course mth 272 иl·—~”ˆÓèÇ£ÿãN‚ëýëòyassignment #016
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12:59:52 Query problem 6.2.50 (was 6.2.48) solid of revolution y = x e^x x = 0 to 1 about x axis
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RESPONSE --> dv = e^x dx v =integral of dv = integral of e^x dx = e^x u = x du=dx x e^x = x e^x - integral of e^x dx = x e^x - e^x + C use fundamental theorem at x= 0 and x=1 (1) e^(1) - e^(1) + C =0 (0) e^(0) - e^(0) + C = -1 = -1 confidence assessment: 2
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13:00:36 What expression do you integrate to obtain the desired result?
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RESPONSE --> x e^x - integral of e^x dx confidence assessment: 3
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13:00:46 What is your result?
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RESPONSE --> -1 confidence assessment: 2
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13:00:59 For your first step, what was u and what was dv, what were du and v, and what integral did you obtain?
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RESPONSE --> already answered confidence assessment: 2
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13:02:09 Answer the same questions for your second step.
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RESPONSE --> ? so is this a using integration by parts repeatedly problem then? confidence assessment: 0
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13:05:34 Problem 6.2.58 (was 6.2.56) revenue function 410.5 t^2 e^(-t/30) + 25000
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RESPONSE --> ok confidence assessment:
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13:20:37 What expression did you evaluate to obtain the average daily receipts during the first quarter, and what was your result?
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RESPONSE --> use integral from x= 0 to x=90 of 410.5t^2 e^-t/30 + 25,000 dv= e^-t/30 v =integral of dv = integral of e^-t/30 = -1/30e^-t/30 u = t^2 du= 2dt integral of t^2e^-t/30 dt = -1/30e^-t/30 + -1/30 integral of e^-t/30 dt =-1/30e^-t/30 - 30e^-t/30 = e^t/30 pv = 410.5 integral at x=0 and 90 of (e^-t/30) =20.437 =410.5 20.437 -410.5 = -390.06 confidence assessment:
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13:23:05 What expression did you evaluate to obtain the year's total daily receipts and was your result?
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RESPONSE --> you use the same expression for teh fourth quarter and for the total daily receipts as you did in the first quarter. your x values change however x= 274 and x=365 for total year x=0 and x=365 confidence assessment: 1
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13:35:32 Problem 6.2.64 (was 6.2.62) c = 5000 + 25 t e^(-t/10), r=6%, t1=10 yr, find present value
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RESPONSE --> integral from x=0 and x=10 of 5025 e^.06(-t/10)
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13:35:51 What expression do you integrate to obtain the present value of the income function and what is your result?
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RESPONSE --> already in previous answer confidence assessment: 2
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13:37:48 Explain the meaning of the expression you integrated--why does this function give the present value?
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RESPONSE --> this present value represents the amount of income you have now that is going to grow over the next 10 years. confidence assessment: 2
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