#$&* course MTH272 If your solution to a stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a The volume over an interval `dx will be approximately equal to pi ( x e^x)^2 `dx, so we will be integrating pi x^2 e^(2x) with respect to x from x = 0 to x = 1. Integrating just x^2 e^(2x) we let u = x^2 and dv = e^(2x) dx so that du = 2x dx and v = 1/2 e^(2x). This gives us u v - int(v du) = 2x (1/2 e^(2x)) - int(1/2 * 2x e^(2x) ). The remaining integral is calculated by a similar method and we get a result that simplifies to e^(2x) ( 2 x^2 - 2x + 1) / 4. Our antiderivative is therefore pi e^(2x) (2 x^2 - 2x + 1) / 4. This antiderivative changes between x = 0 and x = 1 by pi ( e^2 / 4 - 1/4) = 5.02, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: ********************************************************** ********************************************* Question: `qProblem 6.2.58 (was 6.2.56) revenue function 410.5 t^2 e^(-t/30) + 25000 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [410.5 [(t^2) e^(-t/30)] + 25000 x= 0 to 90 u=t^2 du=2t dt v=-30e^(-t) dv=e^(-t/30) (t^2) (-30e^(-t/30))- int(-30e^(-t/30))(2t) (t^2) (-30e^(-t/30))- int(-60t)*e^(-t/30)) [(-30t^2-1800t-5400)(e^(-t/30))]/e^(t)^(1/30) 410.5{-30*e^(-t/30)*[t^2+60t+1800]}+25000t t=0 22167000 t=90 7130854 Difference between 0 and 90 =15036146 15036146/90 = approx.167068 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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.. STUDENT QUESTION After integrating this function, how do we come up with those values for t in order to find the first quarter revenue. Is this done by integrating, then accordingly equating to zero, solving for t. Can I use the integration table to successfully integrate this function as well? For example: u^n e^u du = u^n e^u - n [int.g] u^n-1 e^u du. Referred to the eight edition, but doesn’t really provide enough help on this one. INSTRUCTOR RESPONSE There are three problems here. The first is to understand how the revenue function is related to the revenue. The second is to find the antiderivative, and the third is to use it to answer the question. The revenue function can be defined as the rate at which the revenue changes, with respect to time. It follows from the Fundamental Theorem of Calculus that the revenue between two times is found by integrating the revenue function between those two times. To find the antiderivative we use integration by parts. There is a formula in the table of integrals, but the topic of this section is integration by parts. If you let u = t^2 and dv = e^(-t / 30), then what are du and v? What therefore do you get for u v - integral( v du )? You still have an integral to evaluate. If you did the preceding step correctly, the integral will be a constant multiple of t e^(-t) dt. You need another integration by parts to evaluate this integral. What would you use for u and what would you use for dv? What therefore is your antiderivative for the original function? To use the antiderivative, you have to understand that t is time in days. A year is regarded as 360 days, so that a quarter is 90 days. The revenue function is therefore integrated from t = 0 to t = 90. To get average daily revenue you would divide the total revenue by the 90 days. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qProblem 6.1.74 (8th edition) (7th edition 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: c = 5000 + 25 t e^(t/10), r=6%, t1=10 yr, find present value u=25t du=25 v=10*e^(t/10) dv= e^(t/10) (25t)(10*e^(t/10))-int(25*(10*e^(t/10)) (25t)(10*e^(t/10))-int(250*(10*e^(t/10)) (250t-2500)( e^(t/10)) 250(t-10)( e^(t)^(t/10)) 5000+(250t-2500)( e^(t/10))(e^(-.06 t) After 10 years $53091.83 with an initial $5000.00 the present value is 48091.83 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 $39 238. 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 $39 238 for the 10-year period at 6% continuously compounded interest would yield $50,660 at the end of the period. So $39 238 is the present value of the income stream. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!