then divide by 90 we get the average daily revenue. To get an antiderivative of t^2 e^(-t/30) we first substitute u = t^2 and dv = e^(-t/30), obtaining du = 2t dt and v = -30 e^(-t/30). We proceed through the rest of the steps, which are very similar to steps used in preceding problems, to get antiderivative - 30•e^(- t/30)•(t^2 + 60•t + 1800). Our antiderivative of 410.5 t^2 e^(-t/30) + 25000 is therefore 410.5 (- 30•e^(- t/30)•(t^2 + 60•t + 1800) ) + 25000 t. The change in this antiderivative function between t = 0 and t = 90 is found by substitution to be about 15,000,000, representing $15,000,000 in 90 days for an average daily revenue of $15,000,000 / 90 = $167,000, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My solution was a bit off from the given solution because I did not use 15,000,000 for the total value after 90 days. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qProblem 6.2.64 (was 6.2.62) c = 5000 + 25 t e^(-t/10), r=6%, t1=10 yr, find present value YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the present value, we must first put together a present value function. Using the given values, we find that the present value = Int((5000 + 25te^(t/10))e^(-.06t). We can simplify this expression to Int(5000e^(-.06t) + 25te^(.04t)), and then to Int(5000e^(-.06t)) + Int(25te^(.04t)). Using our integral rules and the formula uv-Int(vdu) we can take the integral of both halves of our new function. We end up with -5000(1/.06)e^(-.06t) + C and (t/.04)e^(.04t) – (1/.04)(1/.04)e^(.04t) + C. Combined, our total present value = -5000(1/.06)e^(-.06t) + (t/.04)e^(.04t) – (1/.04)(1/.04)e^(.04t) + C. Finally, we can find the present value at t = 10 by plugging in 10 for t. So = -5000(1/.06)e^(-.06(10)) + (10/.04)e^(.04(10)) – (1/.04)(1/.04)e^(.04(10)) = 37,295.617. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a At 6% continuous interest the factor by which you multiply your principal to find its value after t years is e^(.06 t). So to find the principal you need now to end up with a certain amount after t years, you divide that amount by e^(.06 t), which is the same as multiplying it by e^(-.06 t). *&*& The income during a time interval `dt is ( 5000 + 25 t e^(-t/10) ) `dt. To get this amount after t years we would have to invest ( 5000 + 25 t e^(-t/10) ) `dt * e^(-.06 t). Integrating this expression from t = 0 to t = 10 we obtain int(( 5000 + 25 t e^(-t/10) ) * e^(-.06 t) dt , t from 0 to 10). Our result is $38,063. Note that the entire income stream gives us int(( 5000 + 25 t e^(-t/10) ) dt , t from 0 to 10) = $50,660 over the 10-year period. The meaning of our solution is that an investment of $38,063 for the 10-year period at 6% continuously compounded interest would yield $50,660 at the end of the period. So $38,063 is the present value of the income stream. "