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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):ok ------------------------------------------------ Self-critique rating #$&*:3 ********************************************* Question: `qProblem 6.2.58 (was 6.2.56) revenue function 410.5 t^2 e^(-t/30) + 25000 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v = -30e^-t/30 u=t^2 dv= e^-t/30 du = 2tdt 410.5 (-30e^(-t/30)(t^2 +60t+1800) + 2500t 15000000/90= 167000
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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 = $167,000, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*:3 ********************************************* 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: e^.06t * 5000+25te^(-t/10) from 0-10 is 38063 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 $38,063 is the present value of the income stream. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):had to look to make sure I was doing it right ------------------------------------------------ Self-critique rating #$&*:2"